Unformatted text preview: 5E07(pp 412421) 1/17/06 4:37 PM Page 412 CHAPTER 7 The graphs of the functions of
The graphs of the functions
this chapter appear as reﬂecof this chapter appear
tions of each other––the natural
as reﬂections of each
exponential and logarithmic
other––the natural expofunctions, the restricted tangent
nential and logarithmic
and inverse tangent functions,
functions, the restricted
and the hyperbolic sine function
tangent and inverse tanand its inverse.
gent functions, and the
hyperbolic sine function
and its inverse. I nverse Functions:
Exponential, Logarithmic, and Inverse Trigonometric Functions 5E07(pp 412421) 1/17/06 4:37 PM Page 413 The common theme that links the functions of this chapter is that they
occur as pairs of inverse functions. In particular, two of the most
important functions that occur in mathematics and its applications
are the exponential function f x
logarithmic function t x a x and its inverse function, the log a x. In this chapter we investigate their properties, compute their derivatives, and use them to describe exponential growth and
decay in biology, physics, chemistry, and other sciences. We also study the inverses of
trigonometric and hyperbolic functions. Finally, we look at a method (l’Hospital’s Rule) for
computing difﬁcult limits and apply it to sketching curves.
There are two possible ways of deﬁning the exponential and logarithmic functions and
developing their properties and derivatives. One is to start with the exponential function
(deﬁned as in algebra or precalculus courses) and then deﬁne the logarithm as its inverse.
That is the approach taken in Sections 7.2, 7.3, and 7.4 and is probably the most intuitive
method. The other way is to start by deﬁning the logarithm as an integral and then deﬁne the
exponential function as its inverse. This approach is followed in Sections 7.2*, 7.3*, and
7.4* and, although it is less intuitive, many instructors prefer it because it is more rigorous
and the properties follow more easily. You need only read one of these two approaches
(whichever your instructor recommends).  7.1 Inverse Functions
Table 1 gives data from an experiment in which a bacteria culture started with 100 bacteria in a limited nutrient medium; the size of the bacteria population was recorded at hourly
intervals. The number of bacteria N is a function of the time t : N f t .
Suppose, however, that the biologist changes her point of view and becomes interested
in the time required for the population to reach various levels. In other words, she is thinking of t as a function of N. This function is called the inverse function of f, denoted by f 1,
and read “ f inverse.” Thus, t f 1 N is the time required for the population level to reach
N . The values of f 1 can be found by reading Table 1 from right to left or by consulting
6 because f 6
Table 2. For instance, f 1 550
550.
T ABLE 1 N as a function of t t
(hours)
0
1
2
3
4
5
6
7
8 N ft
population at time t
100
168
259
358
445
509
550
573
586 TABLE 2 t as a function of N t
N
100
168
259
358
445
509
550
573
586 f 1N
time to reach N bacteria
0
1
2
3
4
5
6
7
8 413 5E07(pp 412421) 414 ❙❙❙❙ 1/17/06 4:37 PM Page 414 CHAPTER 7 INVERSE FUNCTIONS Not all functions possess inverses. Let’s compare the functions f and t whose arrow
diagrams are shown in Figure 1.
4 10 4 3 7 3 2 4 2 2 1 1
FIGURE 1 f A B 10
4
2
g A B Note that f never takes on the same value twice (any two inputs in A have different outputs), whereas t does take on the same value twice (both 2 and 3 have the same output, 4).
In symbols,
t2
t3
but f x1 f x2 whenever x 1 x2 Functions that have this property are called onetoone functions.
 In the language of inputs and outputs, this
deﬁnition says that f is onetoone if each output corresponds to only one input. 1 Definition A function f is called a onetoone function if it never takes on the
same value twice; that is, f x1 f x2 whenever x 1 x2 y If a horizontal line intersects the graph of f in more than one point, then we see from
Figure 2 that there are numbers x 1 and x 2 such that f x 1
f x 2 . This means that f is not
onetoone. Therefore, we have the following geometric method for determining whether
a function is onetoone. y=ƒ
ﬂ 0 ‡ ⁄ x ¤ Horizontal Line Test A function is onetoone if and only if no horizontal line intersects its graph more than once. FIGURE 2 This function is not onetoone
because f(⁄)=f(¤). EXAMPLE 1 Is the function f x x 3 onetoone? 3
x 2 , then x 1 x 3 (two different numbers can’t have the same cube).
2
Therefore, by Deﬁnition 1, f x
x 3 is onetoone. SOLUTION 1 If x 1 y SOLUTION 2 From Figure 3 we see that no horizontal line intersects the graph of f x y=˛ more than once. Therefore, by the Horizontal Line Test, f is onetoone.
0 x EXAMPLE 2 Is the function t x x 2 onetoone? SOLUTION 1 This function is not onetoone because, for instance, t1
FIGURE 3 ƒ=˛ is onetoone. and so 1 and 1 have the same output. 1 t 1 x3 5E07(pp 412421) 1/17/06 4:37 PM Page 415 S ECTION 7.1 INVERSE FUNCTIONS y ❙❙❙❙ 415 SOLUTION 2 From Figure 4 we see that there are horizontal lines that intersect the graph of
y=≈ t more than once. Therefore, by the Horizontal Line Test, t is not onetoone.
Onetoone functions are important because they are precisely the functions that possess inverse functions according to the following deﬁnition. 0 x 2 Definition Let f be a onetoone function with domain A and range B. Then its
inverse function f 1 has domain B and range A and is deﬁned by FIGURE 4 ©=≈ is not onetoone. 1 f y x fx &? y for any y in B.
x A
f This deﬁnition says that if f maps x into y, then f 1 maps y back into x. (If f were not
onetoone, then f 1 would not be uniquely deﬁned.) The arrow diagram in Figure 5 indicates that f 1 reverses the effect of f . Note that f –! B y domain of f 1 range of f range of f FIGURE 5 1 domain of f x 3 is f For example, the inverse function of f x
1 f 
A _10 1 in f 1 1 x3 x3 13 x 3, then x for an exponent. Thus 7 8 Do not mistake the 5 3 ■ f x 1 3 because if y x B 1 CAUTION y 1 f 1 x does not mean 1
fx The reciprocal 1 f x could, however, be written as f x 1 . f EXAMPLE 3 If f 1 5, f 3 7, and f 8 10, ﬁnd f 1 7,f A B 1 f 1 7 7 3 because f3 _10 f 1 5 1 because f1 8 because f8 10 . 5 10 1 7 8 5 , and f 5 3 1 f –! SOLUTION From the deﬁnition of f f 1 1 we have 10 FIGURE 6 The inverse function reverses
inputs and outputs. The diagram in Figure 6 makes it clear how f 1 reverses the effect of f in this case. 5E07(pp 412421) 416 ❙❙❙❙ 1/17/06 4:38 PM Page 416 CHAPTER 7 INVERSE FUNCTIONS The letter x is traditionally used as the independent variable, so when we concentrate
on f 1 rather than on f , we usually reverse the roles of x and y in Deﬁnition 2 and write
1 f 3 x y &? fy x By substituting for y in Deﬁnition 2 and substituting for x in (3), we get the following
cancellation equations: f 4 1 fx x for every x in A 1 x for every x in B ff x The ﬁrst cancellation equation says that if we start with x, apply f , and then apply f 1, we
arrive back at x, where we started (see the machine diagram in Figure 7). Thus, f 1 undoes
what f does. The second equation says that f undoes what f 1 does.
x ƒ f f –! x FIGURE 7 For example, if f x x 3, then f
f 1 ff 1 x 1 3 and the cancellation equations become x x3 fx
1 x x 13 x 133 x These equations simply say that the cube function and the cube root function cancel each
other when applied in succession.
Now let’s see how to compute inverse functions. If we have a function y f x and are
able to solve this equation for x in terms of y, then according to Deﬁnition 2 we must have
x f 1 y . If we want to call the independent variable x, we then interchange x and y and
arrive at the equation y f 1 x .
5 How to Find the Inverse Function of a OneToOne Function f STEP 1 Write y f x.
STEP 2 Solve this equation for x in terms of y (if possible).
STEP 3 To express f 1 as a function of x, interchange x and y.
The resulting equation is y f 1 x .
x3 EXAMPLE 4 Find the inverse function of f x 2. SOLUTION According to (5) we ﬁrst write y x3 x3 y 2 Then we solve this equation for x : x 3
sy 2
2 5E07(pp 412421) 1/17/06 4:38 PM Page 417 S ECTION 7.1 INVERSE FUNCTIONS  In Example 4, notice how f 1 reverses the
effect of f . The function f is the rule “Cube,
then add 2”; f 1 is the rule “Subtract 2, then
take the cube root.” ❙❙❙❙ 417 Finally, we interchange x and y :
y
Therefore, the inverse function is f 1 x 3
sx 3
sx 2
2. The principle of interchanging x and y to ﬁnd the inverse function also gives us the
method for obtaining the graph of f 1 from the graph of f . Since f a
b if and only if
f 1b
a, the point a, b is on the graph of f if and only if the point b, a is on the
graph of f 1. But we get the point b, a from a, b by reﬂecting about the line y x. (See
Figure 8.)
Therefore, as illustrated by Figure 9:
The graph of f y 1 is obtained by reﬂecting the graph of f about the line y x. y (b, a) f –!
