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Unformatted text preview: By measuring slopes at points on the sine curve, ofthe sine function is the cosine function. Derivatives we get strong visual evidence that the derivative é,
«iutbcnzﬁvi a e ..i,,o25¥?
._i{§:¥!e§.e§§:§:4 a? 2.53.53 . zisaeiewﬁﬁﬂmiéziﬁiii .» In this chapter we begin our study of differential calculus,
which is concerned with how one quantity changes in rela
tion 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 prob lems involving rates of change and the approximation of functions. §; 3.1 ieHvaHves In Section 2.6 we deﬁned the slope of the tangent to a curve with equation y = f (x) at the
point where x = a to be 3 mzmgw+2—ﬂm Irv>0 We also saw that the velocity of an object with position function 5 = f (t) at time t = a is ”@mere+m—nm hat) h In fact, limits of the form 1. f(a + h) ‘f(a)
1m h~>0 h 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. E Definition The derivative of a function f at a number a, denoted by f ’(a), is + _
till f'(a) is read "f prime of a." f'(a) = Pug W l
_ if this limit exists. l
If we write x = a + h, then h = x — a and h approaches 0 if and only if x approaches 0. Therefore, an equivalent way of stating the deﬁnition of the derivative, as we saw in
ﬁnding tangent lines, is @ . fW=mﬂWW® X—WI x—a l ll 1 3 128 CHAPTER 3 ntmvnnvrs EXAMPL£ I Find the derivative of the function f (x) = )62 ~ 8x + 9 at the numbe:
SOLUTION From Deﬁnition 2 we have . (a + h — a
Try problems like this one. f’(a) = 11111 l__—).f(_l
‘ Resources / Module 3 I140 h /Derivative ataPolnt 'm [(a + h)2 * 8(a + h) + 9] __ [a2 _ 8a + 9] / Problem Wizard —— 11
h—>0 h 2n a2+2ah+h28a~8h+9—a2+8a—9 £133 h
2 h +h2— 8h
=1imi—————=1im(2a+h— 8)
h—>0 h n—>o
=2a — 8 ”H Interpretation of the nerivative as the Stone of a tangent b: In Section 2.6 we deﬁned the tangent line to the curve y = f (x) at the point P(a
‘l 1: .‘i be the line that passes through P and has slope m given by Equation 1. Since, by
' tion 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 ((1)) whose s10]
equal to f ’(a), the derivative of f at a. Thus, the geometric interpretation of a derivative [as deﬁned by either (2) or (
shown in Figure 1. yﬁm >
a + h x
‘ . f(a+h)‘f(al . f(X)—f(a)
: @ﬂaﬂm~——~—— ®’aﬂm—ﬂ~—
; _ FIGURE! H M h f” H, x—a
l j Geometric interpretation = slope of tangent at P = slope of tangent at P
I of the derivative = slope of curve at P = slope of curve at P l
l
l
l
l
t
r
.
l
r If we use the point—slope form of the equation of a line, we can write an equatil
tangent line to the curve y = f (x) at the point (a, f (a)): yﬂ®=fWU®
EXAMPLE 2 Find an equation of the tangent line to the parabola y = x2  8x + 9
point (3, —6). SOLUTION From Example 1 we know that the derivative of f (x) = x2 — 8x + 9 at
number a is f ’(a) = 2a — 8. Therefore, the slope of the tangent line at (3, ~6) it FIGURE 2 0.01 0.001 0.0001
~01
—0.01
~0.001 0.690
0.093
0.093
(1,670
I). (19 I
0.693
0.1103 (a)[*1,1]by [0, 2] SECTION 3.1 DERIVATIVES I29 f’(3) = 2(3) _ 8 = —2. Thus, an equation of the tangent line, shown in Figure 2, is y — (6) = (—2)(x  3) or y = ~2x EXAMPLE 3 Let f(x) = 2". 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". SOLUTION
(a) From Deﬁnition 2 we have f,(0) “ Iim 11m hMo h h—>O fth) f(0) __ . 2h —1
h Since we are not yet able to evaluate this limit exactly, we use a calculator to approxi
mate the values of (2"  l)/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
f ’(0) z 0.69 (b) In Figure 3 we graph the curve y = 2‘ and zoom in toward the point (0, 1). We see
that the closer we get to (0, l), 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 001, we estimate that the slope of this line is 0.14
—— = 0.7
0.20 So our estimate of the derivative is f ’(0) z 0.7. In Chapter 7 we will show that, correct
to six decimal places, f ’(0) z 0.693147. (b) [~0.5, 0.5] by [0.5, 1.5] (c) [—01, 0.1] by [0.9, 1.1] FIGURE 3 Zooming in on the graph of y = 2" near (0, 1) M» H“ Inrerureranun of the ﬂerivalive 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, as the limit of the average rates of change over smaller and smaller intervals. If the
interval is [x1, x2], then the change in x is Ax = x2 — x1, the corresponding change in y is M =f(x2) — f(xi) and I E .__A “——
