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Unformatted text preview: 5E03(pp 126135) 1/17/06 1:49 PM Page 126 CHAPTER 3
By measuring slopes at points on the sine curve,
we get strong visual evidence that the derivative
of the sine function is the cosine function. D erivatives 5E03(pp 126135) 1/17/06 1:49 PM Page 127 In this chapter we begin our study of differential calculus,
which is concerned with how one quantity changes in relation to another quantity. The central concept of differential
calculus is the derivative, which is an outgrowth of the
velocities and slopes of tangents that we considered in
Chapter 2. After learning how to calculate derivatives, we use them to solve problems involving rates of change and the approximation of functions.  3.1 Derivatives
In Section 2.6 we deﬁned the slope of the tangent to a curve with equation y
point where x a to be
1 m lim fa h
h h l0 fa We also saw that the velocity of an object with position function s
va lim fa h l0 f x at the h
h f t at time t a is fa In fact, limits of the form
lim fa h l0 h
h fa arise whenever we calculate a rate of change in any of the sciences or engineering, such as
a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit
occurs so widely, it is given a special name and notation.
2 Definition The derivative of a function f at a number a, denoted by f a , is fa  f a is read “f prime of a.” lim fa h l0 h
h fa if this limit exists.
a
h, then h
x
a and h approaches 0 if and only if x approaches
If we write x
a. Therefore, an equivalent way of stating the deﬁnition of the derivative, as we saw in
ﬁnding tangent lines, is 3 fa lim xla fx
x fa
a 127 5E03(pp 126135) 128 ❙❙❙❙ 1/17/06 1:49 PM Page 128 CHAPTER 3 DERIVATIVES x2 EXAMPLE 1 Find the derivative of the function f x 9 at the number a. 8x SOLUTION From Deﬁnition 2 we have fa Try problems like this one.
Resources / Module 3
/ Derivative at a Point
/ Problem Wizard lim fa h l0 a lim h
h fa 2 8a h h a 2 2 2ah h h2
h 8h 8h
h 9 a2 lim 2a h 9 8 8a h l0 lim 8a h h l0 lim a2 9 2ah h l0 2a h l0 8a 9 8 Interpretation of the Derivative as the Slope of a Tangent
In Section 2.6 we deﬁned the tangent line to the curve y f x at the point P a, f a to
be the line that passes through P and has slope m given by Equation 1. Since, by Deﬁnition 2, this is the same as the derivative f a , we can now say the following. The tangent line to y f x at a, f a is the line through a, f a whose slope is
equal to f a , the derivative of f at a. Thus, the geometric interpretation of a derivative [as deﬁned by either (2) or (3)] is as
shown in Figure 1.
y y y=ƒ y=ƒ f(a+h)f(a) P
h xa 0 0
a F IGURE 1 ƒf(a) P x a+h a f(a+h)f(a)
h
h=0
=slope of tangent at P
=slope of curve at P ƒf(a)
xa
x=a
=slope of tangent at P
=slope of curve at P (a) f ª(a)=lim Geometric interpretation
of the derivative x x (b) f ª(a)=lim If we use the pointslope form of the equation of a line, we can write an equation of the
tangent line to the curve y f x at the point a, f a :
y fa fax a EXAMPLE 2 Find an equation of the tangent line to the parabola y point 3, x2 8x 9 at the 6.
x 2 8x
8. Therefore, the slope of the tangent line at 3, SOLUTION From Example 1 we know that the derivative of f x number a is f a 2a 9 at the
6 is 5E03(pp 126135) 1/17/06 1:49 PM Page 129 SECTION 3.1 DERIVATIVES y f3 23 y
x 0
(3, _6) y=_2x 129 2. Thus, an equation of the tangent line, shown in Figure 2, is 8 y=≈8x+9 ❙❙❙❙ 6 2x or 3 y 2x 2 x. Estimate the value of f 0 in two ways:
(a) By using Deﬁnition 2 and taking successively smaller values of h.
(b) By interpreting f 0 as the slope of a tangent and using a graphing calculator to
zoom in on the graph of y 2 x.
EXAMPLE 3 Let f x SOLUTION (a) From Deﬁnition 2 we have
FIGURE 2 f0 2 h h 1
h 0.1
0.01
0.001
0.0001
0.1
0.01
0.001
0.0001 0.718
0.696
0.693
0.693
0.670
0.691
0.693
0.693 fh lim lim h h l0 2h h l0 1
h Since we are not yet able to evaluate this limit exactly, we use a calculator to approximate the values of 2 h 1 h. From the numerical evidence in the table at the left we
see that as h approaches 0, these values appear to approach a number near 0.69. So our
estimate is
f0 0.69 (b) In Figure 3 we graph the curve y 2 x and zoom in toward the point 0, 1 . We see
that the closer we get to 0, 1 , the more the curve looks like a straight line. In fact, in
Figure 3(c) the curve is practically indistinguishable from its tangent line at 0, 1 . Since
the xscale and the yscale are both 0.01, we estimate that the slope of this line is
0.14
0.20
So our estimate of the derivative is f 0
to six decimal places, f 0
0.693147. (0, 1) 0.7
0.7. In Chapter 7 we will show that, correct (0, 1) (0, 1) (a) _1, 1 by 0, 2
F IGURE 3 f0 (b) _ 0.5, 0.5 by 0.5, 1.5 (c) _ 0.1, 0.1 by 0.9, 1.1 Zooming in on the graph of y=2® near (0, 1) Interpretation of the Derivative as a Rate of Change
In Section 2.6 we deﬁned the instantaneous rate of change of y f x with respect to x at
x x 1 as the limit of the average rates of change over smaller and smaller intervals. If the
interval is x 1, x 2 , then the change in x is x x 2 x 1, the corresponding change in y is
y f x2 f x1 and
4 instantaneous rate of change lim x l0 y
x lim x2 l x1 f x2
x2 f x1
x1 5E03(pp 126135) 130 ❙❙❙❙ 1/17/06 1:49 PM Page 130 CHAPTER 3 DERIVATIVES From Equation 3 we recognize this limit as being the derivative of f at x 1, that is, f x 1 .
This gives a second interpretation of the derivative:
The derivative f a is the instantaneous rate of change of y
x when x a. y Q P x FIGURE 4 The yvalues are changing rapidly
at P and slowly at Q. f x with respect to The connection with the ﬁrst interpretation is that if we sketch the curve y f x , then
the instantaneous rate of change is the slope of the tangent to this curve at the point where
x a. This means that when the derivative is large (and therefore the curve is steep, as at
the point P in Figure 4), the yvalues change rapidly. When the derivative is small, the
curve is relatively ﬂat and the yvalues change slowly.
In particular, if s f t is the position function of a particle that moves along a straight
line, then f a is the rate of change of the displacement s with respect to the time t. In
other words, f a is the velocity of the particle at time t a. (See Section 2.6.) The speed
of the particle is the absolute value of the velocity, that is, f a .
E XAMPLE 4 The position of a particle is given by the equation of motion s ft
1 1 t , where t is measured in seconds and s in meters. Find the velocity
and the speed after 2 seconds.
SOLUTION The derivative of f when t f2 In Module 3.1 you are asked to compare
and order the slopes of tangent and
secant lines at several points on a curve. lim f2 h
h h l0 1
lim 3 h l0 lim h l0 2 is
f2 3 3
33 lim hh Thus, the velocity after 2 seconds is f 2
1
1
f2
9
9 m s. lim h l0 1 2 h
h h h l0 h
33 1
h
h h l0 1
3 h
h lim 1
2 1 1
33
1
9 h 1
9 m s, and the speed is EXAMPLE 5 A manufacturer produces bolts of a fabric with a ﬁxed width. The cost of producing x yards of this fabric is C f x dollars.
(a) What is the meaning of the derivative f x ? What are its units?
(b) In practical terms, what does it mean to say that f 1000
9?
(c) Which do you think is greater, f 50 or f 500 ? What about f 5000 ?
SOLUTION (a) The derivative f x is the instantaneous rate of change of C with respect to x ; that
is, f x means the rate of change of the production cost with respect to the number of
yards produced. (Economists call this rate of change the marginal cost. This idea is discussed in more detail in Sections 3.4 and 4.8.)
Because
C
fx
lim
xl0
x
the units for f x are the same as the units for the difference quotient C x. Since
C is measured in dollars and x in yards, it follows that the units for f x are dollars
per yard. 5E03(pp 126135) 1/17/06 1:49 PM Page 131 SECTION 3.1 DERIVATIVES ❙❙❙❙ 131 (b) The statement that f 1000
9 means that, after 1000 yards of fabric have been
manufactured, the rate at which the production cost is increasing is $9 yard. (When
x 1000, C is increasing 9 times as fast as x.)
Since x 1 is small compared with x 1000, we could use the approximation
 Here we are assuming that the cost function
is well behaved; in other words, C x doesn’t
oscillate rapidly near x 1000. C
x f 1000 C
1 C and say that the cost of manufacturing the 1000th yard (or the 1001st) is about $9.
(c) The rate at which the production cost is increasing (per yard) is probably lower
when x 500 than when x 50 (the cost of making the 500th yard is less than the cost
of the 50th yard) because of economies of scale. (The manufacturer makes more efﬁcient
use of the ﬁxed costs of production.) So
f 50 f 500 But, as production expands, the resulting largescale operation might become inefﬁcient
and there might be overtime costs. Thus, it is possible that the rate of increase of costs
will eventually start to rise. So it may happen that
f 5000 f 500 The following example shows how to estimate the derivative of a tabular function, that
is, a function deﬁned not by a formula but by a table of values.
t Dt 1980
1985
1990
1995
2000 930.2
1945.9
3233.3
4974.0
5674.2 EXAMPLE 6 Let D t be the U.S. national debt at time t. The table in the margin gives
approximate values of this function by providing end of year estimates, in billions of
dollars, from 1980 to 2000. Interpret and estimate the value of D 1990 .
SOLUTION The derivative D 1990 means the rate of change of D with respect to t when t 1990, that is, the rate of increase of the national debt in 1990.
According to Equation 3,
D 1990 lim Dt
t t l1990 D 1990
1990 So we compute and tabulate values of the difference quotient (the average rates of
change) as follows.
t
1980
1985
1995
2000  Another method is to plot the debt function
and estimate the slope of the tangent line when
t 1990. (See Example 5 in Section 2.6.) Dt
t D 1990
1990
230.31
257.48
348.14
244.09 From this table we see that D 1990 lies somewhere between 257.48 and 348.14 billion
dollars per year. [Here we are making the reasonable assumption that the debt didn’t
ﬂuctuate wildly between 1980 and 2000.] We estimate that the rate of increase of the
national debt of the United States in 1990 was the average of these two numbers, namely
D 1990 303 billion dollars per year 5E03(pp 126135) 132 ❙❙❙❙ 1/17/06 1:50 PM Page 132 CHAPTER 3 DERIVATIVES  3.1 Exercises 1. On the given graph of f, mark lengths that represent f 2 , f2 h,f 2 line has slope f 2 , and h. (Choose h
h
f2
?
h h
f2 10. (a) If G x x 1 2 x , ﬁnd G a and use it to ﬁnd an
equation of the tangent line to the curve y x 1 2 x at
1
the point ( 4 , 1 ).
2
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. 0.) What ; y 3 x. Estimate the value of f 1 in two ways:
(a) By using Deﬁnition 2 and taking successively smaller
values of h.
(b) By zooming in on the graph of y 3 x and estimating the
slope. 11. Let f x y=ƒ ;
0 12. Let t x x 2 2. For the function f whose graph is shown in Exercise 1, arrange the following numbers in increasing order and explain your
reasoning:
0 f2 f3 1
2 f2 f4 2 t0 t2  Find f a . 13. f x numbers in increasing order and explain your reasoning:
t ; 13–18 f2 3. For the function t whose graph is given, arrange the following 0 tan x. Estimate the value of t
4 in two ways:
(a) By using Deﬁnition 2 and taking successively smaller
values of h.
(b) By zooming in on the graph of y tan x and estimating the
slope. t4 3
2t
t 15. f t ■ ■ 16. f x ■ ■ ■ x
x 18. f x 2 ■ 5t
2 1
3 sx t4 14. f t 1 17. f x y 4x 2 2x s3x ■ ■ ■ 1
2
1
■ ■ ■ y=©
19–24  Each limit represents the derivative of some function f at
some number a. State such an f and a in each case.
_1 0 1 2 3 4 19. lim x h l0 21. lim
x l5 4. If the tangent line to y f x at (4, 3) passes through the point
(0, 2), ﬁnd f 4 and f 4 . 5. Sketch the graph of a function f for which f 0 f1 0, and f 2
0, and t 2 7. If f x 0, f 0 3, 0, t 0 3, ■ 22. lim
xl cos h 1 24. lim h
■ 4
s16 h t l1 ■ ■ ■ ■ ■ 2 h h l0 32
5 h l0 ■ 20. lim 4 tan x
x t4 1
4 t
t 2
1 ■ ■ ■ ■ 1. 25–26  A particle moves along a straight line with equation of
motion s f t , where s is measured in meters and t in seconds.
Find the velocity when t 2. 2 5x, ﬁnd f 2 and use it to ﬁnd an equation
3x
of the tangent line to the parabola y 3x 2 5x at the
point 2, 2 .
1 x 3, ﬁnd t 0 and use it to ﬁnd an equation of the
tangent line to the curve y 1 x 3 at the point 0, 1 . 8. If t x x 3 5x 1, ﬁnd F 1 and use it to ﬁnd an
equation of the tangent line to the curve y x 3 5x 1
at the point 1, 3 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. 9. (a) If F x ; 23. lim 2x
x 1 1. 6. Sketch the graph of a function t for which t 0 t1 h 10
h 1 25. f t
■ ■ t2
■ 6t
■ 26. f t 5
■ ■ ■ ■ 2t 3
■ t
■ 1
■ 27. The cost of producing x ounces of gold from a new gold mine is C f x dollars.
(a) What is the meaning of the derivative f x ? What are its
units?
(b) What does the statement f 800
17 mean?
(c) Do you think the values of f x will increase or decrease
in the short term? What about the long term? Explain. ■ 5E03(pp 126135) 1/17/06 1:50 PM Page 133 W RITING PROJECT EARLY METHODS FOR FINDING TANGENTS ❙❙❙❙ 133 28. The number of bacteria after t hours in a controlled laboratory 33. The quantity of oxygen that can dissolve in water depends on experiment is n f t .
(a) What is the meaning of the derivative f 5 ? What are its
units?
(b) Suppose there is an unlimited amount of space and
nutrients for the bacteria. Which do you think is larger,
f 5 or f 10 ? If the supply of nutrients is limited, would
that affect your conclusion? Explain. the temperature of the water. (So thermal pollution inﬂuences
the oxygen content of water.) The graph shows how oxygen
solubility S varies as a function of the water temperature T .
(a) What is the meaning of the derivative S T ? What are its
units?
(b) Estimate the value of S 16 and interpret it.
S
(mg/L)
16 29. The fuel consumption (measured in gallons per hour) of a car
traveling at a speed of v miles per hour is c f v .
(a) What is the meaning of the derivative f v ? What are its 12 units?
(b) Write a sentence (in layman’s terms) that explains the
meaning of the equation f 20
0.05. 8
4 30. The quantity (in pounds) of a gourmet ground coffee that is sold by a coffee company at a price of p dollars per pound
is Q f p .
(a) What is the meaning of the derivative f 8 ? What are its
units?
(b) Is f 8 positive or negative? Explain. 0 2 4 6 8 10 12 73 73 70 69 72 81 88 40 T (°C) 32 S
(cm /s)
20 14 T 24 maximum sustainable swimming speed S of Coho salmon.
(a) What is the meaning of the derivative S T ? What are its
units?
(b) Estimate the values of S 15 and S 25 and interpret them. night on June 2, 2001. The table shows values of this function
recorded every two hours. What is the meaning of T 10 ?
Estimate its value.
0 16 34. The graph shows the inﬂuence of the temperature T on the 31. Let T t be the temperature (in F ) in Dallas t hours after mid t 8 91 32. Life expectancy improved dramatically in the 20th century. The table gives values of E t , the life expectancy at birth (in years)
of a male born in the year t in the United States. Interpret and
estimate the values of E 1910 and E 1950 .
t Et t Et 1900
1910
1920
1930
1940
1950 48.3
51.1
55.2
57.4
62.5
65.6 1960
1970
1980
1990
2000 66.6
67.1
70.0
71.8
74.1 0 35–36  10 Determine whether f 0 exists.
x sin 35. f x 1
x if x x 2 sin 36. f x 1
x ■ 0 if x 0
■ 0 if x 0 ■ T (°C) 20 0 if x
■ ■ 0
■ ■ ■ ■ ■ ■ WRITING PROJECT
E arly Methods for Finding Tangents
The ﬁrst person to formulate explicitly the ideas of limits and derivatives was Sir Isaac Newton in
the 1660s. But Newton acknowledged that “If I have seen further than other men, it is because I
have stood on the shoulders of giants.” Two of those giants were Pierre Fermat (1601–1665) and
Newton’s teacher at Cambridge, Isaac Barrow (1630–1677). Newton was familiar with the methods
that these men used to ﬁnd tangent lines, and their methods played a role in Newton’s eventual
formulation of calculus. ■ 5E03(pp 126135) 134 ❙❙❙❙ 1/17/06 1:50 PM Page 134 CHAPTER 3 DERIVATIVES The following references contain explanations of these methods. Read one or more of the
references and write a report comparing the methods of either Fermat or Barrow to modern methods. In particular, use the method of Section 3.1 to ﬁnd an equation of the tangent line to the
curve y x 3 2 x at the point (1, 3) and show how either Fermat or Barrow would have solved
the same problem. Although you used derivatives and they did not, point out similarities between
the methods.
1. Carl Boyer and Uta Merzbach, A History of Mathematics (New York: Wiley, 1989), pp. 389, 432.
2. C. H. Edwards, The Historical Development of the Calculus (New York: SpringerVerlag, 1979), pp. 124, 132.
3. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Saunders, 1990), pp. 391, 395.
4. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxford
University Press, 1972), pp. 344, 346.  3.2 The Derivative as a Function
In the preceding section we considered the derivative of a function f at a ﬁxed number a:
1 fa lim fa hl0 h
h fa Here we change our point of view and let the number a vary. If we replace a in Equation 1
by a variable x, we obtain 2 fx lim hl0 fx h
h fx Given any number x for which this limit exists, we assign to x the number f x . So we can
regard f as a new function, called the derivative of f and deﬁned by Equation 2. We
know that the value of f at x, f x , can be interpreted geometrically as the slope of the
tangent line to the graph of f at the point x, f x .
The function f is called the derivative of f because it has been “derived” from f by
the limiting operation in Equation 2. The domain of f is the set x f x exists and may
be smaller than the domain of f .
EXAMPLE 1 The graph of a function f is given in Figure 1. Use it to sketch the graph of
the derivative f .
y
y=ƒ
1
0 FIGURE 1 1 x 5E03(pp 126135) 1/17/06 1:50 PM Page 135 S ECTION 3.2 THE DERIVATIVE AS A FUNCTION Watch an animation of the relation between a
function and its derivative.
Resources / Module 3
/ Derivatives as Functions
/ Mars Rover
Resources / Module 3
/ SlopeaScope
/ Derivative of a Cubic ❙❙❙❙ 135 SOLUTION We can estimate the value of the derivative at any value of x by drawing the
tangent at the point x, f x and estimating its slope. For instance, for x 5 we draw the
tangent at P in Figure 2(a) and estimate its slope to be about 3 , so f 5
1.5. This
2
allows us to plot the point P 5, 1.5 on the graph of f directly beneath P. Repeating
this procedure at several points, we get the graph shown in Figure 2(b). Notice that the
tangents at A, B, and C are horizontal, so the derivative is 0 there and the graph of f
crosses the xaxis at the points A , B , and C , directly beneath A, B, and C. Between A
and B the tangents have positive slope, so f x is positive there. But between B and C
the tangents have negative slope, so f x is negative there.
y B y=ƒ 1 P A 0 5 1 x C
 Notice that where the derivative is positive
(to the right of C and between A and B), the
function f is increasing. Where f x is negative
(to the left of A and between B and C ), f is
decreasing. In Section 4.3 we will prove that this
is true for all functions. (a) y P ª (5, 1.5)
y=fª(x) 1 Bª
0 FIGURE 2 Aª Cª 1 5 x (b) If a function is deﬁned by a table of values, then we can construct a table of approximate values of its derivative, as in the next example. 5E03(pp 136145) 136 ❙❙❙❙ 1/17/06 2:33 PM Page 136 CHAPTER 3 DERIVATIVES t
1980
1982
1984
1986
1988
1990
1992
1994
1996
1998
2000 EXAMPLE 2 Let B t be the population of Belgium at time t. The table at the left gives
midyear values of B t , in thousands, from 1980 to 2000. Construct a table of values for
the derivative of this function. Bt
9,847
9,856
9,855
9,862
9,884
9,962
10,036
10,109
10,152
10,175
10,186 SOLUTION We assume that there were no wild ﬂuctuations in the population between the
stated values. Let’s start by approximating B 1988 , the rate of increase of the population of Belgium in mid1988. Since B 1988 B 1988 lim h l0 h
h B 1988 we have
B 1988 B 1988 h
h B 1988 for small values of h.
For h 2, we get
B 1988 B 1990 B 1988 9962 9884 2 2 39 (This is the average rate of increase between 1988 and 1990.) For h
B 1988
t B 1988 9862 9884 2 2 11 which is the average rate of increase between 1986 and 1988. We get a more accurate
approximation if we take the average of these rates of change: Bt 1980
1982
1984
1986
1988
1990
1992
1994
1996
1998
2000 B 1986 2, we have 4.5
2.0
1.5
7.3
25.0
38.0
36.8
29.0
16.5
8.5
5.5 B 1988 1
2 39 11 25 This means that in 1988 the population was increasing at a rate of about 25,000 people
per year.
Making similar calculations for the other values (except at the endpoints), we get the
table at the left, which shows the approximate values for the derivative.
y
10,200
10,100 y=B(t) 10,000
9,900
9,800
 Figure 3 illustrates Example 2 by showing
graphs of the population function B t and its
derivative B t . Notice how the rate of population growth increases to a maximum in 1990
and decreases thereafter. 1980 1984 1988 1992 1996 2000 t 1988 1992 1996 2000 t y
30
20 y=Bª(t) 10 FIGURE 3 1980 1984 5E03(pp 136145) 1/17/06 2:33 PM Page 137 S ECTION 3.2 THE DERIVATIVE AS A FUNCTION ❙❙❙❙ 137 EXAMPLE 3 (a) If f x
x 3 x, ﬁnd a formula for f x .
(b) Illustrate by comparing the graphs of f and f .
SOLUTION (a) When using Equation 2 to compute a derivative, we must remember that the variable
is h and that x is temporarily regarded as a constant during the calculation of the limit.
fx fx lim h
h hl0 x lim 3 fx lim 2 3x h 3x 2h 2 h
h h3 lim 3x 2 h2 3xh hl0 3 x 3 x3 h x h h 3xh 3xh 2
h hl0 h hl0 hl0 lim x h 3x 2 1 x x 3 x 1 (b) We use a graphing device to graph f and f in Figure 4. Notice that f x
0 when
f has horizontal tangents and f x is positive when the tangents have positive slope. So
these graphs serve as a check on our work in part (a).
2 2 fª f
_2 FIGURE 4
See more problems like these.
Resources / Module 3
/ How to Calculate
/ The Essential Examples 2 _2 _2 EXAMPLE 4 If f x sx 1, ﬁnd the derivative of f . State the domain of f . SOLUTION fx lim fx h
h hl0 lim
lim h 1
h sx 1 sx h 1
h sx 1 hl0 lim hl0 lim hl0 fx sx hl0 Here we rationalize the numerator. _2 2 xh1
h(sx h 1
sx h 1
1 sx 1 sx 1 h
h 1
1 sx
sx 1
1 x1
sx 1 )
sx 1 sx
sx 1
1
2 sx 1 We see that f x exists if x 1, so the domain of f is 1,
the domain of f , which is 1, . . This is smaller than 5E03(pp 136145) 138 ❙❙❙❙ 1/17/06 2:33 PM Page 138 CHAPTER 3 DERIVATIVES Let’s check to see that the result of Example 4 is reasonable by looking at the graphs of
f and f in Figure 5. When x is close to 1, s x 1 is close to 0, so f x
1 (2 s x 1 )
is very large; this corresponds to the steep tangent lines near 1, 0 in Figure 5(a) and the
large values of f x just to the right of 1 in Figure 5(b). When x is large, f x is very
small; this corresponds to the ﬂatter tangent lines at the far right of the graph of f and the
horizontal asymptote of the graph of f .
y y 1 1 0 x 1 F IGURE 5 SOLUTION fx fx h
h hl0 lim e ad bc
bd lim 1
e fx 1
2 x
x h
h
h 1 x h2
h2 2 x 2h hl0 lim hl0 lim hl0 lim hl0 1
2 œ„„„„
x1 x
.
x hl0 c
d (b) f ª(x)= 1
2 lim x 1 (a) ƒ=œ„„„„
x1 EXAMPLE 5 Find f if f x a
b 0 x 3
h2 x
x x
x 1 x2
h2 x x2
h2 3h
x h2 h2
2 1
2 xh
x 2
h2 x
x
x h
h x2 xh x
3
x 2 x 2 Other Notations
If we use the traditional notation y f x to indicate that the independent variable is x and
the dependent variable is y, then some common alternative notations for the derivative are
as follows:
fx y dy
dx df
dx d
fx
dx Df x Dx f x The symbols D and d d x are called differentiation operators because they indicate the
operation of differentiation, which is the process of calculating a derivative. 5E03(pp 136145) 1/17/06 2:33 PM Page 139 S ECTION 3.2 THE DERIVATIVE AS A FUNCTION  Gottfried Wilhelm Leibniz was born in
Leipzig in 1646 and studied law, theology,
philosophy, and mathematics at the university
there, graduating with a bachelor’s degree at age
17. After earning his doctorate in law at age 20,
Leibniz entered the diplomatic service and spent
most of his life traveling to the capitals of Europe
on political missions. In particular, he worked to
avert a French military threat against Germany
and attempted to reconcile the Catholic and
Protestant churches.
His serious study of mathematics did not
begin until 1672 while he was on a diplomatic
mission in Paris. There he built a calculating
machine and met scientists, like Huygens, who
directed his attention to the latest developments
in mathematics and science. Leibniz sought to
develop a symbolic logic and system of notation
that would simplify logical reasoning. In particular, the version of calculus that he published in
1684 established the notation and the rules for
ﬁnding derivatives that we use today.