(a, b)
0 0
x x y=x y=x FIGURE 8 f FIGURE 9 y y=ƒ
y=x EXAMPLE 5 Sketch the graphs of f x
same coordinate axes. 0
(_1, 0) (0, _1) x y=f –!(x)
FIGURE 10 s1 x and its inverse function using the SOLUTION First we sketch the curve y y2
1
graph of f
f 1 is f 1
y
x2 s 1 x (the top half of the parabola
x, or x
y 2 1) and then we reﬂect about the line y x to get the
1
. (See Figure 10.) As a check on our graph, notice that the expression for
x
x 2 1, x 0. So the graph of f 1 is the right half of the parabola
1 and this seems reasonable from Figure 10. The Calculus of Inverse Functions
Now let’s look at inverse functions from the point of view of calculus. Suppose that f is
both onetoone and continuous. We think of a continuous function as one whose graph
has no break in it. (It consists of just one piece.) Since the graph of f 1 is obtained from
the graph of f by reﬂecting about the line y x, the graph of f 1 has no break in it either
(see Figure 9). Thus, we might expect that f 1 is also a continuous function.
This geometrical argument does not prove the following theorem but at least it makes
the theorem plausible. A proof can be found in Appendix F.
6 Theorem If f is a onetoone continuous function deﬁned on an interval, then
its inverse function f 1 is also continuous. 5E07(pp 412421) 418 ❙❙❙❙ 1/17/06 4:38 PM Page 418 CHAPTER 7 INVERSE FUNCTIONS y ¨
(a, b) y=x
˙
¨ graph of g
(b, a)
0 x graph of f Now suppose that f is a onetoone differentiable function. Geometrically we can think
of a differentiable function as one whose graph has no corner or kink in it. We get the
graph of f 1 by reﬂecting the graph of f about the line y x, so the graph of f 1 has no
corner or kink in it either. We therefore expect that f 1 is also differentiable (except where
i
t
s
tangents are vertical). In fact, we can predict the value of the derivative of f 1 at a given
point by a geometric argument. In Figure 11 the graphs of f and its inverse t f 1 are
a, then t a
f 1a
b and t a is the slope of the tangent to the
shown. If f b
tan . From Figure 11 we see that
graph of t at a, b , which is tan . Likewise, f b
2, so FIGURE 11 ta tan tan 7 t 2 1
fb 1
f ta ta that is, 1
tan Theorem If f is a onetoone differentiable function with inverse function f 1 and f t a 0, then the inverse function is differentiable at a and
1
f ta ta Proof Write the deﬁnition of derivative as in Equation 3.1.3: ta lim xla tx
x ta
a By (3) we have
tx &? fy x ta and y
b &? fb a Since f is differentiable, it is continuous, so t f 1 is continuous by Theorem 6. Thus,
if x l a, then t x l t a , that is, y l b. Therefore ta lim xla tx
x lim
y lb 1
fb ta
a
1 fy
y fb
b
1
f ta lim
y lb y
fy 1
fy
lim
y lb
y b
fb
fb
b 5E07(pp 412421) 1/17/06 4:38 PM Page 419 S ECTION 7.1 INVERSE FUNCTIONS
NOTE 1 ■ ❙❙❙❙ 419 Replacing a by the general number x in the formula of Theorem 7, we get
1
f tx tx 8 If we write y
becomes t x , then f y x, so Equation 8, when expressed in Leibniz notation,
dy
dx 1
dx
dy If it is known in advance that f 1 is differentiable, then its derivative can be
computed more easily than in the proof of Theorem 7 by using implicit differentiation. If
y f 1 x , then f y
x. Differentiating the equation f y
x implicitly with respect to
x, remembering that y is a function of x, and using the Chain Rule, we get
NOTE 2 ■ dy
dx fy dy
dx Therefore y 1
1 1
dx
dy fy x 2, x
, is not onetoone and therefore does
not have an inverse function, we can turn it into a onetoone function by restricting its
domain. For instance, the function f x
x 2, 0 x 2, is onetoone (by the Horizontal Line Test) and has domain 0, 2 and range 0, 4 . (See Figure 12.) Thus, f has an
inverse function t f 1 with domain 0, 4 and range 0, 2 .
Without computing a formula for t we can still calculate t 1 . Since f 1
1, we
have t 1
1. Also f x
2 x. So by Theorem 7 we have
EXAMPLE 6 Although the function y 0 (a) y=≈, x R y
(2, 4) 1
f t1 t1 0 1
2 In this case it is easy to ﬁnd t explicitly. In fact, t x
s x, 0 x 4. [In general, we
1
could use the method given by (5).] Then t x
1 (2 s x ), so t 1
2 , which agrees
with the preceding computation. The functions f and t are graphed in Figure 13. x 2 1
f1 (b) ƒ=≈, 0¯x¯2 EXAMPLE 7 If f x FIGURE 12 2x cos x, ﬁnd f 1 1 .+ SOLUTION Notice that f is onetoone because y
(2, 4) fx 2 sin x 0 f
(4, 2) and so f is increasing. To use Theorem 7 we need to know f
inspection: g
(1, 1)
0 FIGURE 13 f0 1 ? f 1 1 1 1 and we can ﬁnd it by 0 x Therefore f 1 1 1
ff 1 1 1
f0 2 1
sin 0 1
2 5E07(pp 412421) ❙❙❙❙ 420 1/17/06 4:38 PM Page 420 CHAPTER 7 INVERSE FUNCTIONS  7.1 Exercises 1. (a) What is a onetoone function? (b) How can you tell from the graph of a function whether it is
onetoone? ; 17–18 1 2 3 4 5 1.5 2.0 3.6 5.3 2.8 ■ ■ ■ x3 x ■ ■ ■ ■ ■ 9, what 9?
x
x 21. If h x 1 1 cos x, ﬁnd f
sx, ﬁnd h 1. 6. 22. The graph of f is given. (a) Why is f onetoone?
(b) State the domain and range of f
(c) Estimate the value of f 1 1 . 1 . y 6 fx 1 20. If f x A function is given by a table of values, a graph, a formula,
or a verbal description. Determine whether it is onetoone.
x 18. f x
■ x ■ is f  3. ■ 19. If f is a onetoone function such that f 2 range B. How is the inverse function f 1 deﬁned? What is
the domain of f 1? What is the range of f 1?
(b) If you are given a formula for f , how do you ﬁnd a
formula for f 1?
(c) If you are given the graph of f , how do you ﬁnd the graph
of f 1?
3–16 x3 17. f x
■ 2. (a) Suppose f is a onetoone function with domain A and Use a graph to decide whether f is onetoone.  2.0 2
1
1 4. x 1 2 3 4 5 6 fx 1 2 4 8 16 32 _3 _1 0
_1 _2 2 x 3 _2
y 5. y 6. 5
9 23. The formula C
x F 32 , where F
459.67, expresses
the Celsius temperature C as a function of the Fahrenheit
temperature F. Find a formula for the inverse function and
interpret it. What is the domain of the inverse function? x 24. In the theory of relativity, the mass of a particle with speed v is
7. 8. y m y 25–30
9. f x 1
2 10. f x 1 11. t x
12. t x x4 5 14. h x x4 5, 27. f x 5
4x x2 s10 ■ x ■ ■ ■ ■ ■ ■ ■ 30. f x
■ ■ ■ ■ ■ 1
3
3 2x2
■ 8 x,
■ x
■ 2
■  31. f x
■ 2 x3 28. y 3x sx
sx
■ 4x
2x 26. f x 2x Find an explicit formula for f 1 and use it to graph f 1,
f , and the line y x on the same screen. To check your work, see
whether the graphs of f and f 1 are reﬂections about the line. 16. f t is your height at age t.
■ ■ ; 31–32 2 15. f t is the height of a football t seconds after kickoff.
■ 1
1 29. y 0 v2 c2 Find a formula for the inverse of the function.
3 x
13. h x  25. f x sx x s1 where m 0 is the rest mass of the particle and c is the speed of
light in a vacuum. Find the inverse function of f and explain
its meaning. x x m0 fv ■ ■ ■ ■ 1
■ 2 x 2,
■ x
■ 32. f x 0
■ sx 2 ■ ■ ■ 2x,
■ x
■ 0
■ 5E07(pp 412421) 1/17/06 4:39 PM Page 421 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES 33. Use the given graph of f to sketch the graph of f 1 39–42 .  x 40. f x 1 41. f x
x 1 5 x
x sx ■ 2 x,
tan x ■ a 1,
3 2 3 421 a. x 3 42. f x
■ 3 x 39. f x y 1 Find f ❙❙❙❙ 2 1
a 1 x 2, x ■ 2 1, a ■ ■ x 1, a 3 2
■ ■ ■ ■ 43. Suppose t is the inverse function of f and f 4 ■ 5, f 4 ■ 2
3 . Find t 5 .
1 34. Use the given graph of f to sketch the graphs of f and 1 f . 44. Suppose t is the inverse function of a differentiable function f and let G x y
CAS 1 t x . If f 3 2 and f 3 1
9 , ﬁnd G 2 . 45. Use a computer algebra system to ﬁnd an explicit expression for the inverse of the function f x
s x 3 x 2 x 1.
(Your CAS will produce three possible expressions. Explain
why two of them are irrelevant in this context.)
46. Show that h x sin x, x
, is not onetoone, but its
sin x,
2x
2, is onetoone.
restriction f x
Compute the derivative of f 1 sin 1 by the method of
Note 2. 1
x 1 47. (a) If we shift a curve to the left, what happens to its reﬂection 35–38  about the line y x ? In view of this geometric principle,
ﬁnd an expression for the inverse of t x
f x c,
where f is a onetoone function.