[Z] instantaneous rate of change = hm y = lim ﬁn) f(x1)
Ax—ao Ax .r;—>.r. x2 ‘ x1 130 CHAPTER 3 DERIVATIVES FIGURE 4 The yvalues are changing rapidly
at P and slowly at Q. In Module 31 you are asked to compare
and order the slopes of tangent and secant lines at several points on a curve. From Equation 3 we recognize this limit as being the derivative of f at x,, that i
This gives a second interpretation of the derivative: x whenx = a. I‘
I The derivative f ’(a) is the instantaneous rate of change of y = f (x) with respe
L The connection with the ﬁrst interpretation is that if we sketch the curve y = f
the instantaneous rate of change is the slope of the tangent to this curve at the poi
x = a. This means that when the derivative is large (and therefore the curve is ste
the point P in Figure 4), the y‘values change rapidly. When the derivative is St
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
line, then f ’(a) is the rate of change of the displacement s with respect to the ti
other words, f ’(a) is the velocity of the particle at time t = a. (See Section 2.6.) Tl
of the particle is the absolute value of the velocity, that is, I f ’(a) I EXAMPLE 4 The position of a particle is given by the equation of motion
5 = f (t) = l/(l + t), where t is measured in seconds and s in meters. Find the ve
and the speed after 2 seconds. SOLUTION The derivative of f when t = 2 is 1 1
, _. f(2+h),—f(2)_, 1+(2+h)_1+2
f(2)—l1§3 h W333 h
1 _i 3—(3+h)
=1. 3+h 3_1. 3(3+h)
#33) h ~33) h #33 3(3 + h)h #36 3(3 + h) 9 Thus, the velocity after 2 seconds is f ’(2) = — 5 m/s, and the speed is If’(2)l = l—él = ism/s. EXAMPLE 5 A manufacturer produces bolts of a fabric with a ﬁxed width. The cost
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;
is, f ’(x) means the rate of change of the production cost with respect to the numbe
yards produced. (Economists call this rate of change the marginal cost. This idea i:
cussed in more detail in Sections 3.4 and 4.8.) Because f I(x) = lim Ag Air—>0 Ax the units for f '(x) are the same as the units for the difference quotient AC/Ax. Sinc
AC is measured in dollars and Ax in yards, it follows that the units for f ’(x) are do
per yard. 1111 Here we are ass is well behaved; in other w uming that the cost function
ords. C(x) doesn't oscillate rapidly near x = 1000,
f on) r. 1980 930.2 1985 1945.9 1995 4974.0 2000 5674.2 J llll 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.) SECTION 3.] DERIVATIVES I31 (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 Ax = 1 is small compared with x = 1000, we could use the approximation f’(1000)z .92 = E = AC
Ax l 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) W 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. EXAMPLE 6 Let D(t) be the US. 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’(l990) 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(t) — D(1990)
D’ = —————
0990) .3350 t 1990 So we compute and tabulate values of the difference quotient (the average rates of
change) as follows. 1 D0) v D(1990)
r — 1990
1980 230.31
1985 257.48
1995 348.14
2000 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’(l990) z 303 billion dollars per year W 4;: CI‘S‘:*“;‘~:‘ wrung: “cadmium u» .r I32 CHAPTER 3 DERIVATIVES yi3J Unaus I. On the given graph of f, mark lengths that represent f (2), f(2 + h), f(2 + h) —f(2), and h. (Choose 11 > 0.)What
fe+m—fm7
h . line has slope 2. For the function f whose graph is shown in Exercise 1, mange
the following numbers in increasing order and explain your
reasoning: 0 f'(2) f(3) ~ f(2) §'[f(4) — f(2)] 3. For the function 9 whose graph is given, arrange the following
numbers in increasing order and explain your reasoning: 0 til2) g’(0) g'(2) g’(4) 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) = 0, f ’(0) = 3,
f’(1)= 0, and f’(2) = —1. 6. Sketch the graph of a function g for which 9(0) = 0, g'(0) = 3,
g’(1) = O, and g'(2) = l, 7. If f (x) = 3x2 — 5x, ﬁnd f ’(2) and use it to ﬁnd an equation
of the tangent line to the parabola y = 3x2 — 5x at the
point (2, 2). 8. If g(x) = 1 — x3, ﬁnd g’(0) and use it to ﬁnd an equation of the
tangent line to the curve y = 1 — x3 at the point (0, 1). 9. (a) If F(x) = x3 — 5x + 1, ﬁnd F’(1) and use it to ﬁnd an
equation of the tangent line to the curve y = x3  5x + 1
at the point (1. —3).