Unfortunately, a dreadful priority dispute arose
in the 1690s between the followers of Newton
and those of Leibniz as to who had invented
calculus ﬁrst. Leibniz was even accused of
plagiarism by members of the Royal Society in
England. The truth is that each man invented
calculus independently. Newton arrived at his
version of calculus ﬁrst but, because of his fear
of controversy, did not publish it immediately. So
Leibniz’s 1684 account of calculus was the ﬁrst
to be published. ❙❙❙❙ 139 The symbol d y d x, which was introduced by Leibniz, should not be regarded as a ratio
(for the time being); it is simply a synonym for f x . Nonetheless, it is a very useful and
suggestive notation, especially when used in conjunction with increment notation.
Referring to Equation 3.1.4, we can rewrite the deﬁnition of derivative in Leibniz notation
in the form
dy
dx y
x lim xl0 If we want to indicate the value of a derivative dy d x in Leibniz notation at a speciﬁc number a, we use the notation
dy
dx dy
dx or
xa xa which is a synonym for f a . 3 Definition A function f is differentiable at a if f a exists. It is differentiable
on an open interval a, b [or a,
or
, a or
, ] if it is differentiable
at every number in the interval. EXAMPLE 6 Where is the function f x
SOLUTION If x hence x h x differentiable? 0, then x
x and we can choose h small enough that x
x h. Therefore, for x 0 we have
fx x h
h x lim h
h hl0 lim hl0 and so f is differentiable for any x
Similarly, for x 0 we have x
x h 0 and so x h
x
fx lim x lim x x
x lim hl0 h
h lim 1 hl0 x
x lim hl0 h
h lim hl0 and so f is differentiable for any x 0.
For x 0 we have to investigate
f0 lim f0 hl0 lim hl0 1 x and h can be chosen small enough that
h . Therefore, for x 0, h
h hl0 0 and 0. h
h hl0 h 0 h
h
h
h f0
0 if it exists 1 1 5E03(pp 136145) 140 ❙❙❙❙ 1/17/06 2:34 PM Page 140 CHAPTER 3 DERIVATIVES Let’s compute the left and right limits separately:
lim 0 h
h
h
h hl0 and lim 0 hl0 0 lim 0 h
h lim h
h hl0 hl0 h
h lim hl0 lim 1 1 lim 1 hl0 h
h lim hl0 hl0 1 Since these limits are different, f 0 does not exist. Thus, f is differentiable at all x
except 0.
A formula for f is given by
1 fx 1 if x
if x 0
0 and its graph is shown in Figure 6(b). The fact that f 0 does not exist is reﬂected geometrically in the fact that the curve y
x does not have a tangent line at 0, 0 .
[See Figure 6(a).]
y
y
1
x 0
x 0 FIGURE 6 _1 (a) y=ƒ= x  (b) y=fª(x) Both continuity and differentiability are desirable properties for a function to have. The
following theorem shows how these properties are related.
4 Theorem If f is differentiable at a, then f is continuous at a. Proof To prove that f is continuous at a, we have to show that lim x l a f x f a . We do this by showing that the difference f x
f a approaches 0.
The given information is that f is differentiable at a, that is,
fa lim xla fx
x fa
a exists (see Equation 3.1.3). To connect the given and the unknown, we divide and multiply f x
f a by x a (which we can do when x a):
fx fa fx
x fa
a x a Thus, using the Product Law and (3.1.3), we can write
lim f x xla fa lim xla fx
x fa
a x a 5E03(pp 136145) 1/17/06 2:34 PM Page 141 S ECTION 3.2 THE DERIVATIVE AS A FUNCTION xla lim fx
x fa fa
lim x
xla
a 0 ❙❙❙❙ 141 a 0 To use what we have just proved, we start with f x and add and subtract f a :
lim f x lim f a xla fx xla lim f a lim f x xla fa fa
fa xla 0 fa Therefore, f is continuous at a.  NOTE The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function f x
x is continuous at 0 because
■ lim f x lim x xl0 xl0 0 f0 (See Example 7 in Section 2.3.) But in Example 6 we showed that f is not differentiable
at 0. How Can a Function Fail to Be Differentiable?
We saw that the function y
x in Example 6 is not differentiable at 0 and Figure 6(a)
shows that its graph changes direction abruptly when x 0. In general, if the graph of a
function f has a “corner” or “kink” in it, then the graph of f has no tangent at this point
and f is not differentiable there. [In trying to compute f a , we ﬁnd that the left and right
limits are different.]
Theorem 4 gives another way for a function not to have a derivative. It says that if f is
not continuous at a, then f is not differentiable at a. So at any discontinuity (for instance,
a jump discontinuity) f fails to be differentiable.
A third possibility is that the curve has a vertical tangent line when x a; that is, f
is continuous at a and y vertical tangent
line 0 a x lim f x xla F IGURE 7 This means that the tangent lines become steeper and steeper as x l a. Figure 7 shows
one way that this can happen; Figure 8(c) shows another. Figure 8 illustrates the three possibilities that we have discussed.
y 0 y a x 0 y a x 0 a FIGURE 8 Three ways for ƒ not to be
differentiable at a (a) A corner (b) A discontinuity (c) A vertical tangent x 5E03(pp 136145) 142 ❙❙❙❙ 1/17/06 2:34 PM Page 142 CHAPTER 3 DERIVATIVES A graphing calculator or computer provides another way of looking at differentiability.
If f is differentiable at a, then when we zoom in toward the point a, f a the graph
straightens out and appears more and more like a line. (See Figure 9. We saw a speciﬁc
example of this in Figure 3 in Section 3.1.) But no matter how much we zoom in toward a
point like the ones in Figures 7 and 8(a), we can’t eliminate the sharp point or corner (see
Figure 10).
y 0 y a x 0 x a FIGURE 9  3.2 FIGURE 10 ƒ is differentiable at a. ƒ is not differentiable at a. Exercises
3. (a) f 1–3  Use the given graph to estimate the value of each derivative.
Then sketch the graph of f . 3 (c) f 3 (b) f 2 (b) f 2
(d) f 4 (c) f 1 y 1. (a) f 1 y (e) f 1 y=f(x)
1 (d) f 0
0 x 1 (f ) f 2 y=ƒ (g) f 3
■ 1
0 ■ (c) f 2
(e) f 4 ■ ■ ■ ■ ■ ■ ■ ■ ■ 4. Match the graph of each function in (a)–(d) with the graph of
1 x its derivative in I–IV. Give reasons for your choices.
(a) 2. (a) f 0 ■ y (b) f 1
(d) f 3
(f ) f 5 (b) 0 x y 0 x y y=f(x) (c) 0 1
0 y 1 x (d)
x y 0 x 5E03(pp 136145) 1/17/06 2:34 PM Page 143 S ECTION 3.2 THE DERIVATIVE AS A FUNCTION y I II 0 x 13. y 0 y IV 143 y x
0 III ❙❙❙❙ y ■ ■ ■ x
■ ■ ■ ■ ■ ■ ■
■ ■ 14. Shown is the graph of the population function P t for yeast
0 x 0 cells in a laboratory culture. Use the method of Example 1 to
graph the derivative P t . What does the graph of P tell us
about the yeast population? x P (yeast cells) 5–13 Trace or copy the graph of the given function f . (Assume
that the axes have equal scales.) Then use the method of Example 1
to sketch the graph of f below it.
 5. 6. y 0 500 y 0 x
0 5 10 15 t (hours) x 15. The graph shows how the average age of ﬁrst marriage of 7. 8. y Japanese men varied in the last half of the 20th century. Sketch
the graph of the derivative function M t . During which years
was the derivative negative?
y
M
0 0 x 27 x 25 9. 10. y 1960 y 1970 1980 1990 t 16. Make a careful sketch of the graph of the sine function and
0 x
0 x below it sketch the graph of its derivative in the same manner
as in Exercises 5–13. Can you guess what the derivative of the
sine function is from its graph?
x 2.
Estimate the values of f 0 , f ( 1 ), f 1 , and f 2 by
2
using a graphing device to zoom in on the graph of f.
Use symmetry to deduce the values of f ( 1 ), f
1,
2
and f
2.
Use the results from parts (a) and (b) to guess a formula
for f x .
Use the deﬁnition of a derivative to prove that your guess in
part (c) is correct. ; 17. Let f x
(a)
11. 12. y y (b)
(c) 0 x 0 x (d) 5E03(pp 136145) 144 ❙❙❙❙ 1/17/06 2:34 PM Page 144 CHAPTER 3 DERIVATIVES x 3.
(a) Estimate the values of f 0 , f ( 1 ), f 1 , f 2 , and f 3
2
by using a graphing device to zoom in on the graph of f.
(b) Use symmetry to deduce the values of f ( 1 ), f
1,
2
f
2 , and f
3.
(c) Use the values from parts (a) and (b) to graph f .
(d) Guess a formula for f x .
(e) Use the deﬁnition of a derivative to prove that your guess in
part (d) is correct. ; 18. Let f x 19–29  Find the derivative of the function using the deﬁnition of
derivative. State the domain of the function and the domain of its
derivative.
19. f x
21. f x 20. f x 37
1 23. f x x 25. t x s1 27. G t
29. f x
■ ■ x 7x
2 22. f x
24. f x 5 2x 1 5x
x
3
1 x
3x 1
x2 ■ ■ time t. The table gives values of this function in census years
from 1950 to 2000. t t Pt 1950
1960
1970 (a)
(b)
(c)
(d) Pt
31.1
35.7
34.0 1980
1990
2000 28.0
25.7
25.7 What is the meaning of P t ? What are its units?
Construct a table of values for P t .
Graph P and P .
How would it be possible to get more accurate values
for P t ? sx 26. f x 3x 4t
t 12 28. t x 3x
3 2 34. Let P t be the percentage of Americans under the age of 18 at 3x 2 35. The graph of f is given. State, with reasons, the numbers at which f is not differentiable.
y 4 ■ ■ ■ ■ ■ ■ ■ ■ 30. (a) Sketch the graph of f x ; s6 x by starting with the
graph of y s x and using the transformations of Section 1.3.
(b) Use the graph from part (a) to sketch the graph of f .
(c) Use the deﬁnition of a derivative to ﬁnd f x . What are the
domains of f and f ?
(d) Use a graphing device to graph f and compare with your
sketch in part (b). 31. (a) If f x ; x
2 x , ﬁnd f x .
(b) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f . 2 4 6 8 12 x 10 36. The graph of t is given. (a) At what numbers is t discontinuous? Why?
(b) At what numbers is t not differentiable? Why?
y 6 1 t 2 , ﬁnd f t .
(b) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f . 32. (a) If f t ; 33. The unemployment rate U t varies with time. The table (from the Bureau of Labor Statistics) gives the percentage of unemployed in the U.S. labor force from 1991 to 2000.
t Ut t 6.8
7.5
6.9
6.1
5.6 1996
1997
1998
1999
2000 5.4
4.9
4.5
4.2
4.0 x Ut 1991
1992
1993
1994
1995 01 (a) What is the meaning of U t ? What are its units?
(b) Construct a table of values for U t . ; 37. Graph the function f x x s x . Zoom in repeatedly, ﬁrst
toward the point ( 1, 0)
and then toward the origin. What is different about the behavior
of f in the vicinity of these two points? What do you conclude
about the differentiability of f ? ; 38. Zoom in toward the points (1, 0), (0, 1), and ( 1, 0) on the
graph of the function t x
x 2 1 2 3. What do you notice?
Account for what you see in terms of the differentiability of t. 5E03(pp 136145) 1/17/06 2:34 PM Page 145 S ECTION 3.3 DIFFERENTIATION FORMULAS 3
sx.
(a) If a 0, use Equation 3.1.3 to ﬁnd f a .
(b) Show that f 0 does not exist.
3
(c) Show that y sx has a vertical tangent line at 0, 0 .
(Recall the shape of the graph of f . See Figure 13 in Section 1.2.) 0
5 fx x 2 3, show that t 0 does not exist.
(b) If a 0, ﬁnd t a .
(c) Show that y x 2 3 has a vertical tangent line at 0, 0 .
(d) Illustrate part (c) by graphing y x 2 3. 5 x 6 is not differentiable
at 6. Find a formula for f and sketch its graph. x x. water depends on how long the water has been running.
(a) Sketch a possible graph of T as a function of the time t that
has elapsed since the faucet was turned on.
(b) Describe how the rate of change of T with respect to t
varies as t increases.
(c) Sketch a graph of the derivative of T . 44. The lefthand and righthand derivatives of f at a are deﬁned by f and  a fa fa lim hl0 h
h fa h
h 4 46. When you turn on a hotwater faucet, the temperature T of the (b) For what values of x is f differentiable?
(c) Find a formula for f . lim if x 4 x
f x for all x
in its domain and odd if f x
f x for all such x. Prove
each of the following.
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function. x not differentiable? Find a formula for f and sketch its graph. hl0 0
x 45. Recall that a function f is called even if f 42. Where is the greatest integer function f x a x if x
if 0 (b) Sketch the graph of f .
(c) Where is f discontinuous?
(d) Where is f not differentiable? 41. Show that the function f x f x
1 40. (a) If t x 43. (a) Sketch the graph of the function f x 145 if these limits exist. Then f a exists if and only if these onesided derivatives exist and are equal.
(a) Find f 4 and f 4 for the function 39. Let f x ; ❙❙❙❙ fa be the tangent line to the parabola y x 2 at the point
1, 1 . The angle of inclination of is the angle that makes
with the positive direction of the xaxis. Calculate correct to
the nearest degree. 47. Let 3.3 Differentiation Formulas
If it were always necessary to compute derivatives directly from the deﬁnition, as we did
in the preceding section, such computations would be tedious and the evaluation of some
limits would require ingenuity. Fortunately, several rules have been developed for ﬁnding
derivatives without having to use the deﬁnition directly. These formulas greatly simplify
the task of differentiation.
Let’s start with the simplest of all functions, the constant function f x
c. The graph
of this function is the horizontal line y c, which has slope 0, so we must have f x
0.
(See Figure 1.) A formal proof, from the deﬁnition of a derivative, is also easy: y
c y=c fx slope=0 lim fx lim 0 hl0 0 FIGURE 1 The graph of ƒ=c is the
line y=c, so f ª(x)=0. x h
h hl0 fx 0 In Leibniz notation, we write this rule as follows. Derivative of a Constant Function d
c
dx lim hl0 0 c c
h 5E03(pp 146155) 146 ❙❙❙❙ 1/17/06 2:14 PM Page 146 CHAPTER 3 DERIVATIVES Power Functions
y x n, where n is a positive integer. If n
We next look at the functions f x
x is the line y x, which has slope 1 (see Figure 2). So
of f x 1, the graph y=x
slope=1 d
x
dx 1 1 0
x FIGURE 2 The graph of ƒ=x is the
line y=x, so f ª(x)=1. (You can also verify Equation 1 from the deﬁnition of a derivative.) We have already
investigated the cases n 2 and n 3. In fact, in Section 3.2 (Exercises 17 and 18) we
found that
d
d
x2
2x
x3
3x 2
2
dx
dx
For n x 4 as follows: 4 we ﬁnd the derivative of f x
fx lim fx h
h hl0 lim x4 fx 4 x 3h 4 x 3h lim 4 x 3 6 x 2h hl0 x4 4 xh 3 6x 2h 2
h 6x 2h 2
h hl0 h4
h
h4 x4 hl0 hl0 lim x lim 4 xh 3
4 xh 2 h4
h3 4x 3 Thus
d
x4
dx 3 4x 3 Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a
n x n 1. This turns out to
reasonable guess that, when n is a positive integer, d d x x n
be true. We prove it in two ways; the second proof uses the Binomial Theorem.
The Power Rule If n is a positive integer, then d
xn
dx nx n 1 First Proof The formula xn an x a xn 1 x n 2a xa n 2 an 1 can be veriﬁed simply by multiplying out the righthand side (or by summing the second
x n, we can use Equation 3.1.3 for f a and the
factor as a geometric series). If f x
equation above to write
fa lim xla fx
x fa
a lim xla xn
x an
a 5E03(pp 146155) 1/17/06 2:15 PM Page 147 S ECTION 3.3 DIFFERENTIATION FORMULAS lim x n 1 x n 2a xla an 1 na n xa n a n 2a aa n 2 2 an an ❙❙❙❙ 147 1 1 1 Second Proof fx
 The Binomial Theorem is given on
Reference Page 1. fx lim h
h hl0 fx hn
h x lim hl0 xn In ﬁnding the derivative of x 4 we had to expand x h 4. Here we need to expand
x h n and we use the Binomial Theorem to do so:
xn
fx nn n x n 1h 1
2 lim x n 2h 2 1 hn xn h hl0 nn n x n 1h 1
2 lim x n 2h 2 n xh n 1 hn h hl0 lim n x n nn 1 1
2 hl0 nx n n xh n x n 2h n xh n 2 hn 1 1 because every term except the ﬁrst has h as a factor and therefore approaches 0.
We illustrate the Power Rule using various notations in Example 1.
EXAMPLE 1 (a) If f x
(c) If y
(e) Du u m x 6, then f x
t 4, then
mu m dy
dt 6 x 5. 4t 3. (b) If y
(d) x 1000, then y d3
r
dr 1000 x 999. 3r 2 1 New Derivatives from Old
When new functions are formed from old functions by addition, subtraction, multiplication, or division, their derivatives can be calculated in terms of derivatives of the old functions. In particular, the following formula says that the derivative of a constant times a
function is the constant times the derivative of the function.
The Constant Multiple Rule If c is a constant and f is a differentiable function, then d
cf x
dx c d
fx
dx 5E03(pp 146155) 148 ❙❙❙❙ 1/17/06 2:15 PM Page 148 CHAPTER 3 DERIVATIVES  GEOMETRIC INTERPRETATION
OF THE CONSTANT MULTIPLE RULE Proof Let t x c f x . Then
tx y lim tx h
h hl0 y=2ƒ fx lim c hl0 y=ƒ
0 x Multiplying by c 2 stretches the graph vertically by a factor of 2. All the rises have been
doubled but the runs stay the same. So the
slopes are doubled, too. fx c lim tx h
h hl0 h
h cf x fx h
h hl0 cf x lim fx (by Law 3 of limits) cf x
EXAMPLE 2 (a) d
3x 4
dx 3 (b) d
dx d
dx x d
x4
dx 3 4x 3 1x 1 12 x 3
d
x
dx 11 1 The next rule tells us that the derivative of a sum of functions is the sum of the
derivatives.
The Sum Rule If f and t are both differentiable, then
 Using prime notation, we can write the
Sum Rule as
ft
f
t d
fx
dx Proof Let F x d
tx
dx t x . Then fx Fx d
fx
dx tx lim Fx h
h Fx fx h tx hl0 lim h
h hl0 lim fx h
h hl0 lim hl0 fx fx fx h
h fx fx tx h
h tx tx lim tx h
h hl0 tx (by Law 1) tx The Sum Rule can be extended to the sum of any number of functions. For instance,
using this theorem twice, we get
f t h f t h f t h f t h By writing f t as f
1 t and applying the Sum Rule and the Constant Multiple
Rule, we get the following formula. 5E03(pp 146155) 1/17/06 2:16 PM Page 149 S ECTION 3.3 DIFFERENTIATION FORMULAS ❙❙❙❙ 149 The Difference Rule If f and t are both differentiable, then d
fx
dx d
fx
dx tx d
tx
dx The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined
with the Power Rule to differentiate any polynomial, as the following examples demonstrate.
EXAMPLE 3 Try more problems like this one.
Resources / Module 4
/ Polynomial Models
/ Basic Differentiation Rules and Quiz d
x8
dx 12 x 5 4x 4
d
x8
dx 10 x 3
12 8x 7 12 5x 4 8x 7 60 x 4 6x 5 d
x5
dx d
x4
dx 4 4x 3
16 x 3 10 10 3x 2 4 61 30 x 2 6 d
x
dx d
5
dx 0 6
x4 EXAMPLE 4 Find the points on the curve y d
x3
dx 6x 2 4 where the tangent line is horizontal.
SOLUTION Horizontal tangents occur where the derivative is zero. We have dy
dx d
x4
dx
4x 3 6
12 x d
x2
dx
0 d
4
dx 4x x 2 3 Thus, dy d x 0 if x 0 or x 2 3 0, that is, x
s3. So the given curve has
horizontal tangents when x 0, s3, and s3.
The corresponding points are 0, 4 , (s3, 5), and ( s3, 5). (See Figure 3.)
y
(0, 4) 0 x FIGURE 3 The curve y=x$6x@+4 and
its horizontal tangents Resources / Module 4
/ Polynomial Models
/ Product and Quotient Rules {_ œ„, _5}
3 {œ„, _5}
3 Next we need a formula for the derivative of a product of two functions. By analogy
with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three
centuries ago, that the derivative of a product is the product of the derivatives. We can see,
however, that this guess is wrong by looking at a particular example. Let f x
x and
tx
x 2. Then the Power Rule gives f x
1 and t x
2 x. But f t x
x 3, so
 f t x 3x 2. Thus, f t
f t . The correct formula was discovered by Leibniz (soon
after his false start) and is called the Product Rule. 5E03(pp 146155) 150 ❙❙❙❙ 1/17/06 2:17 PM Page 150 CHAPTER 3 DERIVATIVES  We can write the Product Rule in prime notation as
ft
ft
tf The Product Rule If f and t are both differentiable, then d
f xtx
dx Proof Let F x d
tx
dx fx tx d
fx
dx f x t x . Then
Fx Fx h
h fx lim htx hl0 lim Fx h f xtx h hl0 In order to evaluate this limit, we would like to separate the functions f and t as in
the proof of the Sum Rule. We can achieve this separation by subtracting and adding the
term f x h t x in the numerator:
Fx lim fx htx h fx htx
h hl0 lim f x h hl0 lim f x hl0 f xt x h tx lim h
h tx tx h
h hl0 fx tx
tx htx fx h
h lim t x hl0 f xtx fx lim hl0 fx h
h fx txf x Note that lim h l 0 t x
t x because t x is a constant with respect to the variable h.
Also, since f is differentiable at x, it is continuous at x by Theorem 3.2.4, and so
lim h l 0 f x h
f x . (See Exercise 53 in Section 2.5.)
In words, the Product Rule says that the derivative of a product of two functions is the
ﬁrst function times the derivative of the second function plus the second function times the
derivative of the ﬁrst function.
EXAMPLE 5 Find F x if F x 6 x 3 7x 4 . SOLUTION By the Product Rule, we have Fx 6x 3 d
7x 4
dx 6 x 3 28 x 3
168 x 6 7x 4 d
6x 3
dx 7x 4 18 x 2 126 x 6 294 x 6 Notice that we could verify the answer to Example 5 directly by ﬁrst multiplying the
factors:
Fx 6 x 3 7x 4 42 x 7 ? Fx 42 7x 6 294 x 6 5E03(pp 146155) 1/17/06 2:18 PM Page 151 S ECTION 3.3 DIFFERENTIATION FORMULAS 151 x 2 sin x, for which the Product Rule is the But later we will meet functions, such as y
only possible method.
EXAMPLE 6 If h x ❙❙❙❙ x t x and it is known that t 3 5 and t 3 2, ﬁnd h 3 . SOLUTION Applying the Product Rule, we get d
xt x
dx hx xt x
Therefore h3 d
tx
dx x tx d
x
dx 5 11 tx 3t 3 t3 32 The Quotient Rule If f and t are differentiable, then
 In prime notation we can write the Quotient
Rule as
tf
ft
f
t2
t d
dx
Proof Let F x tx fx
tx d
fx
dx d
tx
dx fx
2 tx f x t x . Then Fx lim Fx hl0 fx lim hl0 h
h Fx htx
ht x lim fx
tx h
h
h hl0 f xtx
htx fx
tx h We can separate f and t in this expression by subtracting and adding the term f x t x
in the numerator:
Fx lim fx htx hl0 tx fx h
h lim hl0 fx
tx hl0 lim t x f xtx
ht x lim fx hl0 fx tx f xtx h
h h tx htx h
fx
tx
lim f x lim
hl0
hl0
h
lim t x h lim t x
hl0 txf x f xtx
htx h
h tx hl0 f xt x
tx 2 Again t is continuous by Theorem 3.2.4, so lim h l 0 t x h tx. In words, the Quotient Rule says that the derivative of a quotient is the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator. 5E03(pp 146155) 152 ❙❙❙❙ 1/17/06 2:19 PM Page 152 CHAPTER 3 DERIVATIVES The theorems of this section show that any polynomial is differentiable on and any
rational function is differentiable on its domain. Furthermore, the Quotient Rule and the
other differentiation formulas enable us to compute the derivative of any rational function,
as the next example illustrates.
 We can use a graphing device to check that
the answer to Example 7 is plausible. Figure 4
shows the graphs of the function of Example 7
and its derivative. Notice that when y grows
rapidly (near 2), y is large. And when y grows
slowly, y is near 0. x2 EXAMPLE 7 Let y x
x 3 2
6 Then
x3 d
x2
dx 6 y 1.5 x x2 2
x3 yª x3 6 2x 2x 4 _4 . x3 4
y 12 x x3
_1.5 x4 2x 3 FIGURE 4 6x 2
x
6
3 x 6 d
x3
dx 6 2 3x 2 3x 4 6 2 2 6 x2
62 1
x3 x 3x 3 6x 2 2 12 x 6 2 General Power Functions
The Quotient Rule can also be used to extend the Power Rule to the case where the exponent is a negative integer.
If n is a positive integer, then
d
x
dx d
x
dx d
dx n d
1
dx nx n1 1
xn xn Proof n 1
x nx n
x 2n d
xn
dx 1
dy
, then
x
dx (b) d
dt 6
t3 6 d
t
dt d
x
dx 1 nx n 3 6 1 3t x 4 1 nx n
x 1 2n nx EXAMPLE 8 (a) If y xn 0 n2 1
x2 2 18
t4 n1 2n 1 5E03(pp 146155) 1/17/06 2:19 PM Page 153 S ECTION 3.3 DIFFERENTIATION FORMULAS ❙❙❙❙ 153 5E03(pp 146155) ❙❙❙❙ 154 1/17/06 2:20 PM Page 154 CHAPTER 3 DERIVATIVES So far we know that the Power Rule holds if the exponent n is a positive or negative
integer. If n 0, then x 0 1, which we know has a derivative of 0. Thus, the Power Rule
holds for any integer n. What if the exponent is a fraction? In Example 3 in Section 2.6 we
found, in effect, that 1 1
”1, 2 ’ y= x
œ„
1+≈ d
sx
dx 4 0 1
2 sx which can be written as FIGURE 5 d 12
x
dx xy=12 0 x 12 This shows that the Power Rule is true even when n 1 . In fact, it also holds for any real
2
number n, as we will prove in Chapter 7. (A proof for rational values of n is indicated in
Exercise 40 in Section 3.7.) In the meantime we state the general version and use it in the
examples and exercises. y
(2, 6) 1
2 The Power Rule (General Version) If n is any real number, then x d
xn
dx (_2, _6) nx n 1 3x+y=0
FIGURE 6 EXAMPLE 9 (a) If f x x , then f x x 1 .