(b) Find an expression for the inverse of h x
f c x , where
c 0. (a)
(b)
(c)
(d) Show that f is onetoone.
Use Theorem 7 to ﬁnd t a , where t f 1.
Calculate t x and state the domain and range of t.
Calculate t a from the formula in part (c) and check that it
agrees with the result of part (b).
(e) Sketch the graphs of f and t on the same axes. 35. f x x 3, a 36. f x sx 37. f x 9 38. f x 1x ■ ■ ■  7.2 48. (a) If f is a onetoone, twice differentiable function with inverse function t, show that 8
2, x 2, a 2 0 3, x 1,
■ x
■ f tx
f tx tx
a 1, a
■ 8
2 ■ ■ ■ ■ ■ ■ 3 (b) Deduce that if f is increasing and concave upward, then its
inverse function is concave downward. Exponential Functions and Their Derivatives  If your instructor has assigned Sections 7.2*,
7.3*, and 7.4*, you don’t need to read
Sections 7.2–7.4 (pp. 421–450). The function f x
2 x is called an exponential function because the variable, x, is the
exponent. It should not be confused with the power function t x
x 2, in which the variable is the base.
In general, an exponential function is a function of the form
ax fx where a is a positive constant. Let’s recall what this means.
If x n, a positive integer, then
an aa
n factors a 5E07(pp 422431) 422 ❙❙❙❙ 1/17/06 4:35 PM Page 422 CHAPTER 7 INVERSE FUNCTIONS 0, then a 0 If x 1, and if x n, where n is a positive integer, then
a If x is a rational number, x 1
an n p q, where p and q are integers and q
ax q
(sa ) p q ap q sa p But what is the meaning of a x if x is an irrational number? For instance, what is meant by
2 s3 or 5 ?
To help us answer this question we ﬁrst look at the graph of the function y 2 x, where
x is rational. A representation of this graph is shown in Figure 1. We want to enlarge the
domain of y 2 x to include both rational and irrational numbers.
There are holes in the graph in Figure 1 corresponding to irrational values of x. We want
to ﬁll in the holes by deﬁning f x
, so that f is an increasing function.
2 x, where x
In particular, since the irrational number s3 satisﬁes y 1
1 1.7 x s3 1.8 2 1.7 0 0, then 2 s3 2 1.8 we must have FIGURE 1 Representation of y=2®, x rational and we know what 21.7 and 21.8 mean because 1.7 and 1.8 are rational numbers. Similarly,
if we use better approximations for s3, we obtain better approximations for 2 s3:
1.73 1.74 ? 2 1.73 2 s3 2 1.74 1.732 s3 1.733 ? 2 1.732 2 s3 2 1.733 1.7320 s3 1.7321 ? 2 1.7320 2 s3 2 1.7321 1.73205
.
.
.
 A proof of this fact is given in J. Marsden
and A. Weinstein, Calculus Unlimited (Menlo
Park, CA: Benjamin/Cummings, 1980). s3 s3 1.73206
.
.
. ? 2 1.73205
.
.
. 2 s3 2 1.73206
.
.
. It can be shown that there is exactly one number that is greater than all of the numbers
2 1.7, 2 1.73, 2 1.732, 2 1.7320, 2 1.73205, ... 2 1.733, 2 1.7321, 2 1.73206, ... and less than all of the numbers
2 1.8, 2 1.74, We deﬁne 2 s3 to be this number. Using the preceding approximation process we can compute it correct to six decimal places: y 2 s3 Similarly, we can deﬁne 2 x where x is any irrational number. Figure 2 shows how all the
holes in Figure 1 have been ﬁlled to complete the graph of the function f x
2 x, x
.
In general, if a is any positive number, we deﬁne 1
0 FIGURE 2 y=2®, x real 3.321997 1 x 1 ax lim a r
r lx r rational 5E07(pp 422431) 1/17/06 4:35 PM Page 423 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES ❙❙❙❙ 423 This deﬁnition makes sense because any irrational number can be approximated as
closely as we like by a rational number. For instance, because s3 has the decimal representation s3 1.7320508 . . . , Deﬁnition 1 says that 2 s3 is the limit of the sequence of
numbers
21.7, 21.73, 21.732, 21.7320, 21.73205, 21.732050, 21.7320508, ... 53.1415926, ... Similarly, 5 is the limit of the sequence of numbers
53.1, 53.14, 53.141, 53.1415, 53.14159, 53.141592, It can be shown that Deﬁnition 1 uniquely speciﬁes a x and makes the function f x
ax
continuous.
The graphs of members of the family of functions y a x are shown in Figure 3 for various values of the base a. Notice that all of these graphs pass through the same point 0, 1
because a 0 1 for a 0. Notice also that as the base a gets larger, the exponential function grows more rapidly (for x 0).
y ” ’®
4 ” ’®
2 1 1 10® 4® y 2® y y=2® 1.5® y=2®
200 y=≈ 1® 100 y=≈
10 0 1 0 x FIGURE 3 2 FIGURE 4 Members of the family of
exponential functions 4 x 0 2 4 x 6 FIGURE 5 Figure 4 shows how the exponential function y 2 x compares with the power function
y x 2. The graphs intersect three times, but ultimately the exponential curve y 2 x
grows far more rapidly than the parabola y x 2. (See also Figure 5.)
You can see from Figure 3 that there are basically three kinds of exponential functions
y a x. If 0 a 1, the exponential function decreases; if a 1, it is a constant;
and if a 1, it increases. These three cases are illustrated in Figure 6. Since
1 a x 1 a x a x, the graph of y
1 a x is just the reﬂection of the graph of y a x
about the yaxis.
y y (0, 1) y 1
(0, 1) 0 (a) y=a®, 0<a<1
FIGURE 6 x 0 (b) y=1® x 0 (c) y=a®, a>1 x 5E07(pp 422431) 424 ❙❙❙❙ 1/17/06 4:35 PM Page 424 CHAPTER 7 INVERSE FUNCTIONS The properties of the exponential function are summarized in the following theorem.
0 and a 1, then f x
a x is a continuous function with
x
domain and range 0, . In particular, a
0 for all x. If 0 a 1, f x
ax
is a decreasing function; if a 1, f is an increasing function. If a, b 0 and
x, y
, then
2 Theorem If a 1. a x y a xa y 2. a x ax
ay y 3. a x y a xy 4. ab x a xb x The reason for the importance of the exponential function lies in properties 1–4, which
are called the Laws of Exponents. If x and y are rational numbers, then these laws are well
known from elementary algebra. For arbitrary real numbers x and y these laws can be
deduced from the special case where the exponents are rational by using Equation 1.
The following limits can be read from the graphs shown in Figure 6 or proved from the
deﬁnition of a limit at inﬁnity. (See Exercise 73 in Section 7.3.)
If a a lim a x 1, then If 0 3 lim a x 1, then lim a x and xl 0 xl 0 xl lim a x and xl In particular, if a 1, then the xaxis is a horizontal asymptote of the graph of the exponential function y a x.
EXAMPLE 1 (a) Find lim x l 2 x 1 .
(b) Sketch the graph of the function y 2 x 1. SOLUTION (a) lim 2 xl x 1 lim [( 1 ) x
2 1] xl 0 [by (3) with a 1 1
2 1] 1
(b) We write y ( 1 ) x 1 as in part (a). The graph of y ( 1 ) x is shown in Figure 3, so
2
2
we shift it down one unit to obtain the graph of y ( 1 ) x 1 shown in Figure 7. (For a
2
review of shifting graphs, see Section 1.3.) Part (a) shows that the line y
1 is a horizontal asymptote.
y y=2–® 1 0 y=_1
FIGURE 7 x 5E07(pp 422431) 1/17/06 4:35 PM Page 425 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES ❙❙❙❙ 425 Applications of Exponential Functions
The exponential function occurs very frequently in mathematical models of nature and
society. Here we indicate brieﬂy how it arises in the description of population growth and
radioactive decay. In Chapter 10 we will pursue these and other applications in greater detail.
In Section 3.4 we considered a bacteria population that doubles every hour and saw that
if the initial population is n0 , then the population after t hours is given by the function
ft
n0 2 t. This population function is a constant multiple of the exponential function
t
y 2 , so it exhibits the rapid growth that we observed in Figures 2 and 5. Under ideal
conditions (unlimited space and nutrition and freedom from disease) this exponential
growth is typical of what actually occurs in nature.
What about the human population? Table 1 shows data for the population of the world
in the 20th century and Figure 8 shows the corresponding scatter plot. TABLE 1 Year Population
(millions) 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 P
6x10 ' 1900 1920 1940 1960 1980 2000 t FIGURE 8 Scatter plot for world population growth The pattern of the data points in Figure 8 suggests exponential growth, so we use a
graphing calculator with exponential regression capability to apply the method of least
squares and obtain the exponential model
P 0.008079266 1.013731 t Figure 9 shows the graph of this exponential function together with the original data
points. We see that the exponential curve ﬁts the data reasonably well. The period of relatively slow population growth is explained by the two world wars and the Great Depression
of the 1930s.
P
6x10 ' FIGURE 9 Exponential model for
population growth 1900 1920 1940 1960 1980 2000 t 5E07(pp 422431) 426 ❙❙❙❙ 1/17/06 4:35 PM Page 426 CHAPTER 7 INVERSE FUNCTIONS Exponential functions also occur in the study of the decay of radioactive substances.