(b) Illustrate part (a) by graphing the curve and the tangent line
on the same screen. E E3 ‘0. (a) If G(x) = x/(l + 2x), ﬁnd G’(a) and use it to ﬁr
equation of the tangent line to the curve y = x/(
the point (—41, —%). (b) Illustrate part (a) by graphing the curve and the t
on the same screen. I]. Let ﬁx) = 3X Estimate the value of f’(1) in two way
(a) By using Deﬁnition 2 and taking successively sm
Values of h. (b) By zooming in on the graph ofy = 3‘ and estimz
slope. ' 12. Let 90‘) = tan )6. Estimate the value of g’(7r/4) in tw
(a) By using Deﬁnition 2 and taking successively sm
Values of h. (b) By zooming in on the graph of y = tan x and esti
slope 1318 Ill! Find f’(a).
I3. f(x) ‘ 3 _ 2x + 4x2 14 f(t) = t4 — 5t 2t+1 X2+1
15150)" r+3 16f(X)—x_2 1
17.1%); jy—T—Z Is. f(x) = W D L: n u u u n
a 1924 ml Each limit represents the derivative of some fun some number a. State such an f and a in each case. . (1+h)‘°~l 20 1i“Ii/16+};— " P35 11 ' h~»0 h
2""32 , tanx—l
2" Pig x ’ 5 22. x1324 x — 7r/4
. coS(7T+h)+1 241.mt4+t—2 23 :13}, h . rl—H t—l D u u n r ,
n U
a 25—26 ”H A particle moves along a straight line with equz
motion 5 g f (I), where s is measured in meters and tin 31
Find the velocity when t = 2. 25. f(t)"2‘ 6’” n a u u 26. f(t) = 2t3 — t ‘i n
a 27. The cost 0f producing x ounces of gold from a new g
is C , f(x) dollars.
(a) What is the meaning of the derivative f ’(x)? What
units?
(b) What does the statement f ’(800) = 17 mean?
(0) D0 you think the values of f ’(x) will increase or (
in the short term? What about the long term? Exp If he 81 ber of bacteria after t hours in a controlled laboratory
um = to). .
i ng of the derivative f '(5)? What are its 28. The n .
experiment IS :1 .
(a) What is the meam nits? ‘ .
(b) guppose there is an unlimited amount of space and trients for the bacteria. Which do you think is larger,
r131(5) or f ’(10)? If the supply of nutrients is limited, would that affect your conclusion? Explain. 29 The fuel consumption (measured in gallons per hour) of a car ‘ ' hour is c =f(v),
l at a speed of 11 miles per . . I .
3,312]: is the meaning of the derivatlve f (1))? What are 1:5 units? .
(b) Write a sentence (in layman’s terms) that explains the
meaning of the equation f ’(20) = —0.05. 30 The quantity (in pounds) of a gourmet ground coffee that is
i sold by a coffee company at a price of p dollars per pound is = f (PI . . , .
(a)QWhat is the meaning of the derivative f (8)? What are its units? . .