1
sx 2 y (b) Let
dy
dx Then 3 d
x
dx
2
3 x 23 2
3 x 23 1 53 EXAMPLE 10 Find an equation of the tangent line to the curve y point (1,  3.3
1–20  1
2 x 2 at the sx 1 ). Exercises Differentiate the function. 1. f x 186.5 3. f x 5x 1 5. f x x2 3x 7. f t 1
4 9. V r 4
3 t 4 r3 2. f x 8 4 x 10 6. t x 5x 8 8. f t 16
2 10. R t 14. f t s10
x7 12. R x 6t s30 4. F x 4 9 ( 1 x) 5
2 11. Y t t 5t 13. F x 2x 5
3t 35 4 st 1
st 6
t 15. y x 25 17. y 4 2 16. y
18. t u 3
sx s2 u s3u 5E03(pp 146155) 1/17/06 2:21 PM Page 155 S ECTION 3.3 DIFFERENTIATION FORMULAS ■ 1
4
st 3 t2 19. v
■ 20. u ■ ■ ■ ■ ■ 3
st 2 ■ ■ ■  Estimate the value of f a by zooming in on the graph
of f . Then differentiate f to ﬁnd the exact value of f a and compare with your estimate. ■ x 2 1 x 3 1 in two ways: by
using the Product Rule and by performing the multiplication
ﬁrst. Do your answers agree? 21. Find the derivative of y x ■  2x3 24. Y u u 1
y2 25. F y
26. y 3 u 2u2 y 5y3 1
1 27. t x
29. y 3t
v3 31. y 1 2v sv x4 35. y ax 2 37. y ■ t t 1 ■ 1 ■ ■ ■ ■ 1 x Px an x where a n ; 44–46 C
x2 ; u 2u
u2 ■ 5 ; ax
cx ■ an 1 x n1 a2 x 2 b
d ■ ■ 3x
■ , x s x, 1, 2 ■ ■ ■ ■ 5x
■ ■ 4
■ ■ ■ a1 x 46. f x 3
■ ■ x ■ ■ ■ 1 2 x 2, ■ ■ ■ 1, 9 ■ ■ ■ x 1 x 2 is called a serpentine. Find an
equation of the tangent line to this curve at the point 3, 0.3 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen.
1, f 5
6, t 5
Find the following values.
(a) f t 5
(b) f t 5
4, t 3
lowing numbers.
(a) f t 3 a0 (c) f
t 2, f 3 (d)
3 1
x
■ ■ ■ 59. If f x sx t x , where t 4
4 and h 2 hx
x (c) t f x2 2.
5 5, ﬁnd the fol ft 3
f
f 8 and t 4 3, ﬁnd
d
dx 3, and t 5 6, and t 3
(b) 60. If h 2 3 4, 0.4 1 54. y
■ , x 1 1 x 2 is called a witch of Maria
Agnesi. Find an equation of the tangent line to this curve at
the point ( 1, 1 ).
2
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. 58. If f 3 1 15 ■ sx 52. y 1, 1 57. Suppose that f 5 Find f x . Compare the graphs of f and f and use them
to explain why your answer is reasonable. 45. f x 1 0. Find the derivative of P. x x2 a ■ 56. (a) The curve y
3  44. f x ■ ■ 55. (a) The curve y 43. The general polynomial of degree n has the form
n 1 sx, ■ 2 cx
6 42. f x c
x x cx 40. y x ■ B
x 38. y 3
st t 2 x A x 53. y x 36. y c x2 2x 51. y 1
1 34. y 1 bx 41. f x sx
sx 32. y r2
1 sr 39. y t
2 ■ 1
x2 ■ 51–54  Find an equation of the tangent line to the curve at the
given point. t2 4
t3
t4 30. y v 33. y 48. f x 1 2t 28. f t t2
2t 2 ■ tx
x 2 x 2 1 in the viewing rectangle 4, 4 by
1, 1.5 .
(b) Using the graph in part (a) to estimate slopes, make a rough
sketch, by hand, of the graph of t . (See Example 1 in Section 3.2.)
(c) Calculate t x and use this expression, with a graphing
device, to graph t . Compare with your sketch in part (b). 1 3x
2x 2x
u5 3
y4 sx x a ■ ; 50. (a) Use a graphing calculator or computer to graph the function 3 x4 2 ■ tion f x
x 4 3x 3 6 x 2 7x 30 in the viewing
rectangle 3, 5 by 10, 50 .
(b) Using the graph in part (a) to estimate slopes, make a
rough sketch, by hand, of the graph of f . (See Example 1
in Section 3.2.)
(c) Calculate f x and use this expression, with a graphing
device, to graph f . Compare with your sketch in part (b). 3x sx
sx Differentiate. 23. V x ■ x 3, ; 49. (a) Use a graphing calculator or computer to graph the func in two ways: by using the Quotient Rule and by simplifying
ﬁrst. Show that your answers are equivalent. Which method do
you prefer?
23–42 3x 2 47. f x 22. Find the derivative of the function Fx 155 ; 47–48 2 st 3 ■ ❙❙❙❙ t 3
7, ﬁnd f 4 . 5E03(pp 156165) 156 ❙❙❙❙ 1/17/06 2:09 PM Page 156 CHAPTER 3 DERIVATIVES 61. If f and t are the functions whose graphs are shown, let
ux
f x t x and v x
f x tx. (a) Find u 1 . (b) Find v 5 . 69. How many tangent lines to the curve y x x 1) pass
through the point 1, 2 ? At which points do these tangent lines
touch the curve? 70. Draw a diagram to show that there are two tangent lines to the y parabola y x 2 that pass through the point 0, 4 . Find the
coordinates of the points where these tangent lines intersect the
parabola. f
g 71. Show that the curve y 1 6x 3 3 has no tangent line 5x with slope 4. 0 x 1 72. Find equations of both lines through the point 2, x2 tangent to the parabola y
62. Let P x F x G x and Q x
F x G x , where F and G
are the functions whose graphs are shown.
(a) Find P 2 .
(b) Find Q 7 . 3 that are x. 73. (a) Use the Product Rule twice to prove that if f , t, and h are differentiable, then
f th f th ft h f th y (b) Use part (a) to differentiate y F 74. (a) Taking f 0 x 1 ative of each of the following functions.
x
(a) y x t x
(b) y
tx 3 3fx (b) Use part (a) to differentiate y 63. If t is a differentiable function, ﬁnd an expression for the deriv (c) y tx
x x 1 2x 3. h in Exercise 73, show that t d
fx
dx G 1 sx x 4 2 fx x4 3x 3 17x 82 3. 75. Find a cubic function y ax 3 bx 2 cx d whose graph has horizontal tangents at the points
and 2, 0 . 2, 6 64. If f is a differentiable function, ﬁnd an expression for the
76. A telephone company wants to estimate the number of new derivative of each of the following functions.
(a) y x2f x (c) y x2
fx fx
x2 (b) y
(d) y 1 xf x
sx 65. The normal line to a curve C at a point P is, by deﬁnition, the line that passes through P and is perpendicular to the tangent
line to C at P. Find an equation of the normal line to the
parabola y 1 x 2 at the point (2, 3). Sketch the parabola
and its normal line.
x x 2 at the
point (1, 0) intersect the parabola a second time? Illustrate with
a sketch. 66. Where does the normal line to the parabola y 67. Find the points on the curve y x3 x2 x tangent is horizontal. 1 where the residential phone lines that it will need to install during the
upcoming month. At the beginning of January the company
had 100,000 subscribers, each of whom had 1.2 phone lines,
on average. The company estimated that its subscribership was
increasing at the rate of 1000 monthly. By polling its existing
subscribers, the company found that each intended to install an
average of 0.01 new phone lines by the end of January.
(a) Let s t be the number of subscribers and let n t be the
number of phone lines per subscriber at time t, where t is
measured in years and t 0 corresponds to the beginning
of January. What are the values of s 0 and n 0 ? What are
the company’s estimates for s 0 and n 0 ?
(b) Estimate the number of new lines the company will have to
install in January by using the Product Rule to calculate the
rate of increase of lines at the beginning of the month.
77. In this exercise we estimate the rate at which the total personal 68. Find equations of the tangent lines to the curve y
that are parallel to the line x x
x 1
1
2y 2. income is rising in the RichmondPetersburg, Virginia, metropolitan area. In 1999, the population of this area was 961,400,
and the population was increasing at roughly 9200 people per
year. The average annual income was $30,593 per capita, and
this average was increasing at about $1400 per year (a little
above the national average of about $1225 yearly). Use the 5E03(pp 156165) 1/17/06 2:10 PM Page 157 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES Product Rule and these ﬁgures to estimate the rate at which
total personal income was rising in the RichmondPetersburg
area in 1999. Explain the meaning of each term in the Product
Rule.
78. A manufacturer produces bolts of a fabric with a ﬁxed width. The quantity q of this fabric (measured in yards) that is sold is
a function of the selling price p (in dollars per yard), so we can
write q f p . Then the total revenue earned with selling price
p is R p
pf p .
(a) What does it mean to say that f 20
10,000 and
f 20
350?
(b) Assuming the values in part (a), ﬁnd R 20 and interpret
your answer.
79. Let 2
x2 fx x
2x 2 if x
if x 1
1 Is f differentiable at 1? Sketch the graphs of f and f . 1
x2
x 2 x if x
1
if 1 x
if x 1 82. Where is the function h x x1
x 2 differentiable? Give a formula for h and sketch the graphs of h and h . 83. For what values of a and b is the line 2 x y b tangent to 2? 84. Let x2
mx fx b if x
if x 2
2 Find the values of m and b that make f differentiable
everywhere.
85. An easy proof of the Quotient Rule can be given if we make the prior assumption that F x exists, where F f t. Write
f Ft; then differentiate using the Product Rule and solve the
resulting equation for F . 1 c at a point P.
(a) Show that the midpoint of the line segment cut from this
tangent line by the coordinate axes is P.
(b) Show that the triangle formed by the tangent line and the
coordinate axes always has the same area, no matter where
P is located on the hyperbola. xl1 Give a formula for t and sketch the graphs of t and t .
x 2 9 differentiable? Find a formula for f .
(b) Sketch the graphs of f and f .  3.4 a x 2 when x the parabola y 87. Evaluate lim
81. (a) For what values of x is the function f x 157 86. A tangent line is drawn to the hyperbola xy 80. At what numbers is the following function t differentiable? tx ❙❙❙❙ x 1000 1
.
x1 88. Draw a diagram showing two perpendicular lines that intersect on the yaxis and are both tangent to the parabola y
Where do these lines intersect? x 2. Rates of Change in the Natural and Social Sciences
Recall from Section 3.1 that if y f x , then the derivative dy d x can be interpreted as the
rate of change of y with respect to x. In this section we examine some of the applications
of this idea to physics, chemistry, biology, economics, and other sciences.
Let’s recall from Section 2.6 the basic idea behind rates of change. If x changes from
x 1 to x 2, then the change in x is
x y y f x1 f x2
x2 f x1
x1 The difference quotient Îx
¤ mPQ average rate of change
m=fª(⁄)= instantaneous rate
of change
F IGURE 1 f x2 y
x Îy P { ⁄, ﬂ} ⁄ x1 and the corresponding change in y is Q { ¤, ‡} 0 x2 x is the average rate of change of y with respect to x over the interval x 1, x 2 and can be
interpreted as the slope of the secant line PQ in Figure 1. Its limit as x l 0 is the derivative f x 1 , which can therefore be interpreted as the instantaneous rate of change of y 5E03(pp 156165) 158 ❙❙❙❙ 1/17/06 2:10 PM Page 158 CHAPTER 3 DERIVATIVES with respect to x or the slope of the tangent line at P x 1, f x 1 . Using Leibniz notation, we
write the process in the form
dy
y
lim
xl0
dx
x
Whenever the function y f x has a speciﬁc interpretation in one of the sciences, its
derivative will have a speciﬁc interpretation as a rate of change. (As we discussed in Section 2.6, the units for d y d x are the units for y divided by the units for x.) We now look at
some of these interpretations in the natural and social sciences. Physics
Resources / Module 4
/ Polynomial Models
/ Start of Polynomial Models If s f t is the position function of a particle that is moving in a straight line, then s t
represents the average velocity over a time period t, and v ds dt represents the instantaneous velocity (the rate of change of displacement with respect to time). This was discussed in Sections 2.6 and 3.1, but now that we know the differentiation formulas, we are
able to solve velocity problems more easily.
EXAMPLE 1 The position of a particle is given by the equation s t3 ft 6t 2 9t where t is measured in seconds and s in meters.
(a) Find the velocity at time t.
(b) What is the velocity after 2 s? After 4 s?
(c) When is the particle at rest?
(d) When is the particle moving forward (that is, in the positive direction)?
(e) Draw a diagram to represent the motion of the particle.
(f ) Find the total distance traveled by the particle during the ﬁrst ﬁve seconds.
SOLUTION (a) The velocity function is the derivative of the position function.
ft t3 ds
dt s 3t 2 vt 6t 2 9t 12 t 9 (b) The velocity after 2 s means the instantaneous velocity when t
ds
dt v2 32 2 2 12 4 12 2 9 2, that is, 3m s t2 The velocity after 4 s is
v4 34 (c) The particle is at rest when v t
3t 2 12 t 9 9 9m s 0, that is, 3 t2 4t 3 3t 1t 3 0 and this is true when t 1 or t 3. Thus, the particle is at rest after 1 s and after 3 s.
(d) The particle moves in the positive direction when v t
0, that is,
3t 2 12 t 9 3t 1t 3 0 5E03(pp 156165) 1/17/06 2:11 PM Page 159 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES t=3
s=0 t=0
s=0 t=1
s=4 s 159 This inequality is true when both factors are positive t 3 or when both factors are
negative t 1 . Thus, the particle moves in the positive direction in the time intervals
t 1 and t 3. It moves backward (in the negative direction) when 1 t 3.
(e) Using the information from part (d), we make a schematic sketch as shown in
Figure 2 of the motion of the particle back and forth along a line (the saxis).
(f ) Because of what we learned in parts (d) and (e), we need to calculate the distances
traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately.
The distance traveled in the ﬁrst second is
f1 FIGURE 2 ❙❙❙❙ From t 1 to t f0 3 to t 0 4m 0 4 4m 20 0 20 m 3 the distance traveled is
f3 From t 4 f1 5 the distance traveled is
f5 The total distance is 4 4 20 f3
28 m. EXAMPLE 2 If a rod or piece of wire is homogeneous, then its linear density is uniform
and is deﬁned as the mass per unit length
m l and measured in kilograms per
meter. Suppose, however, that the rod is not homogeneous but that its mass measured
from its left end to a point x is m f x , as shown in Figure 3. x
x¡
FIGURE 3 x™ This part of the rod has mass ƒ . The mass of the part of the rod that lies between x x 1 and x x 2 is given by
m f x2
f x 1 , so the average density of that part of the rod is
average density m
x f x2
x2 f x1
x1 If we now let x l 0 (that is, x 2 l x 1 ), we are computing the average density over
smaller and smaller intervals. The linear density at x 1 is the limit of these average
densities as x l 0; that is, the linear density is the rate of change of mass with respect
to length. Symbolically,
m
dm
lim
xl0
x
dx
Thus, the linear density of the rod is the derivative of mass with respect to length.
For instance, if m f x
s x, where x is measured in meters and m in kilograms,
then the average density of the part of the rod given by 1 x 1.2 is
m
x
while the density right at x f 1.2
1.2 f1
1 s1.2 1
0.2 0.48 kg m 1 is
dm
dx x1 1
2 sx 0.50 kg m
x1 5E03(pp 156165) 160 ❙❙❙❙ 1/17/06 2:11 PM Page 160 CHAPTER 3 DERIVATIVES F IGURE 4 EXAMPLE 3 A current exists whenever electric charges move. Figure 4 shows part of a
wire and electrons moving through a shaded plane surface. If Q is the net charge that
passes through this surface during a time period t, then the average current during this
time interval is deﬁned as Q
t average current Q2
t2 Q1
t1 If we take the limit of this average current over smaller and smaller time intervals, we
get what is called the current I at a given time t1 :
I lim tl0 Q
t dQ
dt Thus, the current is the rate at which charge ﬂows through a surface. It is measured in
units of charge per unit time (often coulombs per second, called amperes).
Velocity, density, and current are not the only rates of change that are important in
physics. Others include power (the rate at which work is done), the rate of heat ﬂow, temperature gradient (the rate of change of temperature with respect to position), and the rate
of decay of a radioactive substance in nuclear physics. Chemistry
EXAMPLE 4 A chemical reaction results in the formation of one or more substances
(called products) from one or more starting materials (called reactants). For instance, the
“equation” 2H2 O2 l 2 H2 O indicates that two molecules of hydrogen and one molecule of oxygen form two molecules of water. Let’s consider the reaction
A BlC where A and B are the reactants and C is the product. The concentration of a reactant
A is the number of moles (1 mole 6.022 10 23 molecules) per liter and is denoted by
A . The concentration varies during a reaction, so A , B , and C are all functions of
time t . The average rate of reaction of the product C over a time interval t1 t t2 is
C
t C t2
t2 C t1
t1 But chemists are more interested in the instantaneous rate of reaction, which is
obtained by taking the limit of the average rate of reaction as the time interval t
approaches 0:
rate of reaction lim tl0 C
t dC
dt Since the concentration of the product increases as the reaction proceeds, the derivative
d C dt will be positive. (You can see intuitively that the slope of the tangent to the 5E03(pp 156165) 1/17/06 2:11 PM Page 161 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ❙❙❙❙ 161 graph of an increasing function is positive.) Thus, the rate of reaction of C is positive.
The concentrations of the reactants, however, decrease during the reaction, so, to make
the rates of reaction of A and B positive numbers, we put minus signs in front of the
derivatives d A dt and d B dt. Since A and B each decrease at the same rate that
C increases, we have
dC
dt rate of reaction dA
dt dB
dt More generally, it turns out that for a reaction of the form
aA bB l c C dD we have
1dA
a dt 1dB
b dt 1dC
c dt 1dD
d dt The rate of reaction can be determined by graphical methods (see Exercise 22). In some
cases we can use the rate of reaction to ﬁnd explicit formulas for the concentrations as
functions of time (see Exercises 10.3).
E XAMPLE 5 One of the quantities of interest in thermodynamics is compressibility. If a
given substance is kept at a constant temperature, then its volume V depends on its pressure P. We can consider the rate of change of volume with respect to pressure—namely,
the derivative dV dP. As P increases, V decreases, so dV dP 0. The compressibility
is deﬁned by introducing a minus sign and dividing this derivative by the volume V : 1 dV
V dP isothermal compressibility Thus, measures how fast, per unit volume, the volume of a substance decreases as the
pressure on it increases at constant temperature.
For instance, the volume V (in cubic meters) of a sample of air at 25 C was found to
be related to the pressure P (in kilopascals) by the equation
5.3
P V The rate of change of V with respect to P when P
dV
dP 5.3
P2 P 50 50 kPa is P 50 5.3
2500 0.00212 m 3 kPa The compressibility at that pressure is
1 dV
V dP P 50 0.00212
5.3
50 0.02 m 3 kPa m 3 5E03(pp 156165) 162 ❙❙❙❙ 1/17/06 2:12 PM Page 162 CHAPTER 3 DERIVATIVES Biology
f t be the number of individuals in an animal or plant population
EXAMPLE 6 Let n
at time t. The change in the population size between the times t t1 and t t2 is
n f t2
f t1 , and so the average rate of growth during the time period t1 t t2
is
n
f t2
f t1
average rate of growth
t
t2 t1
The instantaneous rate of growth is obtained from this average rate of growth by letting the time period t approach 0:
growth rate lim tl0 n
t dn
dt Strictly speaking, this is not quite accurate because the actual graph of a population
function n f t would be a step function that is discontinuous whenever a birth or
death occurs and, therefore, not differentiable. However, for a large animal or plant
population, we can replace the graph by a smooth approximating curve as in Figure 5.
n FIGURE 5
t 0 A smooth curve approximating
a growth function To be more speciﬁc, consider a population of bacteria in a homogeneous nutrient
medium. Suppose that by sampling the population at certain intervals it is determined
that the population doubles every hour. If the initial population is n0 and the time t is
measured in hours, then
f1
2f 0
2 n0
f2 2f 1 2 2n0 f3 2f 2 2 3n0 and, in general,
ft 2 tn0 The population function is n n0 2 t.
This is an example of an exponential function. In Chapter 7 we will discuss exponential functions in general; at that time we will be able to compute their derivatives and
thereby determine the rate of growth of the bacteria population. 5E03(pp 156165) 1/17/06 2:12 PM Page 163 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ❙❙❙❙ 163 EXAMPLE 7 When we consider the ﬂow of blood through a blood vessel, such as a vein or
artery, we can take the shape of the blood vessel to be a cylindrical tube with radius R
and length l as illustrated in Figure 6. R r FIGURE 6 l Blood flow in an artery Because of friction at the walls of the tube, the velocity v of the blood is greatest
along the central axis of the tube and decreases as the distance r from the axis increases
until v becomes 0 at the wall. The relationship between v and r is given by the law of
laminar ﬂow discovered by the French physician JeanLouisMarie Poiseuille in 1840.
This states that
P
R2
4l v 1 r2 where is the viscosity of the blood and P is the pressure difference between the ends
of the tube. If P and l are constant, then v is a function of r with domain 0, R . [For
more detailed information, see W. Nichols and M. O’Rourke (eds.), McDonald’s Blood
Flow in Arteries: Theoretic, Experimental, and Clinical Principles, 4th ed. (New York:
Oxford University Press, 1998).]
The average rate of change of the velocity as we move from r r1 outward to r r2
is given by
v v r2 r v r1 r2 r1 and if we let r l 0, we obtain the velocity gradient, that is, the instantaneous rate of
change of velocity with respect to r :
velocity gradient lim rl0 dv
dr v r Using Equation 1, we obtain
dv
dr P
0
4l Pr
2l 2r For one of the smaller human arteries we can take
and P 4000 dynes cm2, which gives
v 4000
0.000064
4 0.027 2
1.85 At r 0.027, R 10 4 6.4 10 0.008 cm, l r2
5 r2 6 4 0.002 cm the blood is ﬂowing at a speed of
v 0.002 1.85 10 4 64 1.11 cm s 10 10 6 2 cm, 5E03(pp 156165) 164 ❙❙❙❙ 1/17/06 2:12 PM Page 164 CHAPTER 3 DERIVATIVES and the velocity gradient at that point is
dv
dr 4000 0.002
2 0.027 2 r 0.002 74 cm s cm To get a feeling for what this statement means, let’s change our units from centimeters to micrometers (1 cm 10,000 m). Then the radius of the artery is 80 m. The
velocity at the central axis is 11,850 m s, which decreases to 11,110 m s at a distance
of r 20 m. The fact that dv dr
74 ( m s) m means that, when r 20 m, the
velocity is decreasing at a rate of about 74 m s for each micrometer that we proceed
away from the center. Economics
EXAMPLE 8 Suppose C x is the total cost that a company incurs in producing x units of
a certain commodity. The function C is called a cost function. If the number of items
produced is increased from x 1 to x 2 , the additional cost is C C x 2
C x 1 , and the
average rate of change of the cost is C
x C x2
x2 C x1
x1 C x1 x
x C x1 The limit of this quantity as x l 0, that is, the instantaneous rate of change of cost
with respect to the number of items produced, is called the marginal cost by economists:
marginal cost lim xl0 C
x dC
dx [Since x often takes on only integer values, it may not make literal sense to let x
approach 0, but we can always replace C x by a smooth approximating function as in
Example 6.]
Taking x 1 and n large (so that x is small compared to n), we have
Cn Cn 1 Cn Thus, the marginal cost of producing n units is approximately equal to the cost of producing one more unit [the n 1 st unit].
It is often appropriate to represent a total cost function by a polynomial
Cx a bx cx 2 dx 3 where a represents the overhead cost (rent, heat, maintenance) and the other terms
represent the cost of raw materials, labor, and so on. (The cost of raw materials may be
proportional to x, but labor costs might depend partly on higher powers of x because of
overtime costs and inefﬁciencies involved in largescale operations.)
For instance, suppose a company has estimated that the cost (in dollars) of producing
x items is
Cx 10,000 5x 0.01x 2 Then the marginal cost function is
Cx 5 0.02 x 5E03(pp 156165) 1/17/06 2:13 PM Page 165 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ❙❙❙❙ 165 The marginal cost at the production level of 500 items is
C 500 5 0.02 500 $15 i tem This gives the rate at which costs are increasing with respect to the production level
when x 500 and predicts the cost of the 501st item.
The actual cost of producing the 501st item is
C 501 C 500 10,000 5 501
10,000 0.01 501
5 500 2 0.01 500 2 $15.01
Notice that C 500 C 501 C 500 . Economists also study marginal demand, marginal revenue, and marginal proﬁt, which
are the derivatives of the demand, revenue, and proﬁt functions. These will be considered
in Chapter 4 after we have developed techniques for ﬁnding the maximum and minimum
values of functions. Other Sciences
Rates of change occur in all the sciences. A geologist is interested in knowing the rate at
which an intruded body of molten rock cools by conduction of heat into surrounding rocks.
An engineer wants to know the rate at which water ﬂows into or out of a reservoir. An
urban geographer is interested in the rate of change of the population density in a city as
the distance from the city center increases. A meteorologist is concerned with the rate of
change of atmospheric pressure with respect to height (see Exercise 17 in Section 10.4).
In psychology, those interested in learning theory study the socalled learning curve,
which graphs the performance P t of someone learning a skill as a function of the training time t. Of particular interest is the rate at which performance improves as time passes,
that is, dP dt.
In sociology, differential calculus is used in analyzing the spread of rumors (or innovations or fads or fashions). If p t denotes the proportion of a population that knows a rumor
by time t, then the derivative dp dt represents the rate of spread of the rumor (see Exercise 57 in Section 7.2). Summary
Velocity, density, current, power, and temperature gradient in physics, rate of reaction and
compressibility in chemistry, rate of growth and blood velocity gradient in biology, marginal cost and marginal proﬁt in economics, rate of heat ﬂow in geology, rate of improvement of performance in psychology, rate of spread of a rumor in sociology—these are all
special cases of a single mathematical concept, the derivative.
This is an illustration of the fact that part of the power of mathematics lies in its
abstractness. A single abstract mathematical concept (such as the derivative) can have different interpretations in each of the sciences. When we develop the properties of the mathematical concept once and for all, we can then turn around and apply these results to all of
the sciences. This is much more efﬁcient than developing properties of special concepts in
each separate science. The French mathematician Joseph Fourier (1768–1830) put it succinctly: “Mathematics compares the most diverse phenomena and discovers the secret
analogies that unite them.” 5E03(pp 166175) ❙❙❙❙ 166 1/17/06 t
(a)
(b)
(c)
(d)
(e)
(f ) Exercises
(b) Show that the rate of change of the volume of a cube with
respect to its edge length is equal to half the surface area of
the cube. Explain geometrically why this result is true by
arguing by analogy with Exercise 11(b).  A particle moves according to a law of motion s
f t,
0, where t is measured in seconds and s in feet.