For instance, when physicists say the halflife of strontium90, 90Sr, is 25 years they mean
that half of any given quantity of 90Sr will disintegrate in 25 years. So if the initial mass of
a sample of 90Sr is 24 mg and the mass that remains after t years is m t , then
m 25 1
24
2 m 50 1
24
22 m 75 1
24
23 m 100 1
24
24 From this pattern, we see that the mass remaining after t years is
1 mt 4 24 t 25 2 t 25 24 2 This is an exponential function with base a 1 25 2 . (See Exercise 55 and Section 10.4.) Derivatives of Exponential Functions
a x using the deﬁni Let’s try to compute the derivative of the exponential function f x
tion of a derivative:
fx lim fx h
h hl0
xh lim aa a fx ax h x lim aa h ax
h hl0
x h hl0 lim 1 h hl0 The factor a x doesn’t depend on h, so we can take it in front of the limit:
a x lim fx ah hl0 1
h Notice that the limit is the value of the derivative of f at 0, that is,
lim ah hl0 1
h f0 Therefore, we have shown that if the exponential function f x
then it is differentiable everywhere and
fx 5 h
0.1
0.01
0.001
0.0001 2h 1 3h 1 h h 0.7177
0.6956
0.6934
0.6932 1.1612
1.1047
1.0992
1.0987 a x is differentiable at 0, f 0 ax This equation says that the rate of change of any exponential function is proportional to
the function itself. (The slope is proportional to the height.)
Numerical evidence for the existence of f 0 is given in the table at the left for the
cases a 2 and a 3. (Values are stated correct to four decimal places. For the case
a 2, see also Example 3 in Section 3.1.) It appears that the limits exist and
for a 2, f0 for a 3, f0 lim 2h lim hl0 1
h hl0 3h 1
h 0.69 1.10 5E07(pp 422431) 1/17/06 4:35 PM Page 427 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES ❙❙❙❙ 427 In fact, it can be proved that these limits exist and, correct to six decimal places, the values
are
6 d
2x
dx d
3x
dx 0.693147
x0 1.098612
x0 Thus, from Equation 5 we have
7 d
2x
dx d
3x
dx 0.69 2 x 1.10 3 x Of all possible choices for the base a in Equation 5, the simplest differentiation formula
occurs when f 0
1. In view of the estimates of f 0 for a 2 and a 3, it seems reasonable that there is a number a between 2 and 3 for which f 0
1. It is traditional to
denote this value by the letter e. Thus, we have the following deﬁnition.
y y=3® 8
y=2® Definition of the Number e e is the number such that lim hl0 eh 1
h 1 y=e®
1 x 0 FIGURE 10
y Geometrically, this means that of all the possible exponential functions y a x, the
function f x
e x is the one whose tangent line at (0, 1 has a slope f 0 that is exactly 1
(see Figures 10 and 11).
If we put a e and, therefore, f 0
1 in Equation 5, it becomes the following
important differentiation formula. 9 Derivative of the Natural Exponential Function slope=e® d
ex
dx {x, e ®} 1 Thus, the exponential function f x
e x has the property that it is its own derivative. The
geometrical signiﬁcance of this fact is that the slope of a tangent line to the curve y e x
at any point is equal to the ycoordinate of the point (see Figure 11). slope=1 y=e®
0 x EXAMPLE 2 Differentiate the function y
FIGURE 11 ex e tan x. SOLUTION To use the Chain Rule, we let u dy
dx dy du
du dx tan x. Then we have y
eu du
dx e u, so e tan x sec2x In general if we combine Formula 9 with the Chain Rule, as in Example 2, we get 10 d
eu
dx eu du
dx 5E07(pp 422431) 428 ❙❙❙❙ 1/17/06 4:35 PM Page 428 CHAPTER 7 INVERSE FUNCTIONS EXAMPLE 3 Find y if y e 4x sin 5x. SOLUTION Using Formula 10 and the Product Rule, we have y e 4x cos 5x 5 sin 5x e 4x 4 e 4x 5 cos 5x 4 sin 5x We have seen that e is a number that lies somewhere between 2 and 3, but we can
use Equation 5 to estimate the numerical value of e more accurately. Let e 2 c. Then
e x 2 cx. If f x
2 x, then from Equation 5 we have f x
k 2 x, where the value of k is
f0
0.693147. Thus, by the Chain Rule,
d
ex
dx ex
Putting x 0, we have 1 d
2 cx
dx ck, so c
e 21 k k 2 cx d
cx
dx ck 2 cx 1 k and
2 1 0.693147 2.71828 It can be shown that the approximate value to 20 decimal places is
e 2.71828182845904523536 The decimal expansion of e is nonrepeating because e is an irrational number.
EXAMPLE 4 In Example 6 in Section 3.4 we considered a population of bacteria cells in a
homogeneous nutrient medium. We showed that if the population doubles every hour,
then the population after t hours is n0 2 t n where n0 is the initial population. Now we can use (5) and (6) to compute the growth
rate:
dn
dt  The rate of growth is proportional to the size
of the population. n0 0.693147 2 t For instance, if the initial population is n0
hours is
dn
dt 1000 cells, then the growth rate after two 1000 0.693147 2 t t2 t2 4000 0.693147 2773 cells h EXAMPLE 5 Find the absolute maximum value of the function f x xe x. SOLUTION We differentiate to ﬁnd any critical numbers: fx xe x 1 e x 1 e x 1 x Since exponential functions are always positive, we see that f x
0 when 1 x 0,
that is, when x 1. Similarly, f x
0 when x 1. By the First Derivative Test for
Absolute Extreme Values, f has an absolute maximum value when x 1 and the value is
f1 1e 1 1
e 0.37 5E07(pp 422431) 1/17/06 4:36 PM Page 429 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES ❙❙❙❙ 429 Exponential Graphs
y The exponential function f x
e x is one of the most frequently occurring functions in
calculus and its applications, so it is important to be familiar with its graph (Figure 12) and
properties. We summarize these properties as follows, using the fact that this function
is just a special case of the exponential functions considered in Theorem 2 but with base
a e 1. y=´ 1 11
0 1 FIGURE 12 x ex Properties of the Natural Exponential Function The exponential function f x is an increasing continuous function with domain and range 0,
for all x. Also
lim e x 0
lim e x
xl . Thus, e x 0 xl The natural exponential function e x. So the xaxis is a horizontal asymptote of f x EXAMPLE 6 Find lim
xl e 2x
e 2x 1 . SOLUTION We divide numerator and denominator by e 2 x : lim xl e 2x
e 2x lim 1 xl 1 1
e 1
1
We have used the fact that t lim e as x l
2x xl 1
lim e 1 2x xl 1 0 2x l 2x lim e and so t 0 tl e 1 x, together with asymptotes, EXAMPLE 7 Use the ﬁrst and second derivatives of f x to sketch its graph.
SOLUTION Notice that the domain of f is x x
0 , so we check for vertical asymptotes
by computing the left and right limits as x l 0. As x l 0 , we know that t 1 x l ,
so
lim e 1 x lim e t
xl0 and this shows that x
so tl 0 is a vertical asymptote. As x l 0 , we have t
lim e 1 x xl0 As x l lim e t 0 tl , we have 1 x l 0 and so
lim e 1 x xl e0 1 This shows that y 1 is a horizontal asymptote.
Now let’s compute the derivative. The Chain Rule gives
fx e1 x
x2 1 xl , 5E07(pp 422431) 430 ❙❙❙❙ 1/17/06 4:36 PM Page 430 CHAPTER 7 INVERSE FUNCTIONS 0 for all x 0. Thus, f is
Since e 1 x 0 and x 2 0 for all x 0, we have f x
, 0 and on 0, . There is no critical number, so the function has no
decreasing on
maximum or minimum. The second derivative is
x 2e 1 x fx 1 x2
x4 e 1 x 2x e 1 x 2x
x4 1 1
0 when x
0 and f x
0
Since e 1 x 0 and x 4 0, we have f x
2x
1
1
when x
. So the curve is concave downward on ( , 2 ) and concave upward on
2
( 1 , 0) and on 0, . The inﬂection point is ( 1 , e 2).
2
2
To sketch the graph of f we ﬁrst draw the horizontal asymptote y 1 (as a dashed
line), together with the parts of the curve near the asymptotes in a preliminary sketch
[Figure 13(a)]. These parts reﬂect the information concerning limits and the fact that f
, 0 and 0, . Notice that we have indicated that f x l 0 as
is decreasing on both
x l 0 even though f 0 does not exist. In Figure 13(b) we ﬁnish the sketch by incorporating the information concerning concavity and the inﬂection point. In Figure 13(c) we
check our work with a graphing device. y y
4 y=‰ inﬂection
point
y=1
0 FIGURE 13 y=1
0 x (a) Preliminary sketch _3 x 3
0 (b) Finished sketch (c) Computer confirmation Integration
e x has a simple derivative, its integral is also simple: Because the exponential function y ye 12 EXAMPLE 8 Evaluate 2 x3 yx e x 2 x3 yx e C dx.
x 3. Then du SOLUTION We substitute u ex dx 1
3 dx ye u 1
3 3x 2 dx, so x 2 dx
1
3 du EXAMPLE 9 Find the area under the curve y e eu
3x 1
3 C ex du and 3 C from 0 to l. SOLUTION The area is A y 1 0 e 3x dx 1
3 e 3x 1
0 1
3 1 e 3 5E07(pp 422431) 1/17/06 4:36 PM Page 431 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES  7.2 431 Exercises 1. (a) Write an equation that deﬁnes the exponential function with 17–18 base a 0.
(b) What is the domain of this function?
(c) If a 1, what is the range of this function?
(d) Sketch the general shape of the graph of the exponential
function for each of the following cases.