(b) Is f ’(8) positive or negative? Explain. 1‘ a Let T(t) be the temperature (in °F) in Dallas t hours after mid 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. [J
.L‘
3‘
0C
5
1'3
3: 73 70 69 72 81 88 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). r em r em I 1900 48.3 19610 66.6 1910 51.1 1970 67.1 1920 55.2 1980 70.0 1930 57.4 1990 71.8 1940 62.5 2000 74.1
1950 65.6 7' K'T'NQ._FBO.J.E,C.T, . WRITING PROJECT EARLY METHODS FOR FINDING TANGENIS I33 33. The quantity of oxygen that can dissolve in water depends on 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 12 t 1‘ a >
0 s 16 24 32 40 T(°C) 34. The graph shows the inﬂuence of the temperature T on the 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. S (cm /s)
20
——1 1 t~ 4 1 >
0 10 20 T (“’0 3536 1111 Determine whether f ’(0) exists. xsin— ifx¢0 35. f(x)= x
0 ifx=0
chsinl if xvéO
36. f(x)= x
0 ifx=0 :1 Early 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 “IfI 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. I34 CHAPTER 3 DERIVATIVES The following references contain explanations of these methods. Read one or more of
references and write a report comparing the methods of either Fermat or Barrow to mode]
ods. In particular, use the method of Section 3.1 to ﬁnd an equation of the tangent line to
curve y = x3 + 2x at the point (1, 3) and show how either Fermat or Barrow would have
the same problem. Although you used derivatives and they did not, point out similarities l:
the methods. I. 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: Springer—Ver
1979), pp. 124, 132. 3. Howard Eves, An Introduction to the History of Mathematics, 6th ed. (New York: Sau
1990), pp. 391, 395. 4. Morris Kline, Mathematical Thought from Ancient to Modern Times (New York: Oxfo
University Press, 1972), pp. 344, 346. FIGURE I 3.2 The iierivaiive as a Funtiinn In the preceding section we considered the derivative of a function f at a ﬁxed nu [I f’(a) '= hm f(a + 11) —f(a) h—>0 h Here we change our point of view and let the number a vary. If we replace a in eq
by a variable x, we obtain , *. f(x+h)—f(x)
ii f(X)—}igrg)—*———~h } Given any numberx for which this limit exists, we assign to x the number f ’(x). So
regard f ’ as a new function, called the derivative of f and deﬁned by Equation
know that the value of f ’ at x, f ’(x), can be interpreted geometrically as the slope
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” frm
the limiting operation in Equation 2. The domain of f’ is the set {x] f '(x) exists} at
be smaller than the domain of f. EXAMPLE I The graph of a function f is given in Figure 1. Use it to sketch the grat
the derivative f ’. ",7, , . » , im__ SECTION 3.2 THE DERIVATIVE AS A FUNCTION I35 SOLUTION We can estimate the value of the derivative at any value of x by drawing the
ﬁll its derivative» tangent at the point (x, f (x)) and estimating its slope. For instance, for x = 5 we draw the
Resources/ Module3 . tangent at P In Figure 2(a) and estimate its slope to be about % , so f ’(5) z 1.5. This
/Derivatives as Functions allows us to plot the point P’(5, 1.5) on the graph of f ’ directly beneath P. Repeating
/Mars/Rhi;\$me 3 this procedure at several points, we get the graph shown in Figure 2(b). Notice that the
RisgigngaScope tangents at A, B, and C are horizontal, so the derivative is 0 there and the graph of f ’
/ Derivative of a Cubic crosses the x—axis 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. animation of the relation between a ll llil Notice that where the derivative is positive (a)
" (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. Ne
he at FIGURE 2 (b) If a function is deﬁned by a table of values, then we can construct a table of approxi
mate values of its derivative, as in the next example. I36 CHAPTER 3 DERIVATIVES 1 r BU) ; 1980 9,847
5 1982 9.856
E 1984 9855
1986 9,862
1988 9884
1990 9,962
3 1992 1(1036
1994 10.109
3 1996 IOJ52
1 1998 10J75
1 2000 10J86 1 B’(r)
_,_.__.+__._._A
1980 4.5
1982 2.0
1984 105
1986 7.3
1988 25.0
1990 38.0
1992 36.8
1994 29.0
1996 16.5
1998 815
2000 515 llll Figure 3 illustrates Example 2 by showing
graphs of the population function B(t) and its
derivative B’(t). Notice how the rate of popu
lation growth increases to a maximum in 1990
and ...
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 Tuffaha
 Calculus

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