Find the velocity at time t.
What is the velocity after 3 s?
When is the particle at rest?
When is the particle moving in the positive direction?
Find the total distance traveled during the ﬁrst 8 s.
Draw a diagram like Figure 2 to illustrate the motion of the
particle. 1. f t t2 3. f t 3 t 10t
12t 2. f t 2 36t t2
■ t3 4. f t 12 t 5. s
■ Page 166 CHAPTER 3 DERIVATIVES  3.4
1–6 2:03 PM 4 t ■ ■ ■ ■ ■ ■ 15t 4t st 3t 2 6. s 1 9t 2 13. (a) Find the average rate of change of the area of a circle with 10 1
35t ■ ■ 90
■ ■ 7. The position function of a particle is given by s t3 4.5t 2 7t t 0 When does the particle reach a velocity of 5 m s?
8. If a ball is given a push so that it has an initial velocity of 5 m s down a certain inclined plane, then the distance it has
rolled after t seconds is s 5 t 3 t 2.
(a) Find the velocity after 2 s.
(b) How long does it take for the velocity to reach 35 m s?
9. If a stone is thrown vertically upward from the surface of the moon with a velocity of 10 m s, its height (in meters) after
t seconds is h 10 t 0.83 t 2.
(a) What is the velocity of the stone after 3 s?
(b) What is the velocity of the stone after it has risen 25 m?
10. If a ball is thrown vertically upward with a velocity of 80 ft s, then its height after t seconds is s 80 t 16 t 2.
(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 96 ft above the
ground on its way up? On its way down?
11. (a) A company makes computer chips from square wafers of silicon. It wants to keep the side length of a wafer very
close to 15 mm and it wants to know how the area A x of a
wafer changes when the side length x changes. Find A 15
and explain its meaning in this situation.
(b) Show that the rate of change of the area of a square with
respect to its side length is half its perimeter. Try to explain
geometrically why this is true by drawing a square whose
side length x is increased by an amount x. How can you
approximate the resulting change in area A if x is small?
12. (a) Sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate
to evaporate slowly. If V is the volume of such a cube with
side length x, calculate dV dx when x 3 mm and explain
its meaning. respect to its radius r as r changes from
(i) 2 to 3
(ii) 2 to 2.5
(iii) 2 to 2.1
(b) Find the instantaneous rate of change when r 2.
(c) Show that the rate of change of the area of a circle with
respect to its radius (at any r) is equal to the circumference
of the circle. Try to explain geometrically why this is true
by drawing a circle whose radius is increased by an amount
r. How can you approximate the resulting change in area
A if r is small?
14. A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm s. Find the rate at which
the area within the circle is increasing after (a) 1 s, (b) 3 s, and
(c) 5 s. What can you conclude?
15. A spherical balloon is being inﬂated. Find the rate of increase of the surface area S 4 r 2 with respect to the radius r
when r is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can
you make?
4
3 r 3, where
the radius r is measured in micrometers (1 m 10 6 m).
Find the average rate of change of V with respect to r when
r changes from
(i) 5 to 8 m
(ii) 5 to 6 m
(iii) 5 to 5.1 m
(b) Find the instantaneous rate of change of V with respect to r
when r 5 m.
(c) Show that the rate of change of the volume of a sphere with
respect to its radius is equal to its surface area. Explain
geometrically why this result is true. Argue by analogy with
Exercise 13(c). 16. (a) The volume of a growing spherical cell is V 17. The mass of the part of a metal rod that lies between its left end and a point x meters to the right is 3x 2 kg. Find the linear
density (see Example 2) when x is (a) 1 m, (b) 2 m, and
(c) 3 m. Where is the density the highest? The lowest?
18. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli’s Law gives
the volume V of water remaining in the tank after t minutes as
V 5000 1 t
40 2 0 t 40 Find the rate at which water is draining from the tank after
(a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what time
is the water ﬂowing out the fastest? The slowest? Summarize
your ﬁndings.
19. The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is 5E03(pp 166175) 1/17/06 2:03 PM Page 167 S ECTION 3.4 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES given by Q t
t 3 2 t 2 6 t 2. Find the current when
(a) t 0.5 s and (b) t 1 s. [See Example 3. The unit of current is an ampere (1 A 1 C s).] At what time is the current
lowest?
20. Newton’s Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is
F GmM
r2 where G is the gravitational constant and r is the distance
between the bodies.
(a) Find dF dr and explain its meaning. What does the minus
sign indicate?
(b) Suppose it is known that Earth attracts an object with
a force that decreases at the rate of 2 N km when
r 20,000 km. How fast does this force change when
r 10,000 km?
21. Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the product of the pressure and the volume remains constant: PV C.
(a) Find the rate of change of volume with respect to
pressure.
(b) A sample of gas is in a container at low pressure and is
steadily compressed at constant temperature for 10 minutes.
Is the volume decreasing more rapidly at the beginning or
the end of the 10 minutes? Explain.
(c) Prove that the isothermal compressibility (see Example 5)
is given by
1 P.
22. The data in the table concern the lactonization of hydroxy valeric acid at 25 C. They give the concentration C t of this
acid in moles per liter after t minutes.
t 0 2 4 6 8 C(t ) 0.0800 0.0570 0.0408 0.0295 0.0210 (a) Find the average rate of reaction for the following time
intervals:
(i) 2 t 6
(ii) 2 t 4
(iii) 0 t 2
(b) Plot the points from the table and draw a smooth curve
through them as an approximation to the graph of the concentration function. Then draw the tangent at t 2 and use
it to estimate the instantaneous rate of reaction when t 2. ; 23. The table gives the population of the world in the 20th century. Year Population
(in millions) 1900
1910
1920
1930
1940
1950 1650
1750
1860
2070
2300
2560 Year Population
(in millions) 1960
1970
1980
1990
2000 3040
3710
4450
5280
6080 ❙❙❙❙ 167 (a) Estimate the rate of population growth in 1920 and in 1980
by averaging the slopes of two secant lines.
(b) Use a graphing calculator or computer to ﬁnd a cubic function (a thirddegree polynomial) that models the data. (See
Section 1.2.)
(c) Use your model in part (b) to ﬁnd a model for the rate of
population growth in the 20th century.
(d) Use part (c) to estimate the rates of growth in 1920 and
1980. Compare with your estimates in part (a).
(e) Estimate the rate of growth in 1985. ; 24. The table shows how the average age of ﬁrst marriage of
Japanese women varied in the last half of the 20th century.
t At t At 1950
1955
1960
1965
1970 23.0
23.8
24.4
24.5
24.2 1975
1980
1985
1990
1995 24.7
25.2
25.5
25.9
26.3 (a) Use a graphing calculator or computer to model these data
with a fourthdegree polynomial.
(b) Use part (a) to ﬁnd a model for A t .
(c) Estimate the rate of change of marriage age for women
in 1990.
(d) Graph the data points and the models for A and A .
25. If, in Example 4, one molecule of the product C is formed from one molecule of the reactant A and one molecule of
the reactant B, and the initial concentrations of A and B have
a common value A
B
a moles L, then
a 2kt akt C 1 where k is a constant.
(a) Find the rate of reaction at time t.
(b) Show that if x
C , then
dx
dt ka x 2 26. If f is the focal length of a convex lens and an object is placed at a distance p from the lens, then its image will be at a distance q from the lens, where f , p, and q are related by the
lens equation
1
f 1
p 1
q Find the rate of change of p with respect to q.
27. Refer to the law of laminar ﬂow given in Example 7. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure difference 3000 dynes cm2, and viscosity
0.027.
(a) Find the velocity of the blood along the centerline r 0, at
radius r 0.005 cm, and at the wall r R 0.01 cm. 5E03(pp 166175) 168 ❙❙❙❙ 1/17/06 2:04 PM Page 168 CHAPTER 3 DERIVATIVES when the brightness x of a light source is increased, the eye
reacts by decreasing the area R of the pupil. The experimental
formula (b) Find the velocity gradient at r 0, r 0.005, and
r 0.01.
(c) Where is the velocity the greatest? Where is the velocity
changing most? R 28. The frequency of vibrations of a vibrating violin string is given 40
1 24 x 0.4
4 x 0.4 by
1
2L f T where L is the length of the string, T is its tension, and is its
linear density. [See Chapter 11 in Donald E. Hall, Musical
Acoustics, 3d ed. (Paciﬁc Grove, CA: Brooks/Cole, 2002).]
(a) Find the rate of change of the frequency with respect to
(i) the length (when T and are constant),
(ii) the tension (when L and are constant), and
(iii) the linear density (when L and T are constant).
(b) The pitch of a note (how high or low the note sounds) is
determined by the frequency f . (The higher the frequency,
the higher the pitch.) Use the signs of the derivatives in
part (a) to determine what happens to the pitch of a note
(i) when the effective length of a string is decreased by
placing a ﬁnger on the string so a shorter portion of
the string vibrates,
(ii) when the tension is increased by turning a tuning peg,
(iii) when the linear density is increased by switching to
another string. ; has been used to model the dependence of R on x when R is
measured in square millimeters and x is measured in appropriate units of brightness.
(a) Find the sensitivity.
(b) Illustrate part (a) by graphing both R and S as functions
of x. Comment on the values of R and S at low levels of
brightness. Is this what you would expect?
33. The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters)
is PV nRT , where n is the number of moles of the gas
and R 0.0821 is the gas constant. Suppose that, at a
certain instant, P 8.0 atm and is increasing at a rate of
0.10 atm min and V 10 L and is decreasing at a rate of
0.15 L min. Find the rate of change of T with respect to time
at that instant if n 10 mol.
34. In a ﬁsh farm, a population of ﬁsh is introduced into a pond and harvested regularly. A model for the rate of change of the
ﬁsh population is given by the equation
dP
dt 29. Suppose that the cost (in dollars) for a company to produce x pairs of a new line of jeans is
Cx 2000 3x 0.01x 2 0.0002 x 3 (a) Find the marginal cost function.
(b) Find C 100 and explain its meaning. What does it predict?
(c) Compare C 100 with the cost of manufacturing the 101st
pair of jeans.
30. The cost function for a certain commodity is Cx 84 0.16 x 0.0006 x 2 0.000003x 3 (a) Find and interpret C 100 .
(b) Compare C 100 with the cost of producing the 101st item.
31. If p x is the total value of the production when there are x workers in a plant, then the average productivity of the workforce at the plant is
Ax px
x (a) Find A x . Why does the company want to hire more
workers if A x
0?
(b) Show that A x
0 if p x is greater than the average
productivity.
32. If R denotes the reaction of the body to some stimulus of strength x, the sensitivity S is deﬁned to be the rate of change
of the reaction with respect to x. A particular example is that r0 1 Pt
Pc Pt Pt where r0 is the birth rate of the ﬁsh, Pc is the maximum population that the pond can sustain (called the carrying capacity),
and is the percentage of the population that is harvested.
(a) What value of dP dt corresponds to a stable population?
(b) If the pond can sustain 10,000 ﬁsh, the birth rate is 5%, and
the harvesting rate is 4%, ﬁnd the stable population level.
(c) What happens if is raised to 5%?
35. In the study of ecosystems, predatorprey models are often used to study the interaction between species. Consider populations of tundra wolves, given by W t , and caribou, given by
C t , in northern Canada. The interaction has been modeled by
the equations
dC
dt aC bCW dW
dt cW dCW (a) What values of dC dt and dW dt correspond to stable
populations?
(b) How would the statement “The caribou go extinct” be
represented mathematically?
(c) Suppose that a 0.05, b 0.001, c 0.05, and
d 0.0001. Find all population pairs C, W that lead to
stable populations. According to this model, is it possible
for the two species to live in balance or will one or both
species become extinct? 5E03(pp 166175) 1/17/06 2:04 PM Page 169 S ECTION 3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS  3.5 ❙❙❙❙ 169 Derivatives of Trigonometric Functions  A review of the trigonometric functions is
given in Appendix D. Before starting this section, you might need to review the trigonometric functions. In particular, it is important to remember that when we talk about the function f deﬁned for all
real numbers x by
fx
sin x
it is understood that sin x means the sine of the angle whose radian measure is x. A similar convention holds for the other trigonometric functions cos, tan, csc, sec, and cot. Recall
from Section 2.5 that all of the trigonometric functions are continuous at every number in
their domains.
If we sketch the graph of the function f x
sin x and use the interpretation of f x
as the slope of the tangent to the sine curve in order to sketch the graph of f (see Exercise 16 in Section 3.2), then it looks as if the graph of f may be the same as the cosine
curve (see Figure 1 and also page 126). See an animation of Figure 1.
Resources / Module 4
/ Trigonometric Models
/ SlopeAScope for Sine ƒ=sin x 0 π
2 π π
2 π 2π x fª(x) 0 x FIGURE 1 Let’s try to conﬁrm our guess that if f x
ition of a derivative, we have
fx lim fx hl0 lim sin x hl0  We have used the addition formula for sine.
See Appendix D. lim h
h lim h
h sin x sin x cos h cos x sin h
h sin x cos h
h lim sin x hl0 1 hl0 sin x cos h
h lim sin x lim hl0 cos x. From the deﬁn fx hl0 hl0 sin x, then f x cos h
h 1 sin x cos x sin h
h
cos x 1 sin h
h lim cos x lim hl0 hl0 sin h
h 5E03(pp 166175) 170 ❙❙❙❙ 1/17/06 2:05 PM Page 170 CHAPTER 3 DERIVATIVES Two of these four limits are easy to evaluate. Since we regard x as a constant when computing a limit as h l 0, we have
lim sin x hl0 sin x and lim cos x hl0 cos x The limit of sin h h is not so obvious. In Example 3 in Section 2.2 we made the guess,
on the basis of numerical and graphical evidence, that
lim 2
D
B 1 We now use a geometric argument to prove Equation 2. Assume ﬁrst that lies between
0 and
2. Figure 2(a) shows a sector of a circle with center O, central angle , and
radius 1. BC is drawn perpendicular to OA. By the deﬁnition of radian measure, we have
arc AB
. Also, BC
OB sin
sin . From the diagram we see that
BC E Therefore
O sin l0 AB
so sin arc AB
sin 1 ¨
1 A C (a) Let the tangent lines at A and B intersect at E. You can see from Figure 2(b) that the
circumference of a circle is smaller than the length of a circumscribed polygon, and so
arc AB
AE
EB . Thus B arc AB AE ED AD A O EB AE E OA tan tan
(b)
FIGURE 2 (In Appendix F the inequality
tan is proved directly from the deﬁnition of the length
of an arc without resorting to geometric intuition as we did here.) Therefore, we have
sin
cos We know that lim sin cos so
l0 1 1 and lim l0 cos lim sin l0 But the function sin
Hence, we have 1
1, so by the Squeeze Theorem, we have
1 is an even function, so its right and left limits must be equal. lim l0 so we have proved Equation 2. sin 1 5E03(pp 166175) 1/17/06 2:05 PM Page 171 S ECTION 3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ❙❙❙❙ 171 We can deduce the value of the remaining limit in (1) as follows:
 We multiply numerator and denominator by
1 in order to put the function in a form
cos
in which we can use the limits we know. lim cos 1 cos lim l0 1 cos
cos l0 sin 2
cos
1 lim l0 sin lim lim l0 0 1 1 sin
cos
0 1 lim 3 sin l0 lim l0 1
1 cos lim l0 cos2
cos sin
cos 1
1 1 1
(by Equation 2) 1 0 l0 If we now put the limits (2) and (3) in (1), we get
fx lim sin x lim hl0 sin x hl0 0 cos h
h cos x 1 lim cos x lim hl0 1 hl0 sin h
h cos x So we have proved the formula for the derivative of the sine function: d
sin x
dx 4  Figure 3 shows the graphs of the function of
Example 1 and its derivative. Notice that y
0
whenever y has a horizontal tangent. _4 x 2 sin x. EXAMPLE 1 Differentiate y SOLUTION Using the Product Rule and Formula 4, we have 5
yª cos x dy
dx y x2 d
sin x
dx x 2 cos x 4 sin x d
x2
dx 2 x sin x Using the same methods as in the proof of Formula 4, one can prove (see Exercise 20)
that
_5 FIGURE 3 5 d
cos x
dx sin x The tangent function can also be differentiated by using the deﬁnition of a derivative, 5E03(pp 166175) 172 ❙❙❙❙ 1/17/06 2:06 PM Page 172 CHAPTER 3 DERIVATIVES but it is easier to use the Quotient Rule together with Formulas 4 and 5:
d
tan x
dx d
dx sin x
cos x cos x d
sin x
dx sin x d
cos x
dx cos2x
cos x cos x sin x
cos2x sin x cos2x sin2x
cos2x
1
cos2x sec2x d
tan x
dx 6 sec2x The derivatives of the remaining trigonometric functions, csc, sec, and cot , can also be
found easily using the Quotient Rule (see Exercises 17–19). We collect all the differentiation formulas for trigonometric functions in the following table. Remember that they are
valid only when x is measured in radians.
Derivatives of Trigonometric Functions  When you memorize this table, it is helpful
to notice that the minus signs go with the derivatives of the “cofunctions,” that is, cosine,
cosecant, and cotangent. d
sin x
dx
d
cos x
dx
d
tan x
dx d
csc x
dx
d
sec x
dx
d
cot x
dx cos x
sin x
sec2x EXAMPLE 2 Differentiate f x have a horizontal tangent? 1 csc x cot x
sec x tan x
csc 2x sec x
. For what values of x does the graph of f
tan x SOLUTION The Quotient Rule gives d
d
sec x
sec x
1
dx
dx
1 tan x 2 1 tan x 1 tan x sec x tan x
1 tan x fx sec x tan x
1 tan2x
tan x 2 sec x tan x 1
1 tan x 2 sec x sec2x
2 sec2x tan x 5E03(pp 166175) 1/17/06 2:06 PM Page 173 S ECTION 3.5 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 3 ❙❙❙❙ 173 In simplifying the answer we have used the identity tan2x 1 sec2x.
Since sec x is never 0, we see that f x
0 when tan x 1, and this occurs when
xn
4, where n is an integer (see Figure 4). _3 5 Trigonometric functions are often used in modeling realworld phenomena. In particular, vibrations, waves, elastic motions, and other quantities that vary in a periodic manner
can be described using trigonometric functions. In the following example we discuss an
instance of simple harmonic motion. _3 FIGURE 4 The horizontal tangents in Example 2 EXAMPLE 3 An object at the end of a vertical spring is stretched 4 cm beyond its rest
position and released at time t 0. (See Figure 5 and note that the downward direction
is positive.) Its position at time t is s
0 SOLUTION The velocity is s v FIGURE 5 π ds
dt d
4 cos t
dt 4 d
cos t
dt 4 sin t The object oscillates from the lowest point s 4 cm to the highest point
4 cm . The period of the oscillation is 2 , the period of cos t.
The speed is v
4 sin t , which is greatest when sin t
1, that is, when
cos t 0. So the object moves fastest as it passes through its equilibrium position
s 0 . Its speed is 0 when sin t 0, that is, at the high and low points. See the graphs
in Figure 6.
s √ 2
0 4 cos t Find the velocity at time t and use it to analyze the motion of the object. 4 s ft 2π t _2 Our main use for the limit in Equation 2 has been to prove the differentiation formula
for the sine function. But this limit is also useful in ﬁnding certain other trigonometric limits, as the following two examples show. FIGURE 6 EXAMPLE 4 Find lim xl0 sin 7x
.
4x SOLUTION In order to apply Equation 2, we ﬁrst rewrite the function by multiplying and
dividing by 7:
Note that sin 7x 7
4 sin 7x
4x 7 sin x. sin 7x
7x Notice that as x l 0, we have 7x l 0, and so, by Equation 2 with
lim xl0 Thus lim xl0 sin 7x
7x sin 7x
4x lim xl0 lim 7x l 0 7
4 sin 7x
7x 1 sin 7x
7x 7
sin 7x
lim
4 x l 0 7x 7
4 1 7
4 7x, 5E03(pp 166175) ❙❙❙❙ 174 1/17/06 2:07 PM Page 174 CHAPTER 3 DERIVATIVES EXAMPLE 5 Calculate lim x cot x.
xl0 SOLUTION Here we divide numerator and denominator by x : x cos x
x l 0 sin x
cos x
lim
x l 0 sin x
x lim x cot x lim xl0 cos 0
1
1  3.5
1–16  Differentiate.
2. f x 10 tan x 4. y 3 5. t t t cos t 7. h csc sec
■ tan t u a cos u (by the continuity of cosine and Equation 2) ; 1
x tan x 1
sec x
csc 16. y sin x
cos x x sin x cos x ■ ; d
sec x
dx 19. Prove that d
cot x
dx ■ 28. (a) If f x cot ■ ■ ■ ■ csc x cot x. sx sin x, ﬁnd f x .
(b) Check to see that your answer to part (a) is reasonable by
graphing both f and f for 0 x 2 . 29. For what values of x does the graph of f x
30. Find the points on the curve y x sec x tan x. cos x 2 sin x at which 31. A mass on a spring vibrates horizontally on a smooth level surface (see the ﬁgure). Its equation of motion is x t
8 sin t,
where t is in seconds and x in centimeters.
(a) Find the velocity at time t.
(b) Find the position and velocity of the mass at time
t 2 3. In what direction is it moving at that time? csc 2x.
cos x, sin x. 21–24  Find an equation of the tangent line to the curve at the
given point. 21. y tan x, 23. y x ■ ■ ■ 22. y 4, 1 cos x,
■ 1 ■ ■ x cos x,
1 24. y 0, 1
■ 2 sin x have the tangent is horizontal. 20. Prove, using the deﬁnition of derivative, that if f x then f x 2 x cot x, ﬁnd f x .
(b) Check to see that your answer to part (a) is reasonable by
graphing both f and f for 0 x
. a horizontal tangent? d
csc x
dx 18. Prove that ■ y sec x 2 cos x at the point
3, 1 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen.
27. (a) If f x ;
■ y x cos x at the point ,
.
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen.
26. (a) Find an equation of the tangent line to the curve b cot u 14. y ■ 17. Prove that 4 sec t 12. y tan
■ 5 cos x 10. y sin x
x2 15. y ; 2 csc x 8. y cot sec
1 sec 13. y x sin x 6. t t x
cos x 11. f ■ sin x
x 25. (a) Find an equation of the tangent line to the curve 3 sin x x
sin x 9. y lim xl0 Exercises 1. f x
3. y lim cos x xl0 sin x
■ ■ cos x
■ equilibrium
position 0, 1
, 0, 1
■ ■ 0 x x 5E03(pp 166175) 1/17/06 2:08 PM Page 175 S ECTION 3.6 THE CHAIN RULE ; 32. An elastic band is hung on a hook and a mass is hung on the
lower end of the band. When the mass is pulled downward and
then released, it vibrates vertically. The equation of motion is
s 2 cos t 3 sin t, t 0, where s is measured in centimeters and t in seconds. (We take the positive direction to be
downward.)
(a) Find the velocity at time t.
(b) Graph the velocity and position functions.
(c) When does the mass pass through the equilibrium position
for the ﬁrst time?
(d) How far from its equilibrium position does the mass travel?
(e) When is the speed the greatest? 43. lim l0 ■ ■ sin
tan
■ 44. lim xl1 ■ ■ ■ (c) sin x 1 cos x ; ■ ■ ■ ■ ■ cot x
csc x 46. A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like an icecream cone, as shown
in the ﬁgure. If A is the area of the semicircle and B is
the area of the triangle, ﬁnd
A
B lim by a force acting along a rope attached to the object. If the rope
makes an angle with the plane, then the magnitude of the
force is l0 A(¨ ) W
sin sin x 1
x2 x 2 (or familiar) identity.
sin x
(a) tan x
cos x
1
(b) sec x
cos x 34. An object with weight W is dragged along a horizontal plane F 175 45. Differentiate each trigonometric identity to obtain a new 33. A ladder 10 ft long rests against a vertical wall. Let be the
angle between the top of the ladder and the wall and let x be
the distance from the bottom of the ladder to the wall. If the
bottom of the ladder slides away from the wall, how fast does
x change with respect to when
3? ■ ❙❙❙❙ cos P Q
B(¨ ) where is a constant called the coefﬁcient of friction.
(a) Find the rate of change of F with respect to .
(b) When is this rate of change equal to 0?
(c) If W 50 lb and
0.6, draw the graph of F as a function of and use it to locate the value of for which
0. Is the value consistent with your answer to
dF d
part (b)? ¨
R
47. The ﬁgure shows a circular arc of length s and a chord of 35–44  length d, both subtended by a central angle . Find Find the limit. 35. lim sin 3 x
x 36. lim sin 4 x
sin 6 x 37. lim tan 6 t
sin 2 t 38. lim cos
sin 39. lim sin cos
sec 40. lim sin2 3 t
t2 41. lim cot 2 x
csc x 42. lim xl0 tl0 l0 xl0  3.6 xl0 l0 tl0 xl 4 lim l0 1
d s
d
s ¨ sin x cos x
cos 2 x The Chain Rule
Suppose you are asked to differentiate the function
Fx sx 2 1 The differentiation formulas you learned in the previous sections of this chapter do not
enable you to calculate F x . 5E03(pp 176185) 176 ❙❙❙❙ 1/17/06 1:58 PM Page 176 CHAPTER 3 DERIVATIVES  See Section 1.3 for a review of
composite functions. Resources / Module 4
/ Trigonometric Models
/ The Chain Rule Observe that F is a composite function. In fact, if we let y f u
su and let
u tx
x 2 1, then we can write y F x
f t x , that is, F f t. We know
how to differentiate both f and t, so it would be useful to have a rule that tells us how to
ﬁnd the derivative of F f t in terms of the derivatives of f and t.
It turns out that the derivative of the composite function f t is the product of the derivatives of f and t. This fact is one of the most important of the differentiation rules and is
called the Chain Rule. It seems plausible if we interpret derivatives as rates of change.
Regard du d x as the rate of change of u with respect to x, dy du as the rate of change of
y with respect to u, and dy d x as the rate of change of y with respect to x. If u changes
twice as fast as x and y changes three times as fast as u, then it seems reasonable that y
changes six times as fast as x, and so we expect that
dy
dx dy du
du dx The Chain Rule If f and t are both differentiable and F f t is the composite funcf t x , then F is differentiable and F is given by the tion deﬁned by F x
product Fx
In Leibniz notation, if y f tx t x f u and u
dy
dx Comments on the Proof of the Chain Rule Let t x are both differentiable functions, then
dy du
du dx u be the change in u corresponding to a change of x in x, that is,
u tx x tx fu u fu Then the corresponding change in y is
y
It is tempting to write
dy
dx y
x lim y
u u
x lim y
u lim u
x lim 1 lim y
u lim u
x xl0 xl0 xl0 ul0 xl0 xl0 (Note that u l 0 as x l 0
since t is continuous.) dy du
du dx
The only ﬂaw in this reasoning is that in (1) it might happen that u 0 (even when 5E03(pp 176185) 1/17/06 1:58 PM Page 177 SECTION 3.6 THE CHAIN RULE ❙❙❙❙ 177 x 0) and, of course, we can’t divide by 0. Nonetheless, this reasoning does at least
suggest that the Chain Rule is true. A full proof of the Chain Rule is given at the end of
this section.