(i) a 1
(ii) a 1
(iii) 0 a 1 Ca x whose graph is given. Find the exponential function f x  17. 2
(1, 6)  Graph the given functions on a common screen. How are
these graphs related? y e x, 4. y e x, y e x, y 8 x, 5. y 3 x, y 10 x, y ( 1 ) x,
3 6. y 0.9 x, ■ ■ 0.6 x, y ■ 5 x, y ■ ■ y ■ ■ ■ ■ ■ ■ 7–12  Make a rough sketch of the graph of the function. Do not
use a calculator. Just use the graphs given in Figures 3 and 12 and,
if necessary, the transformations of Section 1.3. 4x 7. y
9. y ■ 10. y 1 12. y 2 51 ■ ■ ■ ■ 13. Starting with the graph of y ■ ■ 3 ■ e
■ graph that results from
(a) reﬂecting about the line y
(b) reﬂecting about the line x
15–16  ■ 15. (a) f x
16. (a) t t
■ sin e
■ ■ t
■ ex ■ ■ ■ ■ 1,000,000,000.
 Find the limit. 25. lim
xl e 3x
e 3x ■ ■ 1 ■ ■ ■ ■ x 2e x
e ax 3 ■ ■ ■ ■ 1u e t sin 2t
s1 30. y
32. y e 37. y 2x ■ ■ ■ Differentiate the function. 35. F t 2t 2 x l2 33. f u ex lim e tan x xl 28. lim e 3 ■ 31. y ■ 26. 2x ■  x2 3x
3x x l2 29. f x s1 24. lim e
xl e
e 27. lim e 3 29–42 1 x xl 4
2 ■ ■ x 10 and t x
e x by graphing
both f and t in several viewing rectangles. When does the
graph of t ﬁnally surpass the graph of f ? 23. lim 1.001 e x, ﬁnd the equation of the (b) t t
■ ■ ; 21. Compare the functions f x 23–28 e x, write the equation of the (b) f x ex 1 ■ x 5 and
tx
5 x by graphing both functions in several viewing rectangles. Find all points of intersection of the graphs correct to
one decimal place. ■ Find the domain of each function.
1 ■ ; 22. Use a graph to estimate the values of x such that x graph that results from
(a) shifting 2 units downward
(b) shifting 2 units to the right
(c) reﬂecting about the xaxis
(d) reﬂecting about the yaxis
(e) reﬂecting about the xaxis and then about the yaxis
14. Starting with the graph of y ■ ; 20. Compare the rates of growth of the functions f x 2e x ex 3
■ 4x x 2 11. y 8. y 3 ■ 0.1x y ■ ■ x 0 2 x are drawn on a
x 2 and t x
coordinate grid where the unit of measurement is 1 inch. Show
that, at a distance 2 ft to the right of the origin, the height of
the graph of f is 48 ft but the height of the graph of t is about
265 mi. 1
(10 ) x y x 19. Suppose the graphs of f x
x 8 0.3 x, y ■ 20 x y 2 ”2, 9 ’ 0 ; 3–6 2 x, y (3, 24) (b) What is an approximate value for e ?
(c) What is the natural exponential function? 3. y 18. y 2. (a) How is the number e deﬁned? ■ ❙❙❙❙ 2e3x 34. t x ex
1 x e u cos u
sx e 36. y e k tan sx 38. y cos e x x cu ■ 5E07(pp 432441) ❙❙❙❙ 432 1/17/06 Page 432 CHAPTER 7 INVERSE FUNCTIONS
x 39. y ee 41. y ae x
ce x ■ 4:29 PM ■ 40. y ■ ■ ■ ■ xe 42. y b
d s1
ex
ex e
e ■ ■ ■ 57. Under certain circumstances a rumor spreads according to the 2x equation x pt x ■ ■  Find an equation of the tangent line to the curve at the
given point. e 2 x cos x, ■ ■ ■ ■ 45. Find y if e x 2 ■ y e x x, 44. y 0, 1
■ ■ ■ ■ 1, e
■ ■ 46. Find an equation of the tangent line to the curve x e y ye x equation 2 y y ex
0. y Ae x
y 0. 48. Show that the function y tial equation y x/2 e 1 ; 58. An object is attached to the end of a vibrating spring and 2y satisﬁes the differential Bxe 49. For what values of r does the function y tion y 6y 8y 50. Find the values of y y 51. If f x x satisﬁes the differene rx satisfy the equa 0?
for which y x e satisﬁes the equation y.
e 2 x, ﬁnd a formula for f n x. 52. Find the thousandth derivative of f x where p t is the proportion of the population that knows the
rumor at time t and a and k are positive constants. [In Section 10.5 we will see that this is a reasonable model for p t .]
(a) Find lim t l p t .
(b) Find the rate of spread of the rumor.
(c) Graph p for the case a 10, k 0.5 with t measured in
hours. Use the graph to estimate how long it will take for
80% of the population to hear the rumor. ; at the point 0, 1 .
47. Show that the function y kt ■ y. x 1
ae ■ 43–44 43. y 1 xe x. 53. (a) Use the Intermediate Value Theorem to show that there is a root of the equation e x x 0.
(b) Use Newton’s method to ﬁnd the root of the equation in
part (a) correct to six decimal places. ; 54. Use a graph to ﬁnd an initial approximation (to one decimal
2 place) to the root of the equation e x
x 3 x 3. Then use
Newton’s method to ﬁnd the root correct to six decimal places. its displacement from its equilibrium position is
y 8e t 2 sin 4 t, where t is measured in seconds and y is
measured in centimeters.
(a) Graph the displacement function together with the
functions y 8e t 2 and y
8e t 2. How are these
graphs related? Can you explain why?
(b) Use the graph to estimate the maximum value of the
displacement. Does it occur when the graph touches the
graph of y 8e t 2 ?
(c) What is the velocity of the object when it ﬁrst returns to its
equilibrium position?
(d) Use the graph to estimate the time after which the displacement is no more than 2 cm from equilibrium. ; 59. The ﬂash unit on a camera operates by storing charge on a
capacitor and releasing it suddenly when the ﬂash is set off.
The following data describe the charge Q remaining on the
capacitor (measured in microcoulombs, C) at time t (measured in seconds).
t 0.00 0.02 0.04 0.06 0.08 0.10 Q 100.00 81.87 67.03 54.88 44.93 36.76 55. If the initial mass of a sample of 90Sr is 24 mg, then the mass after t years, from Equation 4, is
mt ; 24 2 t 25 (a) Find the mass remaining after 40 years.
(b) Use Equation 7 and the Chain Rule to estimate the rate at
which the mass decays after 40 years.
(c) Use a graphing device to estimate the time required for the
mass to be reduced to 5 mg. (a) Use a graphing calculator or computer to ﬁnd an exponential model for the charge. (See Section 1.2.)
(b) The derivative Q t represents the electric current
(measured in microamperes, A) ﬂowing from the capacitor to the ﬂash bulb. Use part (a) to estimate the current
when t 0.04 s. Compare with the result of Example 2 in
Section 2.1. ; 60. The table gives the U.S. population from 1790 to 1860. ; 56. For the period from 1980 to 2000, the percentage of households in the United States with at least one VCR has been
modeled by the function
Vt 1 85
53 e 0.5 t where the time t is measured in years since midyear 1980, so
0 t 20. Use a graph to estimate the time at which the
number of VCRs was increasing most rapidly. Then use derivatives to give a more accurate estimate. Year Population Year Population 1790
1800
1810
1820 3,929,000
5,308,000
7,240,000
9,639,000 1830
1840
1850
1860 12,861,000
17,063,000
23,192,000
31,443,000 (a) Use a graphing calculator or computer to ﬁt an exponential
function to the data. Graph the data points and the exponential model. How good is the ﬁt? 5E07(pp 432441) 1/17/06 4:29 PM Page 433 S ECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES (b) Estimate the rates of population growth in 1800 and 1850
by averaging slopes of secant lines.
(c) Use the exponential model in part (a) to estimate the rates
of growth in 1800 and 1850. Compare these estimates with
the ones in part (b).
(d) Use the exponential model to predict the population in
1870. Compare with the actual population of 38,558,000.
Can you explain the discrepancy? 71–78
5 fx y 73. y e s1 75. y 77. e x. x y 62. Find the absolute minimum value of the function e x x, x tx e ■ 0. Find (a) the intervals of increase or decrease, (b) the
intervals of concavity, and (c) the points of inﬂection.
 xe x 63. f x
■ 65–67 ■ e 67. y e
■ 64. f x
■ ■ ■ ■ ■ ■ dx 76. y ex 78. ye esx
dx
sx
■ ■ ■ ■ xe 0 dx 2 ■ tan x dx e1 x
dx
x2
x sin e x d x ■ ■ ■ ■ ■ 1x1
3x e
■ e2 x 66. y ■ ■ ■ ■ ■ ■ 68. f x
■ e cos x ■ ■ 69. f x
■ ■ ■ ■ ex
■ ■ ■ ■ x
■ ■ ■ 1
s2 e 1, and f 0
e x, y xaxis the region bounded by the curves y
and x 1. 83. If f x 3 84. Evaluate lim
xl 2. 0, x 0, x
e e x, ﬁnd f sin x 1 x 1 x2 ,y e 0, 4. . 85. (a) Show that e x 1 x if x 0.
[Hint: Show that f x
1
ex
for x 0.]
2
(b) Deduce that 4 x01 e x dx e.