The Chain Rule can be written either in the prime notation
ftx 2 or, if y f u and u f tx t x t x , in Leibniz notation:
dy du
du dx dy
dx 3 Equation 3 is easy to remember because if dy du and du d x were quotients, then we could
cancel du. Remember, however, that du has not been deﬁned and du d x should not be
thought of as an actual quotient.
sx 2 EXAMPLE 1 Find F x if F x 1. SOLUTION 1 (using Equation 2): At the beginning of this section we expressed F as Fx f tx f t x where f u
fu we have 1
2 1
2 su 12 u Fx x2 su and t x
and 1. Since tx 2x f tx t x
1
2 sx 2 1 x2 S OLUTION 2 (using Equation 3): If we let u dy du
du dx Fx 1
2 sx 2 x
sx 2 2x 1 and y 1
su, then 1
2x
2 su
1 2x x
sx 2 1 When using Formula 3 we should bear in mind that dy d x refers to the derivative of y
when y is considered as a function of x (called the derivative of y with respect to x),
whereas dy du refers to the derivative of y when considered as a function of u (the derivative of y with respect to u). For instance, in Example 1, y can be considered as a function
of x ( y s x 2 1 ) and also as a function of u ( y su ). Note that
dy
dx
NOTE Fx x
sx 2 1 whereas dy
du fu 1
2 su In using the Chain Rule we work from the outside to the inside. Formula 2 says
that we differentiate the outer function f [at the inner function t x ] and then we multiply
by the derivative of the inner function.
■ d
dx f tx outer
function evaluated
at inner
function f
derivative
of outer
function tx tx evaluated
at inner
function derivative
of inner
function 5E03(pp 176185) 178 ❙❙❙❙ 1/17/06 1:58 PM Page 178 CHAPTER 3 DERIVATIVES sin x 2 and (b) y EXAMPLE 2 Differentiate (a) y sin2x. SOLUTION (a) If y sin x 2 , then the outer function is the sine function and the inner function is
the squaring function, so the Chain Rule gives
dy
dx d
dx sin x2 cos x2 2x outer
function evaluated
at inner
function derivative
of outer
function evaluated
at inner
function derivative
of inner
function 2 x cos x 2 (b) Note that sin2x
sin x 2. Here the outer function is the squaring function and the
inner function is the sine function. So
dy
dx d
sin x
dx 2 2 inner
function  See Reference Page 2 or Appendix D. sin x
evaluated
at inner
function derivative
of outer
function cos x
derivative
of inner
function The answer can be left as 2 sin x cos x or written as sin 2 x (by a trigonometric identity
known as the doubleangle formula).
In Example 2(a) we combined the Chain Rule with the rule for differentiating the sine
function. In general, if y sin u, where u is a differentiable function of x, then, by the
Chain Rule,
dy
dy du
du
cos u
dx
du dx
dx
d
sin u
dx Thus cos u du
dx In a similar fashion, all of the formulas for differentiating trigonometric functions can
be combined with the Chain Rule.
Let’s make explicit the special case of the Chain Rule where the outer function f is a
power function. If y
t x n, then we can write y f u
u n where u t x . By using
the Chain Rule and then the Power Rule, we get
dy
dx
4 dy du
du dx nu n 1 du
dx ntx n1 tx The Power Rule Combined with the Chain Rule If n is any real number and u tx is differentiable, then
d
un
dx
Alternatively, d
tx
dx n nu n
ntx 1 du
dx n1 tx Notice that the derivative in Example 1 could be calculated by taking n 1
2 in Rule 4. 5E03(pp 176185) 1/17/06 1:59 PM Page 179 SECTION 3.6 THE CHAIN RULE x3 E XAMPLE 3 Differentiate y
SOLUTION Taking u dy
dx 100 . 1 x3 tx 1 and n d
x3
dx 1 100 x 3 1 99 100 in (4), we have 100 SOLUTION First rewrite f : 100 x 3
3x 2 3 2 sx x x2 fx 1
x 99 d
x3
dx
1 1 99 .
13 . Thus 1 1
3 x2 x 1 43 1
3 fx 1 300 x 2 x 3 1 EXAMPLE 4 Find f x if f x ❙❙❙❙ x2 x 1 43 d
x2
dx
2x x 1 1 EXAMPLE 5 Find the derivative of the function t
2t tt 2
1 9 S OLUTION Combining the Power Rule, Chain Rule, and Quotient Rule, we get t
2t 2
1 8 9 t
2t 2
1 8 9 tt E XAMPLE 6 Differentiate y d
dt t
2t 2
1 2t 1 1 2t 2t
2x 1 5 x3 2 45 t
2t 2 1 28
1 10 1 4. x SOLUTION In this example we must use the Product Rule before using the Chain Rule:  The graphs of the functions y and y in
Example 6 are shown in Figure 1. Notice that y
is large when y increases rapidly and y
0
when y has a horizontal tangent. So our answer
appears to be reasonable. 2x 1 5 d
x3
dx 2x dy
dx 1 5 4 x3 x3 10 x
x x 1 4 1
1 4 3 x3 x 1 d
x3
dx x d
2x
dx 1 5 2x 1 1 3 3x 2 4 4 d
2x
dx x 5 1 5 x3 1 1 yª 4 2x
_2 1 5 x3 x 1 4 2x 1 4 2 1 Noticing that each term has the common factor 2 2 x
factor it out and write the answer as y
_10 FIGURE 1 dy
dx 2 2x 1 4 x3 x 1 3 17x 3 1 4 x3 x 1 3, we could 6x 2 9x 3 179 5E03(pp 176185) 180 ❙❙❙❙ 1/17/06 1:59 PM Page 180 CHAPTER 3 DERIVATIVES The reason for the name “Chain Rule” becomes clear when we make a longer chain by
adding another link. Suppose that y f u , u t x , and x h t , where f , t, and h are
differentiable functions. Then, to compute the derivative of y with respect to t, we use the
Chain Rule twice:
dy
dt
EXAMPLE 7 If f x dy dx
dx dt dy du dx
du dx dt sin cos tan x , then
cos cos tan x d
cos tan x
dx cos cos tan x fx sin tan x d
tan x
dx cos cos tan x sin tan x sec2x
Notice that the Chain Rule has been used twice.
ssec x 3. EXAMPLE 8 Differentiate y SOLUTION Here the outer function is the square root function, the middle function is the
secant function, and the inner function is the cubing function. So we have dy
dx 1
d
sec x 3
3 dx
2 ssec x
1
d
sec x 3 tan x 3
x3
3
2 ssec x
dx
3x 2 sec x 3 tan x 3
2 ssec x 3 How to Prove the Chain Rule
Recall that if y f x and x changes from a to a
y fa x, we deﬁned the increment of y as x fa According to the deﬁnition of a derivative, we have
y
x lim xl0 So if we denote by
we obtain the difference between the difference quotient and the derivative, lim lim xl0 But fa xl0 y
x fa y
x fa ? fa y fa fa 0 x x 5E03(pp 176185) 1/17/06 2:00 PM Page 181 S ECTION 3.6 THE CHAIN RULE If we deﬁne to be 0 when x 0, then becomes a continuous function of
for a differentiable function f, we can write
y 5 fa x x where l 0 as ❙❙❙❙ 181 x. Thus, xl0 and is a continuous function of x. This property of differentiable functions is what
enables us to prove the Chain Rule.
Proof of the Chain Rule Suppose u
t x is differentiable at a and y f u is differentiable at b t a . If x is an increment in x and u and y are the corresponding increments in u and y, then we can use Equation 5 to write u 6 where 1 x 1 x ta 1 x u ta 2 u fb 2 u l 0 as x l 0. Similarly
y 7 fb where 2 l 0 as u l 0. If we now substitute the expression for u from Equation 6
into Equation 7, we get
y fb
y
x so ta 2 fb ta 2 As x l 0, Equation 6 shows that u l 0. So both
Therefore
dy
dx lim xl0 y
x lim f bt a 1 1 fb xl0 x 1 2 l 0 and ta 2 l 0 as x l 0. 1 f ta t a This proves the Chain Rule.  3.6
1–6 Exercises
13. y 1. y
3. y
5. y
■  ■ 11. t t ■ ■ x 3 4x 4 s1 2x
1 t 4 1 3 7 x x sin s x 18. h t t4 1 3 t3 1 sin sx
■ 19. y 2x 5 4 8x 2 5 20. y x2 3
1 sx 2 21. y 3 x cos n x 22. y 23. y sin x cos x 24. f x ■ ■ ■ ■ 10. f x
12. f t x 2 1
3
s1 x
x 4 23 tan t 1 x2 4 sec 5x tan sin x 8. F x
3 16. y x 3x Find the derivative of the function. 7. F x
9. F x ■ cot x 2 a3 4x 5 3 6. y
■ 14. y 1 s4 4. y 10 ssin x
■ 7–42 x2 1 x3 17. t x 2. y sin 4 x cos a 3 15. y Write the composite function in the form f t x .
[Identify the inner function u t x and the outer function
y f u .] Then ﬁnd the derivative dy dx.
 cos3x 8 4
3 2 3 25. F z z
z 1
1 26. G y x
s7
y
y2 3x
14
2y 5 5E03(pp 176185) ❙❙❙❙ 182 1/17/06 2:01 PM Page 182 CHAPTER 3 DERIVATIVES r 27. y sr 2 1 53. Suppose that F x cos x
sin x cos 28. y x 54. Suppose that w
u2
4, v 0 2 29. y
31. y sin s1 30. y 33. y cos2x 1 cot x 2 1
x 37. y cot 2 sin 38. y 39. y sx sx 40. y sx 41. y sin(tan ssin x ) 42. y 43–46 ■ ■ ■ ■ cot 2 x sx sx (a) If h x
(b) If H x 2 scos sin x
■ ■ ■ ■ ■ 45. y
■ 2 x 10, 1 sin sin x ,
■ ■ ■ 0, 1 44. y
46. y s5 sin2 x, sin x ,0
■ ■ ■ ■ x 2, ■ ■ tx fx tx 3
1
7 2
8
2 4
5
7 6
7
9 f t x , ﬁnd h 1 .
t f x , ﬁnd H 1 .
f f x , ﬁnd F 2 .
t t x , ﬁnd G 3 . 57. If f and t are the functions whose graphs are shown, let
ux
f t x ,v x
t f x , and w x
t t x . Find each 0, 0 derivative, if it exists. If it does not exist, explain why.
(a) u 1
(b) v 1
(c) w 1 2, 3
■ fx ■ y 47. (a) Find an equation of the tangent line to the curve ; f y tan x 2 4 at the point 1, 1 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. g 1 48. (a) The curve y ; x s2 x 2 is called a bulletnose curve.
Find an equation of the tangent line to this curve at the
point 1, 1 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. 49. (a) If f x ; s1 x 2 x, ﬁnd f x .
(b) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f . 0 x 1 58. If f is the function whose graph is shown, let h x f fx
and t x
f x 2 . Use the graph of f to estimate the value of
each derivative.
(a) h 2
(b) t 2
y y=ƒ ; 50. The function f x sin x sin 2 x , 0 x
, arises in
applications to frequency modulation (FM) synthesis.
(a) Use a graph of f produced by a graphing device to make a
rough sketch of the graph of f .
(b) Calculate f x and use this expression, with a graphing
device, to graph f . Compare with your sketch in part (a). 51. Find all points on the graph of the function fx 2 sin x sin x 52. Find the xcoordinates of all points on the curve sin 2 x 1
0 x 1 59. Use the table to estimate the value of h 0.5 , where hx f tx . x 0 0.1 0.2 0.3 0.4 0.5 0.6 fx 12.6 14.8 18.4 23.0 25.9 27.5 29.1 tx 0.58 0.40 0.37 0.26 0.17 0.10 0.05 2 at which the tangent line is horizontal.
y 3, 56. Let f and t be the functions in Exercise 55. (a) If F x
(b) If G x Find an equation of the tangent line to the curve at the
given point.
 43. y u v and u 0
1, v 0
2, u 0
5, and v 2
6. Find w 0 . 1
2
3 sin sin sin x ■ 4, x sec 2x ■ 6, t 3 55. A table of values for f , t, f , and t is given. 35. y ■ tan2x x sin 36. y 6 tan 2 3 34. y x2 sin x
cos x 32. y tan cos x f t x and t 3
7. Find F 3 . 2, and f 6 f3 2 sin x at which the tangent line is horizontal. 5E03(pp 176185) 1/17/06 2:01 PM Page 183 S ECTION 3.6 THE CHAIN RULE 60. If t x f f x , use the table to estimate the value of t 1 .
x 0.0 0.5 1.0 1.5 2.0 1.7 1.8 2.0 2.4 3.1 4.4 61. Suppose f is differentiable on Gx . Let F x
f cos x and
cos f x . Find expressions for (a) F x and (b) G x . 62. Suppose f is differentiable on Fx
f x and G x
(a) F x and (b) G x . fx and is a real number. Let
. Find expressions for
CAS 63. Suppose L is a function such that L x 1 x for x
an expression for the derivative of each function.
(a) f x
(b) t x
L x4
L 4x
(c) F x
(d) G x
Lx 4
L1 x 64. Let r x t2 f t h x , where h 1
2, t 2
5, and f 3
6. Find r 1 . 1
4 70. (a) Use a CAS to differentiate the function 3, h 1 4, 66. If the equation of motion of a particle is given by , the particle is said to undergo simple
s A cos t
harmonic motion.
(a) Find the velocity of the particle at time t.
(b) When is the velocity 0?
increases and decreases. The most easily visible such star is
Delta Cephei, for which the interval between times of maximum brightness is 5.4 days. The average brightness of this star
is 4.0 and its brightness changes by 0.35. In view of these
data, the brightness of Delta Cephei at time t, where t is measured in days, has been modeled by the function
0.35 sin 2 t 5.4 (a) Find the rate of change of the brightness after t days.
(b) Find, correct to two decimal places, the rate of increase
after one day.
68. In Example 4 in Section 1.3 we arrived at a model for the length of daylight (in hours) in Philadelphia on the t th day of
the year:
Lt 12 2.8 sin 1
1 and to simplify the result.
(b) Where does the graph of f have horizontal tangents?
(c) Graph f and f on the same screen. Are the graphs consistent with your answer to part (b)?
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.
72. Use the Chain Rule and the Product Rule to give an alternative proof of the Quotient Rule.
[Hint: Write f x t x f x t x 2
t
365 80 Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and May 21. .] 1 73. (a) If n is a positive integer, prove that d
sinn x cos n x
dx 67. A Cepheid variable star is a star whose brightness alternately 4.0 x
x 71. Use the Chain Rule to prove the following. sin 10 t where s is measured in centimeters and t in seconds. Find the
velocity of the particle after t seconds. Bt x4
x4 fx the equation
10 69. Computer algebra systems have commands that differentiate 0. Find 65. The displacement of a particle on a vibrating string is given by st 183 functions, but the form of the answer may not be convenient
and so further commands may be necessary to simplify the
answer.
(a) Use a CAS to ﬁnd the derivative in Example 5 and compare
with the answer in that example. Then use the simplify
command and compare again.
(b) Use a CAS to ﬁnd the derivative in Example 6. What happens if you use the simplify command? What happens if
you use the factor command? Which form of the answer
would be best for locating horizontal tangents? 2.5 fx CAS ❙❙❙❙ n sinn 1x cos n 1x (b) Find a formula for the derivative of
y cosn x cos n x that is similar to the one in part (a).
74. Suppose y f x is a curve that always lies above the xaxis
and never has a horizontal tangent, where f is differentiable
everywhere. For what value of y is the rate of change of y 5
with respect to x eighty times the rate of change of y with
respect to x ? 75. Use the Chain Rule to show that if is measured in degrees, then
d
sin
d 180 cos (This gives one reason for the convention that radian measure
is always used when dealing with trigonometric functions in
calculus: The differentiation formulas would not be as simple if
we used degree measure.) 5E03(pp 176185) 184 ❙❙❙❙ 1/17/06 2:02 PM Page 184 CHAPTER 3 DERVIATIVES 76. (a) Write x sin x , ﬁnd t x and sketch the graphs of t
(c) If t x
and t . Where is t not differentiable? s x 2 and use the Chain Rule to show that
d
x
dx x
x 77. Suppose P and Q are polynomials and n is a positive integer. Use mathematical induction to prove that the nth derivative of
the rational function f x
P x Q x can be written as a
rational function with denominator Q x n 1. In other words,
there is a polynomial An such that f n x
An x Q x n 1. sin x , ﬁnd f x and sketch the graphs of f
(b) If f x
and f . Where is f not differentiable?  3.7 Implicit Differentiation
The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable—for example,
y sx 3 or 1 y x sin x or, in general, y f x . Some functions, however, are deﬁned implicitly by a relation
between x and y such as
1 x2 y2 25 2 x3 y3 6 xy or In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x. For instance, if we solve Equation 1 for y, we get y
s25 x 2,
so two of the functions determined by the implicit Equation l are f x
s25 x 2 and
2. The graphs of f and t are the upper and lower semicircles of the
tx
s25 x
circle x 2 y 2 25. (See Figure 1.)
y 0 FIGURE 1 y y 0 x (a) ≈+¥=25 x 25≈
(b) ƒ=œ„„„„„„ 0 x 25≈
(c) ©=_ œ„„„„„„ It’s not easy to solve Equation 2 for y explicitly as a function of x by hand. (A computer
algebra system has no trouble, but the expressions it obtains are very complicated.)
Nonetheless, (2) is the equation of a curve called the folium of Descartes shown in
Figure 2 and it implicitly deﬁnes y as several functions of x. The graphs of three such functions are shown in Figure 3. When we say that f is a function deﬁned implicitly by Equation 2, we mean that the equation
x3 fx is true for all values of x in the domain of f . 3 6x f x 5E03(pp 176185) 1/17/06 2:02 PM Page 185 S ECTION 3.7 IMPLICIT DIFFERENTIATION y y y ❙❙❙❙ 185 y ˛+Á=6xy 0 x FIGURE 2 The folium of Descartes 0 0 x 0 x x FIGURE 3 Graphs of three functions defined by the folium of Descartes Fortunately, we don’t need to solve an equation for y in terms of x in order to ﬁnd the
derivative of y. Instead we can use the method of implicit differentiation. This consists of
differentiating both sides of the equation with respect to x and then solving the resulting
equation for y . In the examples and exercises of this section it is always assumed that the
given equation determines y implicitly as a differentiable function of x so that the method
of implicit differentiation can be applied.
EXAMPLE 1 dy
.
dx
(b) Find an equation of the tangent to the circle x 2
(a) If x 2 y2 25, ﬁnd y2 25 at the point 3, 4 . SOLUTION 1 (a) Differentiate both sides of the equation x 2
d
x2
dx
d
x2
dx y2 25:
d
25
dx y2 d
y2
dx 0 Remembering that y is a function of x and using the Chain Rule, we have
d
dy
y2
dy
dx d
y2
dx
Thus 2x 2y dy
dx 2y dy
dx 0 Now we solve this equation for dy d x :
dy
dx
3 and y 4, so dy
dx (b) At the point 3, 4 we have x x
y 3
4 An equation of the tangent to the circle at 3, 4 is therefore
y 4 3
4 x 3 or 3x 4y 25 5E03(pp 186195) 186 ❙❙❙❙ 1/17/06 1:54 PM Page 186 CHAPTER 3 DERIVATIVES SOLUTION 2 (b) Solving the equation x 2 y 2 25, we get y
s25 x 2. The point 3, 4 lies on
the upper semicircle y s25 x 2 and so we consider the function f x
s25 x 2.
Differentiating f using the Chain Rule, we have So 1
2 25 x2 12 1
2 fx 25 x2 d
25
dx 12 x2
x
s 25 x 2 2x
3 f3 s25 3 3
4 2 and, as in Solution 1, an equation of the tangent is 3x 4y 25. NOTE 1 Example 1 illustrates that even when it is possible to solve an equation explicitly for y in terms of x, it may be easier to use implicit differentiation.
■ NOTE 2 The expression dy d x
x y gives the derivative in terms of both x and y. It
is correct no matter which function y is determined by the given equation. For instance, for
y fx
s25 x 2 we have
■ dy
dx
whereas for y tx x
x2 s25 x 2 we have s25
dy
dx x
y x
y x
s25 x
x2 s25 x2 EXAMPLE 2 (a) Find y if x 3 y 3 6 xy.
(b) Find the tangent to the folium of Descartes x 3 y 3 6 xy at the point 3, 3 .
(c) At what points on the curve is the tangent line horizontal?
S OLUTION (a) Differentiating both sides of x 3 y 3 6 xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y 3 term and the Product Rule on the 6 xy term,
we get
3x 2 3y 2 y
6y 6 xy
x2 y2y 2y 2 xy y 2y 2 xy 2y x2 y2 2x y 2y x2 y 2y
y2 x2
2x or
We now solve for y : (b) When x y 3,
y 2 3 32
32 2 3 1 5E03(pp 186195) 1/17/06 1:54 PM Page 187 S ECTION 3.7 IMPLICIT DIFFERENTIATION y 187 and a glance at Figure 4 conﬁrms that this is a reasonable value for the slope at 3, 3 . So
an equation of the tangent to the folium at 3, 3 is (3, 3) y
0 ❙❙❙❙ 3 1x or 3 x y 6 x (c) The tangent line is horizontal if y
0. Using the expression for y from part (a),
we see that y
0 when 2y x 2 0. Substituting y 1 x 2 in the equation of the curve,
2
we get ( 1 x 2)3
2 x3 FIGURE 4 6 x ( 1 x 2)
2 which simpliﬁes to x 6 16 x 3. So either x 0 or x 3 16. If x 16 1 3 2 4 3, then
y 1 28 3
2 5 3. Thus, the tangent is horizontal at (0, 0) and at 2 4 3, 2 5 3 , which
2
is approximately (2.5198, 3.1748). Looking at Figure 5, we see that our answer is
reasonable. 4 NOTE 3 There is a formula for the three roots of a cubic equation that is like the quadratic formula but much more complicated. If we use this formula (or a computer algebra
system) to solve the equation x 3 y 3 6 xy for y in terms of x, we get three functions
determined by the equation: 4 0 ■ FIGURE 5 y 3
s fx 1
2 s1 x 6
4 x3 3
s 8x 3 1
2 s1 x 6
4 x3 8x 3 and
y
 The Norwegian mathematician Niels Abel
proved in 1824 that no general formula can be
given for the roots of a ﬁfthdegree equation in
terms of radicals. Later the French mathematician Evariste Galois proved that it is impossible
to ﬁnd a general formula for the roots of an
nthdegree equation (in terms of algebraic
operations on the coefﬁcients) if n is any integer
larger than 4. 1
2 [ ( 3
s 3s fx 1
2 s1 x 6
4 x3 3
s 8x 3 1
2 s1 x 6
4 x3 8x 3 )] (These are the three functions whose graphs are shown in Figure 3.) You can see that the
method of implicit differentiation saves an enormous amount of work in cases such as this.
Moreover, implicit differentiation works just as easily for equations such as
y5 3x 2 y 2 5x 4 12 for which it is impossible to ﬁnd a similar expression for y in terms of x.
EXAMPLE 3 Find y if sin x y 2 cos x. y SOLUTION Differentiating implicitly with respect to x and remembering that y is a function
of x, we get cos x cos x
2 _2 FIGURE 6 1 y y2 2yy cos x sin x (Note that we have used the Chain Rule on the left side and the Product Rule and Chain
Rule on the right side.) If we collect the terms that involve y , we get 2 _2 y So y y 2 sin x
y 2y cos x y y 2 sin x
2y cos x cos x
cos x cos x y y y
y Figure 6, drawn with the implicitplotting command of a computer algebra system,
shows part of the curve sin x y
y 2 cos x. As a check on our calculation, notice that
y
1 when x y 0 and it appears from the graph that the slope is approximately
1 at the origin. 5E03(pp 186195) 188 ❙❙❙❙ 1/17/06 1:54 PM Page 188 CHAPTER 3 DERIVATIVES Orthogonal Trajectories
Two curves are called orthogonal if at each point of intersection their tangent lines are
perpendicular. In the next example we use implicit differentiation to show that two families of curves are orthogonal trajectories of each other; that is, every curve in one family
is orthogonal to every curve in the other family. Orthogonal families arise in several areas
of physics. For example, the lines of force in an electrostatic ﬁeld are orthogonal to the
lines of constant potential. In thermodynamics, the isotherms (curves of equal temperature) are orthogonal to the ﬂow lines of heat. In aerodynamics, the streamlines (curves of
direction of airﬂow) are orthogonal trajectories of the velocityequipotential curves.
y EXAMPLE 4 The equation
≈¥ =k xy 3 c c 0 represents a family of hyperbolas. (Different values of the constant c give different
hyperbolas. See Figure 7.) The equation xy=c
x 0 x2 4 y2 k k 0 x. Show that every curve
represents another family of hyperbolas with asymptotes y
in the family (3) is orthogonal to every curve in the family (4); that is, the families are
orthogonal trajectories of each other.
SOLUTION Implicit differentiation of Equation 3 gives
FIGURE 7 x 5 dy
dx y 0 dy
dx so y
x Implicit differentiation of Equation 4 gives
2x 6 2y dy
dx 0 so dy
dx x
y From (5) and (6) we see that at any point of intersection of curves from each family, the
slopes of the tangents are negative reciprocals of each other. Therefore, the curves intersect at right angles; that is, they are orthogonal.  3.7
1–4 Exercises
5–20  (a) Find y by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get y in
terms of x.
(c) Check that your solutions to parts (a) and (b) are consistent by
substituting the expression for y into your solution for part (a).
1. xy
3.
■ 2x
1
y 1
x
■ 3x 2 2. 4 x 2 1
■ ■ ■ ■ ■ 9y 2 4. s x 4 sy ■ ■ Find dy d x by implicit differentiation. 5. x y2 7. x 3 2 9. x 2 y 6. x 2 1 xy 4y xy 2 11. x 2 y 2 36 2 6 8. x 2 10. y 5 3x x sin y 4 12. 1 y2 ■ ■ 1 2 xy y3 c x 2y 3 1 x 4y sin x y 2 x 13. 4 cos x sin y 4
■ 
2 1 14. y sin x 2 15. tan x y y 16. sx x y x sin y 2
1 x2y2 5E03(pp 186195) 1/17/06 1:55 PM Page 189 S ECTION 3.7 IMPLICIT DIFFERENTIATION 17. sxy
19. xy
■ x 2y 1 ■ 21. If 1 ■ fx 22. If t x ■ x 2 y 20. sin x cot x y ■ y 18. tan x cos y ■ 3 fx ■ ■ x 2 and t 1 x sin t x ■ ■ capabilities of computer algebra systems.