3
x x 2 2 x is increasing 0,
ex 2 1 x 1
2 x2
2 occurs in probability and statistics, where it is called the
normal density function. The constant is called the mean and
the positive constant is called the standard deviation. For
simplicity, let’s scale the function so as to remove the factor
1 ( s2 ) and let’s analyze the special case where
0. So
we study the function
fx 5 sin x, f 0 1. 86. (a) Use the inequality of Exercise 85(a) to show that, for 70. The family of bellshaped curves y 3e yaxis the region bounded by the curves y
x 0, and x 1.
■ 3 x x e 3x, and x 82. Find the volume of the solid obtained by rotating about the ex 2x
■ e x, y 81. Find the volume of the solid obtained by rotating about the
■  Draw a graph of f that shows all the important aspects of
the curve. Estimate the local maximum and minimum values and
then use calculus to ﬁnd these values exactly. Use a graph of f to
estimate the inﬂection points. ; y sec x e 80. Find f x if f x 2e x ; 68–69 ■ 74. bounded by the curves y Discuss the curve using the guidelines of Section 4.5.  65. y ■ e x dx
1 ■ x2 y x ex 1 72. dx 79. Find, correct to three decimal places, the area of the region 63–64 ■ 3x 71. 61. Find the absolute maximum value of the function 433 Evaluate the integral.  0 ❙❙❙❙ e x2 2 (b) Use part (a) to improve the estimate of x01 e x dx given in
Exercise 85(b).
87. (a) Use mathematical induction to prove that for x positive integer n,
ex 1 x2
2! x xn
n! 2 (a) Find the asymptote, maximum value, and inﬂection points
of f .
(b) What role does play in the shape of the curve?
(c) Illustrate by graphing four members of this family on the
same screen. (b) Use part (a) to show that e
(c) Use part (a) to show that
lim xl ex
xk for any positive integer k. 2.7. 0 and any 5E07(pp 432441) 434 ❙❙❙❙ 1/17/06 4:29 PM Page 434 CHAPTER 7 INVERSE FUNCTIONS  7.3 Logarithmic Functions
If a 0 and a 1, the exponential function f x
a x is either increasing or decreasing
and so it is onetoone. It therefore has an inverse function f 1, which is called the logarithmic function with base a and is denoted by log a . If we use the formulation of an
inverse function given by (7.1.3),
f 1 x y &? fy x then we have
1 Thus, if x log a x y &? ay x 0, log a x is the exponent to which the base a must be raised to give x. EXAMPLE 1 Evaluate (a) log 3 81, (b) log 25 5, and (c) log 10 0.001.
SOLUTION (a) log 3 81 4 because 3 4 81
(b) log 25 5 1 because 25 1 2 5
2
(c) log 10 0.001
3 because 10 y 2
y=a®, a>1 y=log£ x
1 x y=log∞ x
y=log¡¸ x FIGURE 2 for every x x log a x, for every x x 1 0 The logarithmic function log a has domain 0,
and range and is continuous since it
is the inverse of a continuous function, namely, the exponential function. Its graph is the
reﬂection of the graph of y a x about the line y x.
Figure 1 shows the case where a 1. (The most important logarithmic functions have
base a 1.) The fact that y a x is a very rapidly increasing function for x 0 is
reﬂected in the fact that y log a x is a very slowly increasing function for x 1.
Figure 2 shows the graphs of y log a x with various values of the base a. Since
log a 1 0, the graphs of all logarithmic functions pass through the point 1, 0 .
The following theorem summarizes the properties of logarithmic functions. y=log™ x 1 x a x and f x FIGURE 1 0 log a a x
a log a x y=log a x, a>1 y 0.001 The cancellation equations (7.1.4), when applied to f x
become y=x 0 3 1, the function f x
log a x is a onetoone, continuous,
3 Theorem If a
increasing function with domain 0,
and range . If x, y 0 and r is any real
number, then
x
1. log a xy
2. log a
log a x log a y
log a x log a y
y
r log a x
3. log a x r 5E07(pp 432441) 1/17/06 4:29 PM Page 435 S ECTION 7.3 LOGARITHMIC FUNCTIONS ❙❙❙❙ 435 Properties 1, 2, and 3 follow from the corresponding properties of exponential functions
given in Section 7.2.
EXAMPLE 2 Use the properties of logarithms in Theorem 3 to evaluate the following. (a) log 4 2 log 4 32 (b) log 2 80 log 2 5 SOLUTION (a) Using Property 1 in Theorem 3, we have
log 4 2 log 4 32 log 4 2 32 log 4 64 3 since 43 64.
(b) Using Property 2 we have
log 2 80
since 2 4 log 2( 80 )
5 log 2 5 log2 16 4 16. The limits of exponential functions given in Section 7.2 are reﬂected in the following
limits of logarithmic functions. (Compare with Figure 1.)
4 If a 1, then
lim log a x and xl lim log a x xl0 In particular, the yaxis is a vertical asymptote of the curve y log a x. EXAMPLE 3 Find lim log10 tan2x .
xl0 tan2x l tan2 0
1, we have SOLUTION As x l 0, we know that t tive. So by (4) with a 10 lim log10 tan2x 0 and the values of t are posi lim log10 t xl0 tl0 Natural Logarithms
 NOTATION FOR LOGARITHMS
Most textbooks in calculus and the sciences,
as well as calculators, use the notation ln x for
the natural logarithm and log x for the “common logarithm,” log 10 x. In the more advanced
mathematical and scientiﬁc literature and in
computer languages, however, the notation
log x usually denotes the natural logarithm. Of all possible bases a for logarithms, we will see in the next section that the most convenient choice of a base is the number e, which was deﬁned in Section 7.2. The logarithm
with base e is called the natural logarithm and has a special notation:
log e x ln x If we put a e and replace log e with “ln” in (1) and (2), then the deﬁning properties
of the natural logarithm function become
5 6 y &? e y ln x ln e x
e ln x x x x x x 0 5E07(pp 432441) 436 ❙❙❙❙ 1/17/06 4:29 PM Page 436 CHAPTER 7 INVERSE FUNCTIONS In particular, if we set x 1, we get
ln e EXAMPLE 4 Find x if ln x 1 5. SOLUTION 1 From (5) we see that ln x 5 e5 means x 5 Therefore, x e .
(If you have trouble working with the “ln” notation, just replace it by log e. Then the
equation becomes log e x 5; so, by the deﬁnition of logarithm, e 5 x.)
SOLUTION 2 Start with the equation ln x 5 and apply the exponential function to both sides of the equation:
e ln x e5 But the second cancellation equation in (6) says that e ln x
EXAMPLE 5 Solve the equation e 5 3x x. Therefore, x e 5. 10. SOLUTION We take natural logarithms of both sides of the equation and use (6): ln e 5 3x 3x ln 10 3x 5 ln 10 5 x 1
3 ln 10
5 ln 10 Since the natural logarithm is found on scientiﬁc calculators, we can approximate the
solution to four decimal places: x 0.8991.
EXAMPLE 6 Express ln a 1
2 ln b as a single logarithm. SOLUTION Using Properties 3 and 1 of logarithms, we have
1
2 ln b ln a ln b 1 ln a ln a 2 ln sb ln(a sb )
The following formula shows that logarithms with any base can be expressed in terms
of the natural logarithm.
7 Change of Base Formula For any positive number a a log a x ln x
ln a 1 , we have 5E07(pp 432441) 1/17/06 4:30 PM Page 437 S ECTION 7.3 LOGARITHMIC FUNCTIONS ❙❙❙❙ 437 log a x. Then, from (1), we have a y x. Taking natural logarithms of both
sides of this equation, we get y ln a ln x. Therefore Proof Let y ln x
ln a y Scientiﬁc calculators have a key for natural logarithms, so Formula 7 enables us to use
a calculator to compute a logarithm with any base (as shown in the next example). Similarly, Formula 7 allows us to graph any logarithmic function on a graphing calculator or
computer (see Exercises 20–22).
EXAMPLE 7 Evaluate log 8 5 correct to six decimal places.
SOLUTION Formula 7 gives log 8 5 ln 5
ln 8 0.773976 EXAMPLE 8 In Section 7.2 we showed that the mass of sample after t years is m
interpret it. ft t 25 24 2 SOLUTION We need to solve the equation m 90 Sr that remains from a 24mg
. Find the inverse of this function and
24 2 t 25 for t. We start by taking natural logarithms of both sides:
t 25 ln m ln 24 2 ln m ln 24 t
ln 2
25 ln 24 ln m 25
ln 24
ln 2 ln 2 t 25 t
ln 2
25 t ln 24 ln m So the inverse function is
f 1 m 25
ln 24
ln 2 ln m This function gives the time required for the mass to decay to m milligrams. In particular, the time required for the mass to be reduced to 5 mg is
t y f 1 5 25
ln 24
ln 2 ln 5 56.58 years y=´
y=x 1 y=ln x 0
1 x The graphs of the exponential function y e x and its inverse function, the natural logarithm function, are shown in Figure 3. Because the curve y e x crosses the yaxis with
a slope of 1, it follows that the reﬂected curve y ln x crosses the xaxis with a slope of 1.
In common with all other logarithmic functions with base greater than 1, the natural
logarithm is a continuous, increasing function deﬁned on 0, and the yaxis is a vertical
asymptote.