(a) Graph the curve with equation
y y2 ■ 2, ﬁnd f 1 . ■ x 2y 2
■ 25–30 yx 4 ■ y ■ 24. x 2 1 ■ ■ ■ y2 ■ 2 ax 2 y ■ ■ ■ ■ CAS Use implicit differentiation to ﬁnd an equation of the
tangent line to the curve at the given point.
25. x 2 26. x  2 xy y 2 xy 27. x 2 2 y y2 1, 1 3,
2 x 2x2 2y2 1, 2
x 2 2y 3 3 y2 3 (0, ) ( (cardioid) 4 (astroid) 1
2 3 s3, 1) y2 y5 x 0 25 x 2 y2 30. y 2 y 2 (3, 1)
(lemniscate) 4 x 8 2x 3 x2 x2 x2 y2
b2 1 at the point x 0 , y 0 is
x0 x
a2 5 y0 y
b2 1 37. Find an equation of the tangent line to the hyperbola (0, 2)
(devil’s curve) x2
a2 y y x4 36. Show by implicit differentiation that the tangent to the ellipse x2
a2 2 2 35. Find the points on the lemniscate in Exercise 29 where the y y2 1x tangent is horizontal. y 29. 2 x 2 xx has been likened to a bouncing wagon. Use a computer
algebra system to graph this curve and discover why.
(b) At how many points does this curve have horizontal
tangent lines? Find the xcoordinates of these points. (hyperbola) 28. x 2 2 34. (a) The curve with equation (ellipse) 2, 1y At how many points does this curve have horizontal
tangents? Estimate the xcoordinates of these points.
(b) Find equations of the tangent lines at the points (0, 1)
and (0, 2).
(c) Find the exact xcoordinates of the points in part (a).
(d) Create even more fanciful curves by modifying the
equation in part (a). 0, ﬁnd t 1 . 23–24  Regard y as the independent variable and x as the dependent variable and use implicit differentiation to ﬁnd dx d y.
23. y 4 189 33. Fanciful shapes can be created by using the implicit plotting sin x cos y ■ 0 and f 1 C AS x2 1 ❙❙❙❙ y2
b2 1 at the point x 0 , y 0 .
x 0 38. Show that the sum of the x and yintercepts of any tangent x line to the curve s x sy sc is equal to c. 39. Show, using implicit differentiation, that any tangent line at
■ ■ ■ ■ ■ ■ 31. (a) The curve with equation y ; ■ 2 4 ■ ■ ■ a point P to a circle with center O is perpendicular to the
radius OP. ■ 2 5x
x is called a
kampyle of Eudoxus. Find an equation of the tangent line
to this curve at the point 1, 2 .
(b) Illustrate part (a) by graphing the curve and the tangent line
on a common screen. (If your graphing device will graph
implicitly deﬁned curves, then use that capability. If not,
you can still graph this curve by graphing its upper and
lower halves separately.) 32. (a) The curve with equation y 2 ; ■ x 3 3x 2 is called the
Tschirnhausen cubic. Find an equation of the tangent line
to this curve at the point 1, 2 .
(b) At what points does this curve have a horizontal tangent?
(c) Illustrate parts (a) and (b) by graphing the curve and the
tangent lines on a common screen. 40. The Power Rule can be proved using implicit differentiation for the case where n is a rational number, n p q, and
y fx
x n is assumed beforehand to be a differentiable
function. If y x p q, then y q x p. Use implicit differentiation
to show that
p
x
q y
41–42  41. 2 x 2 42. x 2
■ 1 Show that the given curves are orthogonal.
y2
y2 ■ pq 3, 4x 2 5,
■ y2 x ■ 9y 2
■ 72 ■ ■ ■ ■ ■ ■ ■ 5E03(pp 186195) 190 ❙❙❙❙ 1/17/06 1:55 PM Page 190 CHAPTER 3 DERIVATIVES 46. x 2 A 800
600 a x, x2 y2 c x 2, x2 2y 2 k 48. y points with the same elevation. A ball rolling down a hill
follows a curve of steepest descent, which is orthogonal to the
contour lines. Given the contour map of a hill in the ﬁgure,
sketch the paths of balls that start at positions A and B. y2 47. y 43. Contour lines on a map of a hilly region are curves that join a x 3, x2 3y 2 b ■ ■ ■ ■ ■ by ■ ■ ■ ■ ■ ■ ■ 49. The equation x 2 x y y 2 3 represents a “rotated ellipse,”
that is, an ellipse whose axes are not parallel to the coordinate
axes. Find the points at which this ellipse crosses the xaxis
and show that the tangent lines at these points are parallel. 400
300
200 B 50. (a) Where does the normal line to the ellipse 400 ;
44. TV meteorologists often present maps showing pressure fronts. Such maps display isobars—curves along which the air pressure is constant. Consider the family of isobars shown in the
ﬁgure. Sketch several members of the family of orthogonal
trajectories of the isobars. Given the fact that wind blows from
regions of high air pressure to regions of low air pressure, what
does the orthogonal family represent? x 2 x y y 2 3 at the point 1, 1 intersect the ellipse
a second time? (See page 156 for the deﬁnition of a normal
line.)
(b) Illustrate part (a) by graphing the ellipse and the normal
line.
51. Find all points on the curve x 2 y 2 the tangent line is xy 2 where the slope of 1. 52. Find equations of both the tangent lines to the ellipse x2 4y 2 36 that pass through the point 12, 3 . 53. The ﬁgure shows a lamp located three units to the right of the yaxis and a shadow created by the elliptical region
x 2 4y 2 5. If the point 5, 0 is on the edge of the
shadow, how far above the xaxis is the lamp located?
y ?
45–48 Show that the given families of curves are orthogonal
trajectories of each other. Sketch both families of curves on the
same axes.
45. x 2  y2  3.8 r 2, ax by 0 _5 3 x ≈+4¥=5 0 Higher Derivatives
If f is a differentiable function, then its derivative f is also a function, so f may have a
derivative of its own, denoted by f
f . This new function f is called the second
derivative of f because it is the derivative of the derivative of f . Using Leibniz notation,
we write the second derivative of y f x as
d
dx
Another notation is f x D2f x . dy
dx d2y
dx 2 5E03(pp 186195) 1/17/06 1:55 PM Page 191 SECTION 3.8 HIGHER DERIVATIVES In Module 3.8A you can see how changing the coefﬁcients of a polynomial f
affects the appearance of the graphs of
f , f , and f . E XAMPLE 1 If f x ❙❙❙❙ 191 x cos x, ﬁnd and interpret f x . SOLUTION Using the Product Rule, we have fx x d
cos x
dx
x sin x cos x d
x
dx cos x To ﬁnd f x we differentiate f x :
d
dx x sin x cos x x fx 3 d
sin x
dx sin x f·
fª d
dx x d
cos x
dx f x cos x 3 _3 FIGURE 1 The graphs of ƒ=x cos x and
its first and second derivatives sin x x cos x _3 sin x 2 sin x The graphs of f, f , and f are shown in Figure 1.
We can interpret f x as the slope of the curve y f x at the point x, f x . In
other words, it is the rate of change of the slope of the original curve y f x .
Notice from Figure 1 that f x
0 whenever y f x has a horizontal tangent.
Also, f x is positive when y f x has positive slope and negative when y f x
has negative slope. So the graphs serve as a check on our calculations.
In general, we can interpret a second derivative as a rate of change of a rate of change.
The most familiar example of this is acceleration, which we deﬁne as follows.
If s s t is the position function of an object that moves in a straight line, we know
that its ﬁrst derivative represents the velocity v t of the object as a function of time:
vt st ds
dt The instantaneous rate of change of velocity with respect to time is called the acceleration
a t of the object. Thus, the acceleration function is the derivative of the velocity function
and is therefore the second derivative of the position function:
at vt st or, in Leibniz notation,
a dv
dt d 2s
dt 2 EXAMPLE 2 The position of a particle is given by the equation s ft t3 6t 2 9t where t is measured in seconds and s in meters.
(a) Find the acceleration at time t. What is the acceleration after 4 s?
(b) Graph the position, velocity, and acceleration functions for 0 t
(c) When is the particle speeding up? When is it slowing down? 5. 5E03(pp 186195) 192 ❙❙❙❙ 1/17/06 1:56 PM Page 192 CHAPTER 3 DERIVATIVES SOLUTION (a) The velocity function is the derivative of the position function:
s
vt ft t3 ds
dt 3t 2 6t 2 9t 12 t 9 The acceleration is the derivative of the velocity function:
at 25 √ a
s 0 5 _12 dv
dt 6t a4  The units for acceleration are meters per
second per second, written as m/s2. d 2s
dt 2
64 12 12 m s2 12 (b) Figure 2 shows the graphs of s, v, and a.
(c) The particle speeds up when the velocity is positive and increasing (v and a are both
positive) and also when the velocity is negative and decreasing (v and a are both negative). In other words, the particle speeds up when the velocity and acceleration have the
same sign. (The particle is pushed in the same direction it is moving.) From Figure 2 we
see that this happens when 1 t 2 and when t 3. The particle slows down when v
and a have opposite signs, that is, when 0 t 1 and when 2 t 3. Figure 3 summarizes the motion of the particle. FIGURE 2 a √
In Module 3.8B you can see an animation of Figure 3 with an expression for s
that you can choose yourself. 5 s 0 1 t _5 forward
slows
down F IGURE 3 backward
speeds
up forward slows
down speeds
up The third derivative f is the derivative of the second derivative: f
f . So
f x can be interpreted as the slope of the curve y f x or as the rate of change of
f x . If y f x , then alternative notations for the third derivative are
y f x d
dx d2y
dx 2 d 3y
dx 3 D 3f x The process can be continued. The fourth derivative f is usually denoted by f 4 . In general, the nth derivative of f is denoted by f n and is obtained from f by differentiating n
times. If y f x , we write
dny
yn
fn x
D nf x
dx n 5E03(pp 186195) 1/17/06 1:56 PM Page 193 S ECTION 3.8 HIGHER DERIVATIVES ❙❙❙❙ 193 We can interpret the third derivative physically in the case where the function is the
position function s s t of an object that moves along a straight line. Because
s
s
a , the third derivative of the position function is the derivative of the acceleration function and is called the jerk:
d 3s
dt 3 da
dt j Thus, the jerk j is the rate of change of acceleration. It is aptly named because a large jerk
means a sudden change in acceleration, which causes an abrupt movement in a vehicle.
EXAMPLE 3 If x3 y
y 3x 2 y 6x y 0 for all n
1
, ﬁnd f
x SOLUTION fx n x 12 1 1
x2 2 x fx 2 1x 2
x3 3 4 x 321x 4321x f 4 x f 5 5 x. 1
x fx  The factor 1 n occurs in the formula for
f n x because we introduce another negative
sign every time we differentiate. Since the successive values of 1 n are 1, 1, 1, 1, 1,
1, . . . , the presence of 1 n indicates that the
sign changes with each successive derivative. 12 x 4. EXAMPLE 4 If f x f 3 0 then and in fact y n 5x 6 y4 6x 2 x 5 54321x 6 5! x 6 .
.
.
f
or n x 1 nn n f n x 1 n n!
xn 1 1n 2 21x n1 Here we have used the factorial symbol n! for the product of the ﬁrst n positive integers. n! 123 n 1 n The following example shows how to ﬁnd the second derivative of a function that is
deﬁned implicitly. 5E03(pp 186195) 194 ❙❙❙❙ 1/17/06 1:56 PM Page 194 CHAPTER 3 DERIVATIVES EXAMPLE 5 Find y if x 4 y4 16. SOLUTION Differentiating the equation implicitly with respect to x, we get 4x 3
 Figure 4 shows the graph of the curve
x 4 y 4 16 of Example 5. Notice that it’s
a stretched and ﬂattened version of the circle
x 2 y 2 4. For this reason it’s sometimes
called a fat circle. It starts out very steep on the
left but quickly becomes very ﬂat. This can be
seen from the expression
x3
y3 y
y x
y 4y 3 y 0 Solving for y gives
x3
y3 y 1 To ﬁnd y we differentiate this expression for y using the Quotient Rule and remembering that y is a function of x : 3 y d
dx x3
y3 y 3 3x 2 x$+y$=16 y 3 d dx x 3 x 3 d dx y 3
y3 2 x 3 3y 2 y
y6 2 If we now substitute Equation 1 into this expression, we get
0 3x 2 y 3 2x y 3x 3 y 2 x3
y3 y
3 x2y4 x6
y7 3x 2 y 4 x 4
y7 But the values of x and y must satisfy the original equation x 4
simpliﬁes to
3x 2 16
x2
y
48 7
7
y
y FIGURE 4 y4 16. So the answer EXAMPLE 6 Find D 27 cos x.
SOLUTION The ﬁrst few derivatives of cos x are as follows: D cos x
 Look for a pattern. sin x D 2 cos x cos x D 3 cos x sin x D 4 cos x cos x D 5 cos x sin x We see that the successive derivatives occur in a cycle of length 4 and, in particular,
D n cos x cos x whenever n is a multiple of 4. Therefore
D 24 cos x cos x and, differentiating three more times, we have
D 27 cos x sin x 5E03(pp 186195) 1/17/06 1:56 PM Page 195 ❙❙❙❙ S ECTION 3.8 HIGHER DERIVATIVES 195 We have seen that one application of second and third derivatives occurs in analyzing
the motion of objects using acceleration and jerk. We will investigate another application
of second derivatives in Exercise 60 and in Section 4.3, where we show how knowledge of
f gives us information about the shape of the graph of f. In Chapter 12 we will see how
second and higher derivatives enable us to represent functions as sums of inﬁnite series.  3.8 Exercises
4. The ﬁgure shows the graphs of four functions. One is the 1. The ﬁgure shows the graphs of f , f , and f . Identify each position function of a car, one is the velocity of the car, one is
its acceleration, and one is its jerk. Identify each curve, and
explain your choices. curve, and explain your choices.
y a y b d a
b x c c 0 t 2. The ﬁgure shows graphs of f, f , f , and f . Identify each curve, and explain your choices.
y 5–20  Find the ﬁrst and second derivatives of the function.
x5 5. f x ab c d 7. y 6x 2 1 7t 11. h u 1
1
sx 2
x3 15. y 1 sin 10. t x a ss 14. y 23 17. H t y a ■ b c 20. h x ■ 21. (a) If f x ;
0 18. t s csc
■ t ■ ■ b
ss ■ ■ s 2 cos s
x
x2
■ 3
2x
■ ■ ■ 2 2 cos x sin x, ﬁnd f x and f x .
(b) Check to see that your answers to part (a) are reasonable by
comparing the graphs of f , f , and f .
x x 2 1 , ﬁnd f x and f x .
(b) Check to see that your answers to part (a) are reasonable by
comparing the graphs of f , f , and f . 22. (a) If f x ; ■ 1
1 4x
sx 1 tan 3t 19. t 2t 4 xn 16. y 1 3. The ﬁgure shows the graphs of three functions. One is the posi tion function of a car, one is the velocity of the car, and one is
its acceleration. Identify each curve, and explain your choices. 7t 6 2x
x 12. H s 6 4u
3u 13. h x t8 8. y cos 2 9. F t
x 6. f t 7x 5E03(pp 196205) ❙❙❙❙ 196 1/17/06 Page 196 CHAPTER 3 DERIVATIVES 23–24  23. y Find y . s2 x ■ 1:47 PM ■ ■ x 24. y 3
■ ■ ■ ■ 25. If f t t cos t, ﬁnd f 26. If t x s5 27. If f cot , ﬁnd f
sec x, ﬁnd t 2x
■ s 1
■ ■ ■ ■ 6. 28. If t x velocity and acceleration of the car. What is the acceleration at t 10 seconds? 4. 29–32  29. 9 x 2 31. x
■ y
■ 33–37 3 2. 0 32. x 1
■ ■ ■ 35. f x 1 37. f x ■ n Find a formula for f
xn 33. f x 100 30. s x 9 ■  2 x, ﬁnd t 4 ■ sy
y 4 1  The equation of motion is given for a particle, where s is
in meters and t is in seconds. Find (a) the velocity and acceleration
as functions of t, (b) the acceleration after 1 second, and (c) the
acceleration at the instants when the velocity is 0. ■ 4 ■ ■ ■ 43. s 5x 1
3x 3 ■ ■ 38–40 ■ sx 2t 7t
■ 36 t 2 t 2, 12t,
cos
■ t 6, 4t 2 t 1,
■ 0 0 t
■ 0 t 2 0
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■  An equation of motion is given, where s is in meters and
t in seconds. Find (a) the times at which the acceleration is 0 and
(b) the displacement and velocity at these times. ■ t4 47. s
■ ■ 4t 3
■ 48. s 2
■ ■ ■ ■ 2t 3
■ 9t 2
■ ■ ■ ■ 49. A particle moves according to a law of motion 38. D74 sin x
39. D103 cos 2 x
■ t6 3 ■ 15t 2 47–48  ■ sin ■ Find the given derivative by ﬁnding the ﬁrst few derivatives and observing the pattern that occurs. ■ 3t 46. s 1 2t 3 45. s 1 2t 3 44. s x. 36. f x t 20 43–46 1
a 34. f x x 10 (b) Use the acceleration curve from part (a) to estimate the jerk
at t 10 seconds. What are the units for jerk? Find y by implicit differentiation.
y2 3 0. 40. D 35 x sin x
■ ■ ■ ■ ■ ■ ■ ■ ■ ; 41. A car starts from rest and the graph of its position function is shown in the ﬁgure, where s is measured in feet and t in
seconds. Use it to graph the velocity and estimate the acceleration at t 2 seconds from the velocity graph. Then sketch a
graph of the acceleration function. 50. A particle moves along the xaxis, its position at time t given s ; 120
100
80
60
40
20
0 s ft
t 3 12 t 2 36t, t 0, where t is measured in
seconds and s in meters.
(a) Find the acceleration at time t and after 3 s.
(b) Graph the position, velocity, and acceleration functions
for 0 t 8.
(c) When is the particle speeding up? When is it slowing
down?
by x t
t 1 t 2 , t 0, where t is measured in seconds
and x in meters.
(a) Find the acceleration at time t. When is it 0?
(b) Graph the position, velocity, and acceleration functions
for 0 t 4.
(c) When is the particle speeding up? When is it slowing
down?
51. A mass attached to a vertical spring has position function given 1 t 42. (a) The graph of a position function of a car is shown, where s is measured in feet and t in seconds. Use it to graph the by y t
A sin t, where A is the amplitude of its oscillations
and is a constant.
(a) Find the velocity and acceleration as functions of time.
(b) Show that the acceleration is proportional to the displacement y.
(c) Show that the speed is a maximum when the acceleration
is 0. 5E03(pp 196205) 1/17/06 1:47 PM Page 197 A PPLIED PROJECT WHERE SHOULD A PILOT START DESCENT? 52. A particle moves along a straight line with displacement s t ,
velocity v t , and acceleration a t . Show that 1
xx 3, and P 2 P2 5, 2. C AS 54. Find a thirddegree polynomial Q such that Q 1 3, Q 1 Q1 6, and Q 1 1 1
x 1
x 1 to compute the derivatives much more easily. Then ﬁnd an
expression for f n x . This method of splitting up a fraction
in terms of simpler fractions, called partial fractions, will
be pursued further in Section 8.4. Explain the difference between the meanings of the derivatives
dv dt and dv ds.
53. Find a seconddegree polynomial P such that P 2 197 (b) Use the identity dv
vt
ds at ❙❙❙❙ 62. (a) Use a computer algebra system to compute f , where 1, 12. fx 7x 17
2 x 2 7x 4 55. The equation y y
2y sin x is called a differential
equation because it involves an unknown function y and its
derivatives y and y . Find constants A and B such that the
function y A sin x B cos x satisﬁes this equation. (Differential equations will be studied in detail in Chapter 10.) (b) Find a much simpler expression for f by ﬁrst splitting
f into partial fractions. [In Maple, use the command
convert(f,parfrac,x); in Mathematica, use Apart[f].]
63. Suppose p is a positive integer such that the function f is ptimes differentiable and f p
f . Using mathematical induction, show that f is in fact ntimes differentiable for every positive integer n and that each of its higher derivatives f n equals
one of the p functions f , f , f , . . . , f p 1 . 56. Find constants A, B, and C such that the function Ax 2
y y
y Bx
2y C satisﬁes the differential equation
x 2. 57–59  The function t is a twice differentiable function. Find f
in terms of t, t , and t .
57. f x tx
x 59. f x f x t x , where f and t have derivatives of all
orders, show that xt x 2 58. f x 64. (a) If F x t (s x ) ■ ■ ■ F ft 2f t ft (b) Find similar formulas for F and F 4 .
(c) Guess a formula for F n .
■ ■ ■ ■ ■ ■ ■ ■ ■ 65. If y f u and u t x , where f and t are twice differentiable functions, show that 3x 5 10 x 3 5, graph both f and f . On what
intervals is f x
0? On those intervals, how is the graph of
f related to its tangent lines? What about the intervals where
fx
0? ; 60. If f x 61. (a) Compute the ﬁrst few derivatives of the function d2y
dx 2 d2y
du 2 du
dx 2 dy d 2u
du dx 2 66. If y fx
1 x 2 x until you see that the computations are
becoming algebraically unmanageable. f u and u t x , where f and t possess third
derivatives, ﬁnd a formula for d 3 y d x 3 similar to the one given
in Exercise 65. APPLIED PROJECT
Where Should a Pilot Start Descent?
y y=P (x) 0 An approach path for an aircraft landing is shown in the ﬁgure and satisﬁes the following
conditions:
(i) The cruising altitude is h when descent starts at a horizontal distance from touchdown at
the origin.
(ii) The pilot must maintain a constant horizontal speed v throughout descent.
(iii) The absolute value of the vertical acceleration should not exceed a constant k (which is
much less than the acceleration due to gravity). h x ax 3 b x 2 c x d that satisﬁes condition (i) by
imposing suitable conditions on P x and P x at the start of descent and at touchdown. 1. Find a cubic polynomial P x 5E03(pp 196205) 198 ❙❙❙❙ 1/17/06 1:47 PM Page 198 CHAPTER 3 DERIVATIVES 2. Use conditions (ii) and (iii) to show that 6hv 2
2 k 3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k 860 mi h2. If the cruising altitude of a plane is 35,000 ft and the speed is 300 mi h,
how far away from the airport should the pilot start descent? ; 4. Graph the approach path if the conditions stated in Problem 3 are satisﬁed. APPLIED PROJECT
Building a Better Roller Coaster
L¡ P Suppose you are asked to design the ﬁrst ascent and drop for a new roller coaster. By studying
photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the
slope of the drop 1.6. You decide to connect these two straight stretches y L 1 x and
y L 2 x with part of a parabola y f x
a x 2 b x c, where x and f x are measured
in feet. For the track to be smooth there can’t be abrupt changes in direction, so you want the
linear segments L 1 and L 2 to be tangent to the parabola at the transition points P and Q. (See
the ﬁgure.) To simplify the equations, you decide to place the origin at P. f Q
L™ 1. (a) Suppose the horizontal distance between P and Q is 100 ft. Write equations in a, b, and ; c that will ensure that the track is smooth at the transition points.
(b) Solve the equations in part (a) for a, b, and c to ﬁnd a formula for f x .
(c) Plot L 1, f , and L 2 to verify graphically that the transitions are smooth.
(d) Find the difference in elevation between P and Q.
2. The solution in Problem 1 might look smooth, but it might not feel smooth because the piecewise deﬁned function [consisting of L 1 x for x 0, f x for 0 x 100, and L 2 x
for x 100] doesn’t have a continuous second derivative. So you decide to improve the
design by using a quadratic function q x
a x 2 b x c only on the interval 10 x 90
and connecting it to the linear functions by means of two cubic functions:
tx
hx C AS  3.9 kx3
3 px lx 2
qx 2 mx n 0 x 10 rx s 90 x 100 (a) Write a system of equations in 11 unknowns that ensure that the functions and their ﬁrst
two derivatives agree at the transition points.
(b) Solve the equations in part (a) with a computer algebra system to ﬁnd formulas for
q x , t x , and h x .
(c) Plot L 1, t, q, h, and L 2, and compare with the plot in Problem 1(c). Related Rates Explore an expanding balloon interactively.
Resources / Module 5
/ Related Rates
/ Start of Related Rates If we are pumping air into a balloon, both the volume and the radius of the balloon are
increasing and their rates of increase are related to each other. But it is much easier to measure directly the rate of increase of the volume than the rate of increase of the radius.
In a related rates problem the idea is to compute the rate of change of one quantity in
terms of the rate of change of another quantity (which may be more easily measured). The 5E03(pp 196205) 1/17/06 1:47 PM Page 199 S ECTION 3.9 RELATED RATES ❙❙❙❙ 199 procedure is to ﬁnd an equation that relates the two quantities and then use the Chain Rule
to differentiate both sides with respect to time.
EXAMPLE 1 Air is being pumped into a spherical balloon so that its volume increases at a
rate of 100 cm3 s. How fast is the radius of the balloon increasing when the diameter is
50 cm?
SOLUTION We start by identifying two things:
 According to the Principles of Problem
Solving discussed on page 58, the ﬁrst step is to
understand the problem. This includes reading
the problem carefully, identifying the given and
the unknown, and introducing suitable notation. the given information:
the rate of increase of the volume of air is 100 cm3 s
and the unknown:
the rate of increase of the radius when the diameter is 50 cm
In order to express these quantities mathematically, we introduce some suggestive
notation:
Let V be the volume of the balloon and let r be its radius.
The key thing to remember is that rates of change are derivatives. In this problem, the
volume and the radius are both functions of the time t. The rate of increase of the volume with respect to time is the derivative dV dt, and the rate of increase of the radius is
dr dt. We can therefore restate the given and the unknown as follows:
Given:
Unknown:  The second stage of problem solving is to
think of a plan for connecting the given and the
unknown. dV
dt
dr
dt 100 cm3 s
when r 25 cm In order to connect dV dt and dr dt, we ﬁrst relate V and r by the formula for the
volume of a sphere:
V 4
3 r3 In order to use the given information, we differentiate each side of this equation with
respect to t. To differentiate the right side, we need to use the Chain Rule:
dV
dt dV dr
dr dt 4 r2 dr
dt Now we solve for the unknown quantity:
dr
dt  Notice that, although dV dt is constant,
dr d t is not constant. If we put r 25 and dV dt 1 dV
4 r 2 dt 100 in this equation, we obtain
dr
dt 1
100
4 25 2 1
25 The radius of the balloon is increasing at the rate of 1 25 cm s. 5E03(pp 196205) 200 ❙❙❙❙ 1/17/06 1:47 PM Page 200 CHAPTER 3 DERIVATIVES How high will a ﬁreman get while climbing a
sliding ladder?
Resources / Module 5
/ Related Rates
/ Start of the Sliding Fireman wall EXAMPLE 2 A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft s, how fast is the top of the ladder sliding
down the wall when the bottom of the ladder is 6 ft from the wall?