If we put a e in (4), then we have the following limits:
8 FIGURE 3 lim ln x xl lim ln x xl0 5E07(pp 432441) ❙❙❙❙ 438 1/17/06 4:30 PM Page 438 CHAPTER 7 INVERSE FUNCTIONS EXAMPLE 9 Sketch the graph of the function y ln x 1. 2 SOLUTION We start with the graph of y
ln x as given in Figure 3. Using the transformations of Section 1.3, we shift it 2 units to the right to get the graph of y ln x 2 and
then we shift it 1 unit downward to get the graph of y ln x 2
1. (See Figure 4.)
Notice that the line x 2 is a vertical asymptote since lim ln x 2 x l2 y y 1
y x=2 x=2 y=ln x y=ln(x2)1 y=ln(x2)
0 (1, 0) 0 x 2 0 x (3, 0) x 2
(3, _1) FIGURE 4 We have seen that ln x l as x l . But this happens very slowly. In fact, ln x grows
more slowly than any positive power of x. To illustrate this fact, we compare approximate
values of the functions y ln x and y x 1 2 sx in the following table and we graph
them in Figures 5 and 6.
x 1 2 5 10 50 100 500 1000 10,000 100,000 ln x 0 0.69 1.61 2.30 3.91 4.6 6.2 6.9 9.2 11.5 sx 1 1.41 2.24 3.16 7.07 10.0 22.4 31.6 100 316 ln x
sx 0 0.49 0.72 0.73 0.55 0.46 0.28 0.22 0.09 0.04 y y x
y=œ„
20 x
y=œ„
1 0 FIGURE 5 y=ln x y=ln x
1 x 0 1000 x FIGURE 6 You can see that initially the graphs of y s x and y ln x grow at comparable rates,
but eventually the root function far surpasses the logarithm. In fact, we will be able to show
in Section 7.7 that
ln x
lim p
0
xl
x
for any positive power p. So for large x, the values of ln x are very small compared with
x p. (See Exercise 74.) 5E07(pp 432441) 1/17/06 4:30 PM Page 439 SECTION 7.3 LOGARITHMIC FUNCTIONS  7.3 25. y log a x deﬁned?
(b) What is the domain of this function?
(c) What is the range of this function?
(d) Sketch the general shape of the graph of the function
y log a x if a 1. ln x ■ 29–38 28. y 2 ■ ■ ■ ■ 29. (a) 2 ln x 31. (a) 5 10 (b) log10 x 3x 1 k (b) log 2 m x 35. 2 ln x 1 (b) e ln 15 6. (a) log 10 0.1 (b) log 8 320 (b) ln 5 0 ln 2 5. (a) log 5 25 8. (a) 2
■ ■ 9–12 ■  (b) log 2 5
(b) e ■ ■ 37. e ■ log 2 90 2 log 2 3 ■ ■ ■ ■ ■ ■ xy
z2 10. ln sa b 10 3x 2
12. ln
x1 ■ ■ ■ ■ ■ ■ 2 14. ln x log10 b
y 15. 2 ln 4
17. ln x ■ ■ ■ c ■ ■ ■ ■ 10 ■ ■ 1
e 2 2x 12 ■ ■ ln x
■ ■ 40. ln(1 2 ■ ■ sx ) 1x4 42. 3 3
■ ■ ■ ■ ■ 2 7
■ ■ ■ ■ ■ ■ ■ Solve each inequality for x. 
x 44. (a) 2
■ 10 (b) ln x ln x
■ 9
■ (b) e
■ ■ ■ 1 2 3x
■ 4
■ y 2 ln z
16. ln 3 5 ln x
■ 1 ■ ■ 1
3 18. ln x ln 2
2 log 2 x is drawn on a coordinate
grid where the unit of measurement is an inch. How many
miles to the right of the origin do we have to move before the
height of the curve reaches 3 ft? a ln y ■ ■ ■ ln 8 ■ 46. The velocity of a particle that moves in a straight line under the
inﬂuence of viscous forces is v t
ce k t, where c and k are b ln z
■ ■ ■ (b) log 6 13.54 (c) log 2 20. y log 2 x, 21. y log 1.5 x,
ln x,
■ log 4 x, y ln x, y log 10 x, y
■ ■ y
y
y
■ log 6 x, y log 10 x, y x e, y ■ 10
■ log 8 x
log 50 x
x
■ ■ ■ ■ 23–28  Make a rough sketch of the graph of each function. Do
not use a calculator. Just use the graphs given in Figures 1, 2, and 3
and, if necessary, the transformations of Section 1.3. log 10 x 5 24. y log 2 x 3 positive constants.
(a) Show that the acceleration is proportional to the velocity.
(b) Explain the signiﬁcance of the number c.
(c) At what time is the velocity equal to half the initial velocity?
47. The geologist C. F. Richter deﬁned the magnitude of an  Use Formula 7 to graph the given functions on a common screen. How are these graphs related? 23. y ■ 4
c 45. Suppose that the graph of y ; 20–22 ■ 38. 7e b 100 x 43. (a) e 5 log10 c ln x decimal places.
(a) log 12 e ■ ■ 1 36. ln 2 x
x 3 Find the solution of the equation correct to four decimal 5x 43–44 19. Use Formula 7 to evaluate each logarithm correct to six 22. y ■ 4 x 2x Express the quantity as a single logarithm.  13. log10 a 1
2 ■ 2 ■ 13–18 ■  39. e 2 ■ 3 11. ln uv ln 3 x Ce , where a 41. ln e Use the properties of logarithms to expand the quantity. 9. log 2 34. e e bx ■ 39–42 ■ places. 3 ln 2 ■ ax ■ log 8 5 ■ 5 7 (b) log 3 81 log 12 48 x x3 30. (a) e 100 log 2 5 ■ 2x 3 1 log 2 3 ■ (b) e 1 33. ln ln x 7. (a) log 12 3 ln x ■ Solve each equation for x.  (b) log16 4 4. (a) ln e ln 10 x 32. (a) e Find the exact value of each expression. 3. (a) log10 1000 5 ■ (b) What is the common logarithm?
(c) Sketch the graphs of the natural logarithm function and the
natural exponential function with a common set of axes. 26. y ln x 27. y 2. (a) What is the natural logarithm?  439 Exercises 1. (a) How is the logarithmic function y 3–8 ❙ ❙❙❙ earthquake to be log10 I S , where I is the intensity of the
quake (measured by the amplitude of a seismograph 100 km
from the epicenter) and S is the intensity of a “standard”
earthquake (where the amplitude is only 1 micron 10 4 cm).
The 1989 Loma Prieta earthquake that shook San Francisco
had a magnitude of 7.1 on the Richter scale. The 1906 San
Francisco earthquake was 16 times as intense. What was its
magnitude on the Richter scale?
48. A sound so faint that it can just be heard has intensity I0 10 12 watt m2 at a frequency of 1000 hertz (Hz). The
loudness, in decibels (dB), of a sound with intensity I is then 5E07(pp 432441) ❙❙❙❙ 440 1/17/06 4:30 PM Page 440 CHAPTER 7 INVERSE FUNCTIONS deﬁned to be L 10 log10 I I0 . Ampliﬁed rock music is measured at 120 dB, whereas the noise from a motordriven lawn
mower is measured at 106 dB. Find the ratio of the intensity of
the rock music to that of the mower.
49. If a bacteria population starts with 100 bacteria and doubles  1. Prove, using Deﬁnitions 4.4.6 and 4.4.7, that
(a) lim a x 0
(b) lim a x x 56. lim ln 2 2 x ■ 57–58 ■  57. f x
■ ■ 59–60 59. f x
■ ■ 61–66 x 0.1 and
tx
ln x by graphing both f and t in several viewing
rectangles. When does the graph of f ﬁnally surpass the
graph of t ?
(b) Graph the function h x
ln x x 0.1 in a viewing rectangle
that displays the behavior of the function as x l .
(c) Find a number N such that 61. y ln 1
■ log 2 5x x
■ ■ ■ ■ ■ ■ s3 e
■ 2x ■ ln e t ■ ■ ■ 1 ■ ■ ■ ■ ■ ■ 2
■ and its domain. 60. f x
■ ln 2 ■ ■ ln x
■ ■ ■ Find the inverse function.
2 10 62. y 3
3 ■ ■ ■ ■ ■ 67. On what interval is the function f x
68. On what interval is the curve y 2e x x N 2x 2 0. lim nl ■ e3x ■ e x increasing?
3x concave downward?
69. (a) Show that the function f x function.
(b) Find the inverse function of f . ln( x sx 2 than 1 and itself. The ﬁrst few primes are 2, 3, 5, 7, 11, 13,
17, . . . . We denote by n the number of primes that are less
than or equal to n. For instance, 15
6 because there are
six primes smaller than 15.
(a) Calculate the numbers 25 and 100 .
[Hint: To ﬁnd 100 , ﬁrst compile a list of the primes up
to 100 using the sieve of Eratosthenes: Write the numbers
from 2 to 100 and cross out all multiples of 2. Then cross
out all multiples of 3. The next remaining number is 5, so
cross out all remaining multiples of it, and so on.]
(b) By inspecting tables of prime numbers and tables of logarithms, the great mathematician K. F. Gauss made the guess
in 1792 (when he was 15) that the number of primes up to
n is approximately n ln n when n is large. More precisely,
he conjectured that 1 e
ex
■ e x x 1
1 66. y
■ x ln x 2, 64. y 10
10 x 1
■ ■ 58. G t x 65. y whenever 76. A prime number is a positive integer that has no factors other Find (a) the domain of f and (b) f ex 0.1 75. Solve the inequality ln x 2 x 3 ■ ln x 63. f x ln x
x 0.1 6 Find the domain and range of the function. ■  ln 1 ■ ■  xl ; 74. (a) Compare the rates of growth of f x xl0 55. lim ln 1 ■ 5x ln x xl 54. lim ln sin x xl0 xl xl0 (d) lim ln 2 x xl xl3 53. lim ln cos x
xl 2 has no solution. What can
x 1 ln x ? 73. Let a ta 52. lim log10 x 2 x xl2 you say about the function f x xl0 Find the limit. 51. lim ln 2 ln x xl (The maximum charge capacity is Q 0 and t is measured in
seconds.)