S OLUTION We ﬁrst draw a diagram and label it as in Figure 1. Let x feet be the distance
from the bottom of the ladder to the wall and y feet the distance from the top of the
ladder to the ground. Note that x and y are both functions of t (time).
We are given that dx dt 1 ft s and we are asked to ﬁnd dy dt when x 6 ft (see
Figure 2). In this problem, the relationship between x and y is given by the Pythagorean
Theorem: 10 y x2 y2 100 Differentiating each side with respect to t using the Chain Rule, we have
x ground 2x FIGURE 1 dx
dt dy
dt 2y 0 and solving this equation for the desired rate, we obtain dy
dt dy
dt =? x dx
y dt When x 6, the Pythagorean Theorem gives y
dx dt 1, we have y dy
dt x
dx
dt =1 FIGURE 2 8 and so, substituting these values and 6
1
8 3
ft s
4 The fact that dy dt is negative means that the distance from the top of the ladder to
the ground is decreasing at a rate of 3 ft s. In other words, the top of the ladder is sliding
4
down the wall at a rate of 3 ft s.
4
EXAMPLE 3 A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m3 min, ﬁnd the rate
at which the water level is rising when the water is 3 m deep.
2 r
4 SOLUTION We ﬁrst sketch the cone and label it as in Figure 3. Let V , r, and h be the volume of the water, the radius of the surface, and the height at time t, where t is measured
in minutes.
We are given that dV dt 2 m3 min and we are asked to ﬁnd dh dt when h is 3 m.
The quantities V and h are related by the equation h FIGURE 3 V 1
3 r 2h but it is very useful to express V as a function of h alone. In order to eliminate r, we use
the similar triangles in Figure 3 to write
r
h 2
4 r h
2 and the expression for V becomes
V 1
3 h
2 2 h 12 h3 5E03(pp 196205) 1/17/06 1:47 PM Page 201 S ECTION 3.9 RELATED RATES ❙❙❙❙ 201 Now we can differentiate each side with respect to t :
dV
dt 4 dh
dt so
Substituting h 3 m and dV dt 4 dV
h 2 dt 2 m3 min, we have
dh
dt 4
3 2 The water level is rising at a rate of 8 9
 Look back: What have we learned from
Examples 1–3 that will help us solve future
problems? dh
dt h2 8
9 2 0.28 m min. Strategy It is useful to recall some of the problemsolving principles from page 58 and adapt them to related rates in light of our experience in Examples 1–3:
1. Read the problem carefully.
2. Draw a diagram if possible.
3. Introduce notation. Assign symbols to all quantities that are functions of time.  WARNING: A common error is to substitute the given numerical information (for quantities that vary with time) too early. This should
be done only after the differentiation. (Step 7
follows Step 6.) For instance, in Example 3 we
dealt with general values of h until we ﬁnally
substituted h 3 at the last stage. (If we had
put h 3 earlier, we would have gotten
dV dt 0, which is clearly wrong.) 4. Express the given information and the required rate in terms of derivatives.
5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate one of the variables by substitution (as in
Example 3).
6. Use the Chain Rule to differentiate both sides of the equation with respect to t.
7. Substitute the given information into the resulting equation and solve for the
unknown rate.
The following examples are further illustrations of the strategy.
EXAMPLE 4 Car A is traveling west at 50 mi h and car B is traveling north at 60 mi h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
C
y
B x z A SOLUTION We draw Figure 4, where C is the intersection of the roads. At a given time t,
let x be the distance from car A to C, let y be the distance from car B to C, and let z be
the distance between the cars, where x, y, and z are measured in miles.
We are given that dx dt
50 mi h and dy dt
60 mi h. (The derivatives are
negative because x and y are decreasing.) We are asked to ﬁnd dz dt. The equation that
relates x, y, and z is given by the Pythagorean Theorem: z2
FIGURE 4 x2 y2 Differentiating each side with respect to t, we have
2z dz
dt 2x dx
dt dz
dt 1
z x 2y
dx
dt dy
dt
y dy
dt 5E03(pp 196205) ❙❙❙❙ 202 1/17/06 1:48 PM Page 202 CHAPTER 3 DERIVATIVES When x 0.3 mi and y 0.4 mi, the Pythagorean Theorem gives z
dz
dt 1
0.3
0.5 50 0.4 0.5 mi, so 60 78 mi h
The cars are approaching each other at a rate of 78 mi h.
EXAMPLE 5 A man walks along a straight path at a speed of 4 ft s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is
the searchlight rotating when the man is 15 ft from the point on the path closest to the
searchlight?
SOLUTION We draw Figure 5 and let x be the distance from the man to the point on the
path closest to the searchlight. We let be the angle between the beam of the searchlight
and the perpendicular to the path.
We are given that dx dt 4 ft s and are asked to ﬁnd d dt when x 15. The equation that relates x and can be written from Figure 5: x x
20 20
¨ tan x 20 tan Differentiating each side with respect to t, we get
dx
dt FIGURE 5 d
dt so
When x 1
20 cos2 20 sec2
dx
dt 1
20 cos2 15, the length of the beam is 25, so cos
d
dt 1
5 4
5 2 16
125 d
dt
1
5 4
4
5 cos2 and 0.128 The searchlight is rotating at a rate of 0.128 rad s.  3.9 Exercises 1. If V is the volume of a cube with edge length x and the cube 6. A particle moves along the curve y s1 x 3. As it reaches
the point 2, 3 , the ycoordinate is increasing at a rate of
4 cm s. How fast is the xcoordinate of the point changing at
that instant? expands as time passes, ﬁnd dV dt in terms of dx dt.
2. (a) If A is the area of a circle with radius r and the circle expands as time passes, ﬁnd dA dt in terms of dr dt.
(b) Suppose oil spills from a ruptured tanker and spreads in a
circular pattern. If the radius of the oil spill increases at a
constant rate of 1 m s, how fast is the area of the spill
increasing when the radius is 30 m?
3. If y
4. If x 2
5. If z 2 x x3
y2 2 x and dx dt 5, ﬁnd dy dt when x 25 and dy d t x 2 y 2, dx d t
5 and y 12. 6, ﬁnd dx d t when y 2, and dy d t 2.
4. 3, ﬁnd d z d t when 7–10 (a)
(b)
(c)
(d)
(e)  What quantities are given in the problem?
What is the unknown?
Draw a picture of the situation for any time t.
Write an equation that relates the quantities.
Finish solving the problem. 7. A plane ﬂying horizontally at an altitude of 1 mi and a speed of 500 mi h passes directly over a radar station. Find the rate at
which the distance from the plane to the station is increasing
when it is 2 mi away from the station. 5E03(pp 196205) 1/17/06 1:48 PM Page 203 S ECTION 3.9 RELATED RATES 8. If a snowball melts so that its surface area decreases at a rate of 1 cm2 min, ﬁnd the rate at which the diameter decreases when
the diameter is 10 cm.
6 ft tall walks away from the pole with a speed of 5 ft s along
a straight path. How fast is the tip of his shadow moving when
he is 40 ft from the pole?
10. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km h and ship B is sailing north at 25 km h. How fast is
the distance between the ships changing at 4:00 P.M.?
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 203 17. At noon, ship A is 100 km west of ship B. Ship A is sailing south at 35 km h and ship B is sailing north at 25 km h. How
fast is the distance between the ships changing at 4:00 P.M.? 9. A street light is mounted at the top of a 15fttall pole. A man ■ ❙❙❙❙ ■ 11. Two cars start moving from the same point. One travels south at 60 mi h and the other travels west at 25 mi h. At what rate
is the distance between the cars increasing two hours later?
12. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
of 1.6 m s, how fast is the length of his shadow on the building
decreasing when he is 4 m from the building?
13. A man starts walking north at 4 ft s from a point P. Five min utes later a woman starts walking south at 5 ft s from a point
500 ft due east of P. At what rate are the people moving apart
15 min after the woman starts walking?
14. A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward ﬁrst base with a speed of 24 ft s.
(a) At what rate is his distance from second base decreasing
when he is halfway to ﬁrst base?
(b) At what rate is his distance from third base increasing at
the same moment? 18. A particle is moving along the curve y sx. As the particle
passes through the point 4, 2 , its xcoordinate increases at a
rate of 3 cm s. How fast is the distance from the particle to the
origin changing at this instant? 19. Water is leaking out of an inverted conical tank at a rate of 10,000 cm3 min at the same time that water is being pumped
into the tank at a constant rate. The tank has height 6 m and the
diameter at the top is 4 m. If the water level is rising at a rate
of 20 cm min when the height of the water is 2 m, ﬁnd the rate
at which water is being pumped into the tank.
20. A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft.
If the trough is being ﬁlled with water at a rate of 12 ft 3 min,
how fast is the water level rising when the water is 6 inches
deep?
21. A water trough is 10 m long and a crosssection has the shape of an isosceles trapezoid that is 30 cm wide at the bottom,
80 cm wide at the top, and has height 50 cm. If the trough is
being ﬁlled with water at the rate of 0.2 m3 min, how fast is the
water level rising when the water is 30 cm deep?
22. A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. A crosssection is shown in the ﬁgure. If the pool is being ﬁlled at a
rate of 0.8 ft 3 min, how fast is the water level rising when the
depth at the deepest point is 5 ft?
3
6
6 12 16 6 90 ft 23. Gravel is being dumped from a conveyor belt at a rate of
15. The altitude of a triangle is increasing at a rate of 1 cm min while the area of the triangle is increasing at a rate of
2 cm2 min. At what rate is the base of the triangle changing
when the altitude is 10 cm and the area is 100 cm2 ?
16. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m
higher than the bow of the boat. If the rope is pulled in at a rate
of 1 m s, how fast is the boat approaching the dock when it is
8 m from the dock? 30 ft 3 min, and its coarseness is such that it forms a pile in the
shape of a cone whose base diameter and height are always
equal. How fast is the height of the pile increasing when the
pile is 10 ft high? 5E03(pp 196205) 204 ❙❙❙❙ 1/17/06 1:48 PM Page 204 CHAPTER 3 DERVIATIVES 24. A kite 100 ft above the ground moves horizontally at a speed of 8 ft s. At what rate is the angle between the string and the
horizontal decreasing when 200 ft of string have been let out?
25. Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad s. Find the rate
at which the area of the triangle is increasing when the angle
between the sides of ﬁxed length is 3. how fast is the angle between the top of the ladder and the wall
changing when the angle is 4 rad?
32. Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the ﬁgure). The point Q is on the
ﬂoor 12 ft directly beneath P and between the carts. Cart A is
being pulled away from Q at a speed of 2 ft s. How fast is cart
B moving toward Q at the instant when cart A is 5 ft from Q ? 26. Two sides of a triangle have lengths 12 m and 15 m. The angle between them is increasing at a rate of 2 min. How fast is the
length of the third side increasing when the angle between the
sides of ﬁxed length is 60 ? P 27. Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the
equation PV C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and
the pressure is increasing at a rate of 20 kPa min. At what rate
is the volume decreasing at this instant?
28. When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation
PV 1.4 C, where C is a constant. Suppose that at a certain
instant the volume is 400 cm3 and the pressure is 80 kPa and is
decreasing at a rate of 10 kPa min. At what rate is the volume
increasing at this instant?
29. If two resistors with resistances R1 and R2 are connected in parallel, as in the ﬁgure, then the total resistance R, measured
in ohms ( ), is given by
1
R 1
R1 1
R2 If R1 and R2 are increasing at rates of 0.3 s and 0.2 s,
respectively, how fast is R changing when R1 80 and
R2 100 ? 12 f t
A B
Q 33. A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has
to change at the correct rate in order to keep the rocket in sight.
Also, the mechanism for focusing the camera has to take into
account the increasing distance from the camera to the rising
rocket. Let’s assume the rocket rises vertically and its speed is
600 ft s when it has risen 3000 ft.
(a) How fast is the distance from the television camera to the
rocket changing at that moment?
(b) If the television camera is always kept aimed at the rocket,
how fast is the camera’s angle of elevation changing at that
same moment?
34. A lighthouse is located on a small island 3 km away from the nearest point P on a straight shoreline and its light makes four
revolutions per minute. How fast is the beam of light moving
along the shoreline when it is 1 km from P ?
35. A plane ﬂying with a constant speed of 300 km h passes over a R¡ R™ ground radar station at an altitude of 1 km and climbs at an
angle of 30 . At what rate is the distance from the plane to the
radar station increasing a minute later?
36. Two people start from the same point. One walks east at 30. Brain weight B as a function of body weight W in ﬁsh has been modeled by the power function B 0.007W 2 3, where
B and W are measured in grams. A model for body weight
as a function of body length L (measured in centimeters) is
W 0.12L2.53. If, over 10 million years, the average length of
a certain species of ﬁsh evolved from 15 cm to 20 cm at a
constant rate, how fast was this species’ brain growing when
the average length was 18 cm?
31. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2 ft s, 3 mi h and the other walks northeast at 2 mi h. How fast is
the distance between the people changing after 15 minutes?
37. A runner sprints around a circular track of radius 100 m at a constant speed of 7 m s. The runner’s friend is standing at a
distance 200 m from the center of the track. How fast is the
distance between the friends changing when the distance
between them is 200 m?
38. The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the
hands changing at one o’clock? 5E03(pp 196205) 1/17/06 1:48 PM Page 205 S ECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS  3.10 205 Linear Approximations and Differentials Resources / Module 3
/ Linear Approximation
/ Start of Linear Approximation y y=ƒ {a, f(a)} ❙❙❙❙ We have seen that a curve lies very close to its tangent line near the point of tangency. In
fact, by zooming in toward a point on the graph of a differentiable function, we noticed
that the graph looks more and more like its tangent line. (See Figure 2 in Section 2.6 and
Figure 3 in Section 3.1.) This observation is the basis for a method of ﬁnding approximate
values of functions.
The idea is that it might be easy to calculate a value f a of a function, but difﬁcult (or
even impossible) to compute nearby values of f. So we settle for the easily computed values of the linear function L whose graph is the tangent line of f at a, f a . (See Figure 1.)
In other words, we use the tangent line at a, f a as an approximation to the curve
y f x when x is near a. An equation of this tangent line is y=L(x) y fa fax a and the approximation
fx 1
0 FIGURE 1 x fa fax a is called the linear approximation or tangent line approximation of f at a. The linear
function whose graph is this tangent line, that is,
Lx 2 fa fax a is called the linearization of f at a.
The following example is typical of situations in which we use a linear approximation
to predict the future behavior of a function given by empirical data.
EXAMPLE 1 Suppose that after you stuff a turkey its temperature is 50 F and you then put it in a 325 F oven. After an hour the meat thermometer indicates that the temperature of
the turkey is 93 F and after two hours it indicates 129 F. Predict the temperature of the
turkey after three hours.
SOLUTION If T t represents the temperature of the turkey after t hours, we are given that T0
50, T 1
93, and T 2
129. In order to make a linear approximation with
a 2, we need an estimate for the derivative T 2 . Because
T2 lim
t l2 Tt T2
2 t we could estimate T 2 by the difference quotient with t
T2 T1
1 T2
2 93 1: 129
1 36 This amounts to approximating the instantaneous rate of temperature change by the
average rate of change between t 1 and t 2, which is 36 F h. With this estimate, the
linear approximation (1) for the temperature after 3 h is
T3 T2 T23 129 36 1 2
165 So the predicted temperature after three hours is 165 F. 5E03(pp 206217) 206 ❙❙❙❙ 1/17/06 1:45 PM Page 206 CHAPTER 3 DERIVATIVES T We obtain a more accurate estimate for T 2 by plotting the given data, as in Figure 2, and estimating the slope of the tangent line at t 2 to be 150 T2 33 Then our linear approximation becomes
100 50 L T3 T 0 T2 T2 1 129 33 162 and our improved estimate for the temperature is 162 F.
Because the temperature curve lies below the tangent line, it appears that the actual
temperature after three hours will be somewhat less than 162 F, perhaps closer to 160 F.
1 2 3 t EXAMPLE 2 Find the linearization of the function f x
s x 3 at a 1 and use it to
approximate the numbers s3.98 and s4.05. Are these approximations overestimates or
underestimates? FIGURE 2 SOLUTION The derivative of f x x
1
2 fx
and so we have f 1
the linearization is 3
x f1 is 3 1
4 2 and f 1 Lx 12 12 1
2 sx 3 . Putting these values into Equation 2, we see that f1x 1 2 1
4 x 7
4 1 x
4 The corresponding linear approximation (1) is
sx 7
4 3 x
4 (when x is near 1) In particular, we have
s3.98 7
4 0.98
4 1.995 and s4.05 7
4 1.05
4 2.0125 The linear approximation is illustrated in Figure 3. We see that, indeed, the tangent
line approximation is a good approximation to the given function when x is near l. We
also see that our approximations are overestimates because the tangent line lies above the
curve.
y
7 x y= 4 + 4
(1, 2) FIGURE 3 _3 0 1 y= x+3
œ„„„„
x Of course, a calculator could give us approximations for s3.98 and s4.05, but the
linear approximation gives an approximation over an entire interval.
In the following table we compare the estimates from the linear approximation in
Example 2 with the true values. Notice from this table, and also from Figure 3, that the 5E03(pp 206217) 1/17/06 1:45 PM Page 207 S ECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ❙❙❙❙ 207 tangent line approximation gives good estimates when x is close to 1 but the accuracy of
the approximation deteriorates when x is farther away from 1.
From L x x
s3.9
s3.98
s4
s4.05
s4.1
s5 0.9
0.98
1
1.05
1.1
2
3 s6 Actual value 1.975
1.995
2
2.0125
2.025
2.25
2.5 1.97484176 . . .
1.99499373 . . .
2.00000000 . . .
2.01246117 . . .
2.02484567 . . .
2.23606797 . . .
2.44948974 . . . How good is the approximation that we obtained in Example 2? The next example
shows that by using a graphing calculator or computer we can determine an interval throughout which a linear approximation provides a speciﬁed accuracy.
EXAMPLE 3 For what values of x is the linear approximation sx 7
4 3 x
4 accurate to within 0.5? What about accuracy to within 0.1?
SOLUTION Accuracy to within 0.5 means that the functions should differ by less than 0.5:
4.3
Q sx y= œx+3+0.5
„„„„ L(x) P 10 sx 3 _1 FIGURE 4 Q
y= œx+3+0.1
„„„„ _2 y= œx+30.1
„„„„ FIGURE 5 1 0.5 0.5 7
4 x
4 sx 3 0.5 This says that the linear approximation should lie between the curves obtained by shifting the curve y s x 3 upward and downward by an amount 0.5. Figure 4 shows
the tangent line y
7 x 4 intersecting the upper curve y sx 3 0.5 at P
and Q. Zooming in and using the cursor, we estimate that the xcoordinate of P is about
2.66 and the xcoordinate of Q is about 8.66. Thus, we see from the graph that the
approximation
7
x
sx 3
4
4 3 P x
4 Equivalently, we could write y= œx+30.5
„„„„ _4 7
4 3 5 is accurate to within 0.5 when 2.6 x 8.6. (We have rounded to be safe.)
Similarly, from Figure 5 we see that the approximation is accurate to within 0.1 when
1.1 x 3.9. Applications to Physics
Linear approximations are often used in physics. In analyzing the consequences of an
equation, a physicist sometimes needs to simplify a function by replacing it with its linear
approximation. For instance, in deriving a formula for the period of a pendulum, physics 5E03(pp 206217) 208 ❙❙❙❙ 1/17/06 1:45 PM Page 208 CHAPTER 3 DERIVATIVES textbooks obtain the expression a T
t sin for tangential acceleration and then replace
sin by with the remark that sin is very close to if is not too large. [See, for
example, Physics: Calculus, 2d ed., by Eugene Hecht (Paciﬁc Grove, CA: Brooks/Cole,
2000), p. 431.] You can verify that the linearization of the function f x
sin x at a 0
is L x
x and so the linear approximation at 0 is
sin x x (see Exercise 46). So, in effect, the derivation of the formula for the period of a pendulum
uses the tangent line approximation for the sine function.
Another example occurs in the theory of optics, where light rays that arrive at shallow
angles relative to the optical axis are called paraxial rays. In paraxial (or Gaussian) optics,
both sin and cos are replaced by their linearizations. In other words, the linear approximations
sin and cos 1 are used because is close to 0. The results of calculations made with these approximations became the basic theoretical tool used to design lenses. [See Optics, 4th ed., by Eugene
Hecht (Reading, MA: AddisonWesley, 2002), p. 154.]
In Section 12.12 we will present several other applications of the idea of linear approximations to physics. Differentials
The ideas behind linear approximations are sometimes formulated in the terminology and
notation of differentials. If y f x , where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number.
The differential dy is then deﬁned in terms of dx by the equation  If dx 0, we can divide both sides of
Equation 3 by dx to obtain
dy
dx fx We have seen similar equations before, but now
the left side can genuinely be interpreted as a
ratio of differentials.
y Q R Îy P
dx=Îx 0 x y=ƒ
FIGURE 6 dy 3 dy So dy is a dependent variable; it depends on the values of x and dx. If dx is given a speciﬁc value and x is taken to be some speciﬁc number in the domain of f , then the numerical value of d y is determined.
The geometric meaning of differentials is shown in Figure 6. Let P x, f x and
Qx
x, f x
x be points on the graph of f and let dx
x. The corresponding
change in y is S x+Î x y
x f x dx fx x fx The slope of the tangent line PR is the derivative f x . Thus, the directed distance from S
to R is f x dx dy. Therefore, dy represents the amount that the tangent line rises or
falls (the change in the linearization), whereas y represents the amount that the curve
y f x rises or falls when x changes by an amount dx.
EXAMPLE 4 Compare the values of y and dy if y f x
x changes (a) from 2 to 2.05 and (b) from 2 to 2.01. x3 x2 2x 1 and 5E03(pp 206217) 1/17/06 1:45 PM Page 209 S ECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS  Figure 7 shows the function in Example 4
and a comparison of dy and y when a 2.
The viewing rectangle is 1.8, 2.5 by 6, 18 . (a) We have
23 f2
f 2.05 (2, 9) 209 SOLUTION 22 y=˛+≈2x+1 dy ❙❙❙❙ 2.05 y Îy dy
2 and dx 3 2.05 f 2.05 In general,
When x 22 x 1
2 9
2 2.05 f2 1 9.717625 0.717625
3x 2 f x dx 2x 2 dx 0.05, this becomes FIGURE 7 dy
(b) f 2.01
y When dx x 32 2.01 2 3 22
2.01 f 2.01 2 0.05 2 f2 2 2.01 0.7
1 9.140701 0.140701 0.01,
dy 32 2 22 2 0.01 0.14 Notice that the approximation y d y becomes better as x becomes smaller in
Example 4. Notice also that dy was easier to compute than y. For more complicated functions it may be impossible to compute y exactly. In such cases the approximation by differentials is especially useful.
In the notation of differentials, the linear approximation (1) can be written as
fa
For instance, for the function f x sx dy
If a 1 and dx x fa dy 3 in Example 2, we have
dx
2 sx 3 f x dx 0.05, then
dy and dx s4.05 0.05
2 s1 3
f 1.05 0.0125 f1 dy 2.0125 just as we found in Example 2.
Our ﬁnal example illustrates the use of differentials in estimating the errors that occur
because of approximate measurements.
EXAMPLE 5 The radius of a sphere was measured and found to be 21 cm with a possible
error in measurement of at most 0.05 cm. What is the maximum error in using this value
of the radius to compute the volume of the sphere?
4
3 r 3. If the error in the
measured value of r is denoted by dr
r, then the corresponding error in the calculated value of V is V , which can be approximated by the differential
SOLUTION If the radius of the sphere is r, then its volume is V dV 4 r 2 dr 5E03(pp 206217) 210 ❙❙❙❙ 1/17/06 1:45 PM Page 210 CHAPTER 3 DERIVATIVES When r 21 and dr 0.05, this becomes
dV 21 2 0.05 4 277 The maximum error in the calculated volume is about 277 cm3.
NOTE Although the possible error in Example 5 may appear to be rather large, a better
picture of the error is given by the relative error, which is computed by dividing the error
by the total volume:
■ V
V 4 r 2 dr
4
3
3r dV
V 3 dr
r Thus, the relative error in the volume is about three times the relative error in the radius.
In Example 5 the relative error in the radius is approximately dr r 0.05 21 0.0024
and it produces a relative error of about 0.007 in the volume. The errors could also be
expressed as percentage errors of 0.24% in the radius and 0.7% in the volume.  3.10 Exercises 1. The turkey in Example 1 is removed from the oven when its 4. The table shows the population of Nepal (in millions) as of temperature reaches 185 F and is placed on a table in a room
where the temperature is 75 F. After 10 minutes the temperature of the turkey is 172 F and after 20 minutes it is 160 F.
Use a linear approximation to predict the temperature of the
turkey after half an hour. Do you think your prediction is an
overestimate or an underestimate? Why? June 30 of the given year. Use a linear approximation to
estimate the population at midyear in 1984. Use another linear
approximation to predict the population in 2006.
t temperature of 15 C, the pressure is 101.3 kilopascals (kPa) at
sea level, 87.1 kPa at h 1 km, and 74.9 kPa at h 2 km.
Use a linear approximation to estimate the atmospheric pressure at an altitude of 3 km. 5–8 P
20
Percent
aged 65
and over 1990 1995 2000 15.0 17.0 19.3 22.0 24.9 Find the linearization L x of the function at a.  5. f x x 3, 6. f x 1 s2 x, 7. f x 3. The graph indicates how Australia’s population is aging by showing the past and projected percentage of the population
aged 65 and over. Use a linear approximation to predict the
percentage of the population that will be 65 and over in the
years 2040 and 2050. Do you think your predictions are too
high or too low? Why? 1985 Nt 2. Atmospheric pressure P decreases as altitude h increases. At a 1980 cos x, a 8. f x
■ a 1
a
2 3 sx, a ■ ■ 0 8 ■ ■ ■ ■ ■ ■ ■ ■ ; 9. Find the linear approximation of the function f x s1
at a 0 and use it to approximate the numbers s0.9 and
s0.99. Illustrate by graphing f and the tangent line. ; 10. Find the linear approximation of the function t x x 3
s1 x
3
at a 0 and use it to approximate the numbers s0.95 and
3
s1.1. Illustrate by graphing t and the tangent line. 10 ; 11–14  Verify the given linear approximation at a
0. Then
determine the values of x for which the linear approximation is
accurate to within 0.1. 0 ■ 1900 2000 t 3
11. s1 x 1 1
3 x 12. tan x x 5E03(pp 206217) 1/17/06 1:45 PM Page 211 S ECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS 13. 1 1
14. 1 s4 4 2x
x ■ ■ 15–20 1 ■ 1
16
■ x
■ ■ ■ ■ ■ ■ ■ ■ x4
2 40. The radius of a circular disk is given as 24 cm with a maxi 5x 16. y 17. y x tan x 18. y s1 t 19. y u
u 20. y 1 2r 1
1 ■ cos x mum error in measurement of 0.2 cm.
(a) Use differentials to estimate the maximum error in the calculated area of the disk.
(b) What is the relative error? What is the percentage error? 2 4 41. The circumference of a sphere was measured to be 84 cm with ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 21–26  (a) Find the differential dy and (b) evaluate dy for the
given values of x and dx. 21. y x2 2 x, 22. y x3 6x 2 23. y s4 24. y 1x 1, 25. y tan x, x cos x, x 26. y
■ ■ ■ 7, 5x 5 x, 1
2 3, dx x 0, x x 2, dx 0.04 1, x dx 0.1 dx apply a coat of paint 0.05 cm thick to a hemispherical dome
with diameter 50 m. 3, dx 43. (a) Use differentials to ﬁnd a formula for the approximate vol 0.1
0.05 ■ ■ ■ ■ ■ ■ ■ ■ x 2, 28. y s x, x 6 2 30. y
■ 1, x 1, x, 16 x,
■ x ■ x x
■ F 1 2,
4, x ■ of blood per unit time that ﬂows past a given point) is proportional to the fourth power of the radius R of the blood vessel: 0.5 x x 0.4
1 ■ ■ ■ ■ ■ ■ ■ 31–36  Use differentials (or, equivalently, a linear approximation)
to estimate the given number. 31. 2.001
33. 8.06 5 32. s99.8 23 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■  Explain, in terms of linear approximations or differentials,
why the approximation is reasonable. ■ ■ 38. 1.01 1
■ ■ ■ ■ ■ ■ 6 ■ (c) d u ■ ■ c du
v (d) d uv
(e) d 1.06
■ 0 (b) d cu 37–38 37. sec 0.08 (This is known as Poiseuille’s Law; we will show why it is true
in Section 9.4.) A partially clogged artery can be expanded by
an operation called angioplasty, in which a balloontipped
catheter is inﬂated inside the artery in order to widen it and
restore the normal blood ﬂow.
Show that the relative change in F is about four times the
relative change in R. How will a 5% increase in the radius
affect the ﬂow of blood? (a) dc 36. cos 31.5
■ kR 4 45. Establish the following rules for working with differentials
(where c denotes a constant and u and v are functions of x). 34. 1 1002 35. tan 44
■ ume of a thin cylindrical shell with height h, inner radius r,
and thickness r.
(b) What is the error involved in using the formula from
part (a)?
44. When blood ﬂows along a blood vessel, the ﬂux F (the volume  Compute y and dy for the given values of x and
dx
x. Then sketch a diagram like Figure 6 showing the line
segments with lengths dx, dy, and y. 27. y a possible error of 0.5 cm.
(a) Use differentials to estimate the maximum error in the
calculated surface area. What is the relative error?
(b) Use differentials to estimate the maximum error in the
calculated volume. What is the relative error?
42. Use differentials to estimate the amount of paint needed to 0.01 4, ■ dx 27–30 29. y in measurement of 0.1 cm. Use differentials to estimate the
maximum possible error, relative error, and percentage error in
computing (a) the volume of the cube and (b) the surface area
of the cube. Find the differential of the function.  15. y ■ 211 39. The edge of a cube was found to be 30 cm with a possible error 8x 1
2 ❙❙❙❙ u du
u dv dv
v du v du u dv v v2 (f) d x n nx n 1 dx 5E03(pp 206217) 212 ❙❙❙❙ 1/17/06 1:45 PM Page 212 CHAPTER 3 DERVIATIVES 46. On page 431 of Physics: Calculus, 2d ed., by Eugene Hecht (b) Are your estimates in part (a) too large or too small?
Explain. (Paciﬁc Grove, CA: Brooks/Cole, 2000), in the course of deriving the formula T 2 sL t for the period of a pendulum of
length L, the author obtains the equation a T
t sin for the
tangential acceleration of the bob of the pendulum. He then
says, “for small angles, the value of in radians is very nearly
the value of sin ; they differ by less than 2% out to about 20°.”
(a) Verify the linear approximation at 0 for the sine function:
sin x ; y y=fª(x)
1 x 0 (b) Use a graphing device to determine the values of x for
which sin x and x differ by less than 2%. Then verify
Hecht’s statement by converting from radians to degrees. x 1 48. Suppose that we don’t have a formula for t x but we know that t 2
sx 2 5 for all x.
4 and t x
(a) Use a linear approximation to estimate t 1.95 and t 2.05 .
(b) Are your estimates in part (a) too large or too small?
Explain. 47. Suppose that the only information we have about a function f is that f 1
5 and the graph of its derivative is as shown.
(a) Use a linear approximation to estimate f 0.9 and f 1.1 . LABORATORY PROJECT
; Taylor Polynomials
The tangent line approximation L x is the best ﬁrstdegree (linear) approximation to f x near
x a because f x and L x have the same rate of change (derivative) at a. For a better approximation than a linear one, let’s try a seconddegree (quadratic) approximation P x . In other
words, we approximate a curve by a parabola instead of by a straight line. To make sure that the
approximation is a good one, we stipulate the following:
(i) P a fa (P and f should have the same value at a.) (ii) P a fa (P and f should have the same rate of change at a.) (iii) P a fa (The slopes of P and f should change at the same rate.) A Bx C x 2 to the function f x
cos x that
satisﬁes conditions (i), (ii), and (iii) with a 0. Graph P, f , and the linear approximation
Lx
1 on a common screen. Comment on how well the functions P and L approximate f . 1. Find the quadratic approximation P x 2. Determine the values of x for which the quadratic approximation f x is accurate to within 0.1. [Hint: Graph y
a common screen.] Px,y P x in Problem 1
0.1, and y cos x 0.1 on cos x 3. To approximate a function f by a quadratic function P near a number a, it is best to write P in the form
Px A Bx a Cx a 2 Show that the quadratic function that satisﬁes conditions (i), (ii), and (iii) is
Px fa fax a 1
2 f ax a 2 4. Find the quadratic approximation to f x s x 3 near a 1. Graph f , the quadratic
approximation, and the linear approximation from Example 3 in Section 3.10 on a common
screen. What do you conclude? 5. Instead of being satisﬁed with a linear or quadratic approximation to f x near x a, let’s
try to ﬁnd better approximations with higherdegree polynomials. We look for an nthdegree
polynomial
Tn x c0 c1 x a c2 x a 2 c3 x a 3 cn x a n 5E03(pp 206217) 1/17/06 1:45 PM Page 213 C HAPTER 3 REVIEW ❙❙❙❙ 213 such that Tn and its ﬁrst n derivatives have the same values at x a as f and its ﬁrst n derivatives. By differentiating repeatedly and setting x a, show that these conditions are satisﬁed if c0 f a , c1 f a , c2 1 f a , and in general
2
ck
where k! fa k a k! k. The resulting polynomial 1234 Tn x f fax fa
x
2! a a f 2 n a n! x a n is called the nthdegree Taylor polynomial of f centered at a.
6. Find the 8thdegree Taylor polynomial centered at a 0 for the function f x
cos x.
Graph f together with the Taylor polynomials T2 , T4 , T6 , T8 in the viewing rectangle [ 5, 5]
by [ 1.4, 1.4] and comment on how well they approximate f .  3 Review ■ CONCEPT CHECK 1. Deﬁne the derivative f a . Discuss two ways of interpreting 4. State the derivative of each function. this number. (a)
(c)
(e)
(g) 2. (a) What does it mean for f to be differentiable at a ? (b) What is the relation between the differentiability and continuity of a function?
3. State each of the following differentiation rules both in ■ y
y
y
y xn
cos x
csc x
cot x 6. What are the second and third derivatives of a function f ? If f is the position function of an object, how can you interpret f
and f ?
7. (a) Write an expression for the linearization of f at a. (b) If y f x , write an expression for the differential dy.
(c) If dx
x, draw a picture showing the geometric meanings of y and dy. ■ TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. ■ 4. If f and t are differentiable, then d
f tx
dx 1. If f is continuous at a, then f is differentiable at a. d
sf x
dx fx
.
2 sf x 6. If f is differentiable, then fx d
f (s x )
dx fx
.
2 sx tx 3. If f and t are differentiable, then d
f xtx
dx f tx t x 5. If f is differentiable, then 2. If f and t are differentiable, then tx sin x
tan x
sec x 5. Explain how implicit differentiation works. symbols and in words.
(a) The Power Rule
(b) The Constant Multiple Rule
(c) The Sum Rule
(d) The Difference Rule
(e) The Product Rule
(f) The Quotient Rule
(g) The Chain Rule d
fx
dx (b) y
(d) y
(f) y f xt x 7. d
x2
dx x 2x 1 5E03(pp 206217) 214 ❙❙❙❙ 1/17/06 1:45 PM CHAPTER 3 DERIVATIVES 8. If f r exists, then lim f x x 5, then lim 9. If t x xl2 d2y
dx 2 tx
x t2
2 2, 4 is y
80.
12. 2 dy
dx x 2 at 11. An equation of the tangent line to the parabola y f r. x lr 10. Page 214 ■ EXERCISES 1. For the function f whose graph is shown, arrange the following numbers in increasing order: d
tan2x
dx 4 2x x 2. d
sec 2x
dx ■ 7. The ﬁgure shows the graphs of f , f , and f . Identify each curve, and explain your choices.
y 0 1 f2 f3 f5 f7 a y b
x 0 c 1
0 x 1 8. The total fertility rate at time t, denoted by F t , is an esti 2. Find a function f and a number a such that
6 2 lim h
h h l0 64 mate of the average number of children born to each woman
(assuming that current birth rates remain constant). The graph
of the total fertility rate in the United States shows the ﬂuctuations from 1940 to 1990.
(a) Estimate the values of F 1950 , F 1965 , and F 1987 .
(b) What are the meanings of these derivatives?
(c) Can you suggest reasons for the values of these derivatives? fa 3. The total cost of repaying a student loan at an interest rate of r % per year is C f r .
(a) What is the meaning of the derivative f r ? What are its
units?
(b) What does the statement f 10
1200 mean?
(c) Is f r always positive or does it change sign?
4–6 y 3.0 Trace or copy the graph of the function. Then sketch a
graph of its derivative directly beneath.
 4. 5. y baby
boom 3.5 baby
bust 2.5 baby
boomlet y=F(t) y
2.0
1.5 0 x
x 0 1940 6. 1950 1960 1970 1980 1990 9. Let B t be the total value of U.S. banknotes in circulation at y time t. The table gives values of this function from 1980 to
1998, at year end, in billions of dollars. Interpret and estimate
the value of B 1990 .
x t ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1980 1985 1990 1995 1998 Bt
■ t 124.8 182.0 268.2 401.5 492.2 5E03(pp 206217) 1/17/06 1:46 PM Page 215 C HAPTER 3 REVIEW 10–11  Find f x from ﬁrst principles, that is, directly from the
deﬁnition of a derivative. 4
3 10. f x
■ ■ x
x ■ x3 11. f x 5x 45–48 ■ ■ ■ ■ ■ ■ ; 1
sx 4 16. y 3x
s2 x 2
1 ; 1 18. y x 1
x2 3x 2 15. y sx 17. y 2 x sx 2 5 3 3 t 21. y 20. y t2 1 tan s1 23. x y 4 x 2y x 1 sec 2
tan 2 1 25. y x 27. y 1 sec 1 x2 3
1 sx y 30. y 34. y tan2 sin 36. x tan y 37. y 5
sx tan x 38. y 39. If f t ■ ■ s4 t ■ sin , ﬁnd t 41. Find y if x 6 n x ax y 6 bx
x 2y 2  43. lim 1
x b x c cos2x sin2x sin x y ■ ■ sin x cos a cos x sin a obtain the addition formula for the cosine function. 1 1x
2x a 55. Suppose that h x f x t x and F x
f2
3, t 2
5, t 2
4, f 2
Find (a) h 2 and (b) F 2 . 4
3
■ ■ f t x , where
2, and f 5
11. 56. If f and t are the functions whose graphs are shown, let Px
f x t x,Q x
f x t x , and C x
Find (a) P 2 , (b) Q 2 , and (c) C 2 . f tx . y x. xl0 Find the limit.
44. lim
tl0 t3
tan3 2t 1
0 ■ ■ ■ ■ ■ ■ ■ ■ 2, 1 g sec x
1 sin x x 1 where the tangent f
43–44 ■ obtain the doubleangle formula for the sine function.
(b) By differentiating the addition formula 6. 12 ■ c , show that a 1. x if f x ■ cos x, 0 sin x 1 cos 2 x 4 x
x
■ ■ 54. (a) By differentiating the doubleangle formula 1, ﬁnd f 2 . 40. If t 42. Find f ■ ■ 4 x tan x,
2x
2, ﬁnd f and f .
(b) Check to see that your answers to part (a) are reasonable by
comparing the graphs of f , f , and f . fx
fx xy sin m x
x 35. y x3) ■ 4 x
x4 sin(tan s1 ■ x s5 x, ﬁnd f x .
(b) Find equations of the tangent lines to the curve
y x s5 x at the points 1, 2 and 4, 4 .
(c) Illustrate part (b) by graphing the curve and tangent lines
on the same screen.
(d) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f . 53. If f x sx 32. y 33. y 2, 1 ■ ssin sx 5 ■ 13, ■ line has slope 1. sin 2y 28. y cot 3x 2 ■ y2 ■ 0, 1 52. Find the points on the ellipse x 2 sin x 31. y ■ 4 sin x, is the tangent line horizontal? 1 x2 29. sin x y 1 51. At what points on the curve y 1 26. x 2 cos y 0, 50. (a) If f x ; sin x 24. y 3y s7 sin cos x 22. y x 1
,
1 4 xy
■ 6, 1 49. (a) If f x cos tan x x4 19. y ■ 14. y 13. y s1 48. x 2 Calculate y .  x
x2 47. y s3 5x, use the deﬁnition of a derivative to
ﬁnd f x .
(b) Find the domains of f and f .
(c) Graph f and f on a common screen. Compare the graphs
to see whether your answer to part (a) is reasonable. 13–38 4 sin2 x, 46. y ■ 12. (a) If f x ; Find an equation of the tangent to the curve at the given  45. y 4 ■ 215 point. 2 ■ ❙❙❙❙ ■ ■ ■ ■ 1 x 5E03(pp 206217) 216 ❙❙❙❙ 57–64 1/17/06 1:46 PM Page 216 CHAPTER 3 DERIVATIVES  57. f x 73. The mass of part of a wire is x (1 s x ) kilograms, where x is
measured in meters from one end of the wire. Find the linear
density of the wire when x 4 m. Find f in terms of t .
2 2 58. f x
2 tx 60. f x xtx x at x b 59. f x tx 61. f x ttx 62. f x sin t x 63. f x t sin x 64. f x t(tan s x ) ■ ■ ■ ■ 65–67 ■  ■ ■ ■ ■ 74. The cost, in dollars, of producing x units of a certain commod ity is ■ Cx
■ f xtx
fx
tx 67. h x f t sin 4 x ■ ■ ■ ■ fx
tx 66. h x ■ ■ ; 68. (a) Graph the function f x ■ ■ ■ 0.02 x 2 0.00007x 3 75. The volume of a cube is increasing at a rate of 10 cm3 min. ■ ■ How fast is the surface area increasing when the length of an
edge is 30 cm? ■ 76. A paper cup has the shape of a cone with height 10 cm and 2 sin x in the viewing rect x 2x (a) Find the marginal cost function.
(b) Find C 100 and explain its meaning.
(c) Compare C 100 with the cost of producing the 101st item. Find h in terms of f and t . 65. h x 920 ■ radius 3 cm (at the top). If water is poured into the cup at a rate
of 2 cm3 s, how fast is the water level rising when the water is
5 cm deep? angle 0, 8 by 2, 8 .
(b) On which interval is the average rate of change larger:
1, 2 or 2, 3 ?
(c) At which value of x is the instantaneous rate of change
larger: x 2 or x 5?
(d) Check your visual estimates in part (c) by computing f x
and comparing the numerical values of f 2 and f 5 . 77. A balloon is rising at a constant speed of 5 ft s. A boy is cycling along a straight road at a speed of 15 ft s. When he
passes under the balloon, it is 45 ft above him. How fast is the
distance between the boy and the balloon increasing 3 s later? 69. The graph of f is shown. State, with reasons, the numbers at 78. A waterskier skis over the ramp shown in the ﬁgure at a speed which f is not differentiable. of 30 ft s. How fast is she rising as she leaves the ramp? y 4 ft
_1 0 2 4 6 x 15 ft 79. The angle of elevation of the Sun is decreasing at a rate of 0.25 rad h. How fast is the shadow cast by a 400fttall building increasing when the angle of elevation of the Sun is 6? 70. A particle moves along a horizontal line so that its coordinate at time t is x sb 2 c 2 t 2, t 0, where b and c are positive
constants.
(a) Find the velocity and acceleration functions.
(b) Show that the particle always moves in the positive
direction. ; 80. (a) Find the linear approximation to f x s25 x 2 near 3.
(b) Illustrate part (a) by graphing f and the linear
approximation.
(c) For what values of x is the linear approximation accurate to
within 0.1? 71. A particle moves on a vertical line so that its coordinate at time t is y t 3 12 t 3, t 0.
(a) Find the velocity and acceleration functions.
(b) When is the particle moving upward and when is it moving
downward?
(c) Find the distance that the particle travels in the time interval 0 t 3.
72. The volume of a right circular cone is V r 2h 3, where r is the radius of the base and h is the height.
(a) Find the rate of change of the volume with respect to the
height if the radius is constant.
(b) Find the rate of change of the volume with respect to the
radius if the height is constant. 81. (a) Find the linearization of f x ; 3
s1 3x at a 0. State
the corresponding linear approximation and use it to give
3
an approximate value for s1.03.
(b) Determine the values of x for which the linear approximation given in part (a) is accurate to within 0.1. 82. Evaluate dy if y x3 2x 2 1, x 2, and dx 0.2. 83. A window has the shape of a square surmounted by a semi circle. The base of the window is measured as having width
60 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum error possible in computing
the area of the window. 5E03(pp 206217) 1/17/06 1:46 PM Page 217 C HAPTER 3 REVIEW 84–86 17 84. lim
x l1 86. lim
l ■ ■ 88. Suppose f is a differentiable function such that f t x Express the limit as a derivative and evaluate.  x
x 4 1
1 85. lim cos h and f x 2 h 1 fx ■ 2 . Show that t x d
f 2x
dx
■ ■ ■ ■ ■ 11 217 x x2 . 89. Find f x if it is known that 0.5
3 3
■ s16 hl0 ❙❙❙❙ ■ ■ x2 ■ 90. Show that the length of the portion of any tangent line to the
87. Evaluate lim xl0 s1 tan x s1
x 3 sin x . astroid x 2 3
constant. y2 3 a 2 3 cut off by the coordinate axes is 5E03(pp 218221) 1/17/06 1:40 PM PROBLEMS
PLUS Page 218 Before you look at the example, cover up the solution and try it yourself ﬁrst.
EXAMPLE How many lines are tangent to both of the parabolas y
1 x 2 and
2
y 1 x ? Find the coordinates of the points at which these tangents touch the
parabolas.
SOLUTION To gain insight into this problem, it is essential to draw a diagram. So we sketch
the parabolas y 1 x 2 (which is the standard parabola y x 2 shifted 1 unit upward)
and y
1 x 2 (which is obtained by reﬂecting the ﬁrst parabola about the xaxis). If
we try to draw a line tangent to both parabolas, we soon discover that there are only two
possibilities, as illustrated in Figure 1.
Let P be a point at which one of these tangents touches the upper parabola and let a
be its xcoordinate. (The choice of notation for the unknown is important. Of course we
could have used b or c or x 0 or x1 instead of a. However, it’s not advisable to use x in
place of a because that x could be confused with the variable x in the equation of the
parabola.) Then, since P lies on the parabola y 1 x 2, its ycoordinate must be 1 a 2.
Because of the symmetry shown in Figure 1, the coordinates of the point Q where the
tangent touches the lower parabola must be a, 1 a 2 .
To use the given information that the line is a tangent, we equate the slope of the line
PQ to the slope of the tangent line at P. We have y P
1 x
_1 Q mPQ FIGURE 1 a2
a 1 1
a a2 a2 1
a If f x
1 x 2, then the slope of the tangent line at P is f a
tion that we need to use is that
a2 1 2 a. Thus, the condi 2a a Solving this equation, we get 1 a 2 2a 2, so a 2 1 and a
1. Therefore, the
points are (1, 2) and ( 1, 2). By symmetry, the two remaining points are ( 1, 2)
and (1, 2). P RO B L E M S 1 x 2 so that the triangle ABC formed by the xaxis
and the tangent lines at P and Q is an equilateral triangle. 1. Find points P and Q on the parabola y y x 3 3x 4 and y 3 x 2 x are tangent to each
other, that is, have a common tangent line. Illustrate by sketching both curves and the common
tangent. ; 2. Find the point where the curves y A 3. Suppose f is a function that satisﬁes the equation
P
B fx Q
0 C x y x2y fy for all real numbers x and y. Suppose also that
lim FIGURE FOR PROBLEM 1 x l0 (a) Find f 0 . 218 fx (b) Find f 0 . fx
x 1 (c) Find f x . xy 2 5E03(pp 218221) 1/17/06 1:40 PM Page 219 y 4. A car is traveling at night along a highway shaped like a parabola with its vertex at the origin (see the ﬁgure). The car starts at a point 100 m west and 100 m north of the origin and travels
in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At
what point on the highway will the car’s headlights illuminate the statue?
5. Prove that
x FIGURE FOR PROBLEM 4 dn
sin4 x
dx n cos4 x 4n 1 cos 4 x n 2. xn 1 6. Find the n th derivative of the function f x x. 7. The ﬁgure shows a circle with radius 1 inscribed in the parabola y x 2. Find the center of the circle.
y y=≈ 1 1 0 8. If f is differentiable at a, where a x 0, evaluate the following limit in terms of f a :
fx
sx lim xla fa
sa 9. The ﬁgure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1.2 m. The pin P slides back and forth along the xaxis as the wheel rotates counterclockwise
at a rate of 360 revolutions per minute.
(a) Find the angular velocity of the connecting rod, d dt, in radians per second, when
3.
(b) Express the distance x
OP in terms of .
(c) Find an expression for the velocity of the pin P in terms of .
y A
¨ å
P (x, 0) x O x 2 and they
intersect at a point P. Another tangent line T is drawn at a point between P1 and P2 ; it intersects T1 at Q1 and T2 at Q2. Show that 10. Tangent lines T1 and T2 are drawn at two points P1 and P2 on the parabola y PQ1
PP1 PQ2
PP2 1 219 5E03(pp 218221) 1/17/06 1:41 PM Page 220 y 11. Let T and N be the tangent and normal lines to the ellipse x 2 9 yT y 2 4 1 at any point P on
the ellipse in the ﬁrst quadrant. Let x T and y T be the x and yintercepts of T and x N and yN be
the intercepts of N . As P moves along the ellipse in the ﬁrst quadrant (but not on the axes),
what values can x T , y T , x N , and yN take on? First try to guess the answers just by looking at the
ﬁgure. Then use calculus to solve the problem and see how good your intuition is. T 2 P
xT xN
0 3 N yN 12. Evaluate lim
x sin 3 xl0 x2
x sin 9 . 13. (a) Use the identity for tan x y (see Equation 14b in Appendix D) to show that if two lines
L 1 and L 2 intersect at an angle , then
m2 m1
1 m1 m2 tan FIGURE FOR PROBLEM 11 where m1 and m2 are the slopes of L 1 and L 2 , respectively.
(b) The angle between the curves C1 and C2 at a point of intersection P is deﬁned to be the
angle between the tangent lines to C1 and C2 at P (if these tangent lines exist). Use part (a)
to ﬁnd, correct to the nearest degree, the angle between each pair of curves at each point
of intersection.
(i) y x 2 and y
x 22
2
2
(ii) x
y
3 and x 2 4 x y 2 3 0
14. Let P x 1, y1 be a point on the parabola y 2 4 px with focus F p, 0 . Let be the angle
between the parabola and the line segment FP, and let be the angle between the horizontal
line y y1 and the parabola as in the ﬁgure. Prove that
. (Thus, by a principle of geometrical optics, light from a source placed at F will be reﬂected along a line parallel to the
xaxis. This explains why paraboloids, the surfaces obtained by rotating parabolas about their
axes, are used as the shape of some automobile headlights and mirrors for telescopes.)
y å
0 y=› ∫
P(⁄, ›) x F( p, 0)
¥=4px 15. Suppose that we replace the parabolic mirror of Problem 14 by a spherical mirror. Although
Q P ¨
¨
A R O the mirror has no focus, we can show the existence of an approximate focus. In the ﬁgure,
C is a semicircle with center O. A ray of light coming in toward the mirror parallel to the
axis along the line PQ will be reﬂected to the point R on the axis so that PQO
OQR
(the angle of incidence is equal to the angle of reﬂection). What happens to the point R as P
is taken closer and closer to the axis?
16. If f and t are differentiable functions with f 0 t0 C lim xl0 FIGURE FOR PROBLEM 15
17. Evaluate lim xl0 220 sin a 2x 2 sin a
x2 fx
tx
x f0
t0
sin a . 0 and t 0 0, show that 5E03(pp 218221) 1/17/06 1:41 PM Page 221 18. Given an ellipse x 2 a 2 y 2 b 2 1, where a b, ﬁnd the equation of the set of all points
from which there are two tangents to the curve whose slopes are (a) reciprocals and
(b) negative reciprocals. 19. Find the two points on the curve y x4 2x 2 x that have a common tangent line. x 2 have the property that their normal lines
intersect at a common point. Show that the sum of their xcoordinates is 0. 20. Suppose that three points on the parabola y 21. A lattice point in the plane is a point with integer coordinates. Suppose that circles with radius r are drawn using all lattice points as centers. Find the smallest value of r such that any
line with slope 2 intersects some of these circles.
5
22. A cone of radius r centimeters and height h centimeters is lowered point ﬁrst at a rate of 1 cm s into a tall cylinder of radius R centimeters that is partially ﬁlled with water. How fast
is the water level rising at the instant the cone is completely submerged?
23. A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It is partially ﬁlled with a liquid that oozes through the sides at a rate proportional to the area
of the container that is in contact with the liquid. (The surface area of a cone is rl, where
r is the radius and l is the slant height.) If we pour the liquid into the container at a rate of
2 cm3 min, then the height of the liquid decreases at a rate of 0.3 cm min when the height is
10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should we
pour the liquid into the container?
C AS 24. (a) The cubic function f x x x 2 x 6 has three distinct zeros: 0, 2, and 6. Graph f
and its tangent lines at the average of each pair of zeros. What do you notice?
(b) Suppose the cubic function f x
x a x b x c has three distinct zeros: a, b,
and c. Prove, with the help of a computer algebra system, that a tangent line drawn at the
average of the zeros a and b intersects the graph of f at the third zero. 221 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.
 Fall '09
 hamrick
 The Land

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