(a) Find the inverse of this function and explain its meaning.
(b) How long does it take to recharge the capacitor to 90% of
capacity if a 2 ?
51–56 that is per 8. y (c) lim x 1 x to recharge the ﬂash’s capacitor, which stores electric charge
given by
e 71. Show that the equation x 1 x t x h x , where t x
0,
can be analyzed as a power of e by writing t x
e ln t x so that
fx
e h x ln t x . Using this device, calculate each limit.
(a) lim x ln x
(b) lim x ln x 50. When a camera ﬂash goes off, the batteries immediately begin Q0 1 pendicular to the line 2 x e 72. Any function of the form f x every three hours, then the number of bacteria after t hours is
n ft
100 2 t 3.
(a) Find the inverse of this function and explain its meaning.
(b) When will the population reach 50,000? Qt 70. Find an equation of the tangent to the curve y 1 ) is an odd ■ n
n ln n 1 This was ﬁnally proved, a hundred years later, by Jacques
Hadamard and Charles de la Vallée Poussin and is called
the Prime Number Theorem. Provide evidence for the
truth of this theorem by computing the ratio of n to
n ln n for n 100, 1000, 10 4, 10 5, 10 6, and 10 7. Use the
following data: 1000
168, 10 4
1229,
10 5
9592, 10 6
78,498, 10 7
664,579.
(c) Use the Prime Number Theorem to estimate the number of
primes up to a billion. 5E07(pp 432441) 1/17/06 4:30 PM Page 441 S ECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS  7.4 ❙❙❙❙ 441 Derivatives of Logarithmic Functions
In this section we ﬁnd the derivatives of the logarithmic functions y log a x and the exponential functions y a x. We start with the natural logarithmic function y ln x. We know
that it is differentiable because it is the inverse of the differentiable function y e x.
d
ln x
dx 1 Proof Let y 1
x ln x. Then
ey x Differentiating this equation implicitly with respect to x, we get
dy
dx and so ln x 3 EXAMPLE 1 Differentiate y 1 dy
dx ey 1
ey 1. SOLUTION To use the Chain Rule we let u dy
dx dy du
du dx 1
x x3 1 du
u dx 1. Then y
1
x3 1 ln u, so
3x 2 3x 2 x3 1 In general, if we combine Formula 1 with the Chain Rule as in Example 1, we get 2 EXAMPLE 2 Find d
ln u
dx 1 du
u dx d
ln t x
dx or tx
tx d
ln sin x .
dx SOLUTION Using (2), we have d
ln sin x
dx
EXAMPLE 3 Differentiate f x 1d
sin x
sin x dx 1
cos x
sin x cot x sln x. SOLUTION This time the logarithm is the inner function, so the Chain Rule gives fx 1
2 ln x 12 d
ln x
dx 1
2 sln x 1
x 1
2 x sln x 5E07(pp 442451) 442 ❙❙❙❙ 1/17/06 4:27 PM Page 442 CHAPTER 7 INVERSE FUNCTIONS EXAMPLE 4 Find d
x
ln
dx
sx 1
.
2 SOLUTION 1 d
x
ln
dx
sx 1
2 1
x
sx 1
2 sx
x  Figure 1 shows the graph of the function f
of Example 4 together with the graph of its derivative. It gives a visual check on our calculation.
Notice that f x is large negative when f is
rapidly decreasing and f x
0 when f has a
minimum. 2 sx
1 x 1
2 2
x y 1 d
x
ln
dx
sx x 21 x x
x 1 1x 1
2 1
2
x 2 2x d
[ln x
dx fª 1
x FIGURE 1 1
2 (1) x
2 5
1x 2 12 2 SOLUTION 2 If we ﬁrst simplify the given function using the laws of logarithms, then the
differentiation becomes easier: f 0 dx
dx s x 1
2 1 1
2 1 ln x 2 1
x 2 (This answer can be left as written, but if we used a common denominator we would see
that it gives the same answer as in Solution 1.)
x 2 ln x. EXAMPLE 5 Find the absolute minimum value of f x
SOLUTION The domain is 0, and the Product Rule gives
1
x x2 fx 2 x ln x x1 2 ln x 1
0 when 2 ln x
1, that is, ln x
e 1 2. Also,
Therefore, f x
2 , or x
12
12
fx
0 when x e
0 for 0 x e . So by the First Derivative
and f x
1 2e is the absolute minimum.
Test for Absolute Extreme Values, f (1 se ) EXAMPLE 6 Discuss the curve y x 2 using the guidelines of Section 4.5. ln 4 SOLUTION
A. The domain is x2 x4 x x2 0 B. The yintercept is f 0 4 x x 2 2, 2 ln 4. To ﬁnd the xintercept we set
y ln 4 x2 0 We know that ln 1 log e 1 0 (since e 0 1), so we have 4 x 2 1 ? x 2 3
and therefore the xintercepts are s3.
x
f x , f is even and the curve is symmetric about the yaxis.
C. Since f
x2 l 0
D. We look for vertical asymptotes at the endpoints of the domain. Since 4
as x l 2 and also as x l 2 , we have
lim ln 4 x l2 Thus, the lines x x2 2 and x and lim ln 4 xl 2 2 are vertical asymptotes. x2 5E07(pp 442451) 1/17/06 4:27 PM Page 443 SECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS E.
y
(0, ln 4) x=_2 x=2
0
{_ œ„ , 0}
3 fx 443 2x
x2 4 0 when 0 x 2, f is increasing
0 when 2 x 0 and f x
Since f x
on 2, 0 and decreasing on 0, 2 .
0. Since f changes from positive to negative at 0,
F. The only critical number is x
f0
ln 4 is a local maximum by the First Derivative Test. x
{œ„ , 0}
3 G. FIGURE 2
y=ln(4 ≈) ❙❙❙❙ x2 4 fx 2
4 2x
x 2x 2x 2
x2 2 8
4 22 0 for all x, the curve is concave downward on
Since f x
inﬂection point.
H. Using this information, we sketch the curve in Figure 2.
EXAMPLE 7 Find f x if f x 2, 2 and has no ln x . SOLUTION Since ln x
ln x fx if x
if x 0
0 it follows that
1
x
1
x fx Thus, f x 1 x for all x if x
if x 1
x 1 0
0 0. The result of Example 7 is worth remembering:
d
(ln x
dx 3 ) 1
x The corresponding integration formula is y 4 1
dx
x ln x C Notice that this ﬁlls the gap in the rule for integrating power functions: y
The missing case n x n dx xn 1
n1 C if n 1 is supplied by Formula 4. 1 5E07(pp 442451) 444 ❙❙❙❙ 1/17/06 4:27 PM Page 444 CHAPTER 7 INVERSE FUNCTIONS y EXAMPLE 8 Find, correct to three decimal places, the area of the region under the hyper y=Δ bola xy 1 from x 1 to x 2. SOLUTION The given region is shown in Figure 3. Using Formula 4 (without the absolute
value sign, since x 0), we see that the area is
0 1 t 2 y A area=ln 2 2 1 1
dx
x ln 2 FIGURE 3 EXAMPLE 9 Evaluate y x
x 2 1 ln x ln 1 2 1 ln 2 0.693 d x. x 2 1 because the differential du
occurs (except for the constant factor 2). Thus, x dx 1 du and
2
SOLUTION We make the substitution u x
x2 1 du
u 1
2 dx y 1
2 y 1
2 ln x 2 ln u 1 C
1
2 C ln x 2 Notice that we removed the absolute value signs because x 2
use the properties of logarithms to write the answer as
ln s x 2 1 2 x dx 1 C 1 0 for all x. We could C but this isn’t necessary.
 Since the function f x
ln x x in
Example 10 is positive for x 1, the integral
represents the area of the shaded region in
Figure 4.
y
0.5 y= EXAMPLE 10 Calculate
SOLUTION We let u x 1, u ln 1 FIGURE 4 1 e 1 ln x
d x.
x ln x because its differential du d x x occurs in the integral. When
0; when x e, u ln e 1. Thus ln x
x y e 1 EXAMPLE 11 Calculate
0 y e ln x
dx
x y 1 0 u du u2
2 1 0 1
2 y tan x dx. x SOLUTION First we write tangent in terms of sine and cosine: y tan x dx y
This suggests that we should substitute u
sin x dx
du: y tan x dx y cos x since then du sin x
dx
cos x
ln u sin x
dx
cos x y
C sin x dx and so du
u
ln cos x C 5E07(pp 442451) 1/17/06 4:27 PM Page 445 S ECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS Since ln cos x
written as ln 1 cos x 445 ln sec x , the result of Example 11 can also be y tan x dx 5 ❙❙❙❙ ln sec x C General Logarithmic and Exponential Functions
Formula 7 in Section 7.3 expresses a logarithmic function with base a in terms of the natural logarithmic function:
ln x
log a x
ln a
Since ln a is a constant, we can differentiate as follows:
d ln x
dx ln a d
log a x
dx 1d
ln x
ln a dx d
log a x
dx 6 1
x ln a 1
x ln a EXAMPLE 12 d
log10 2
dx sin x 1
d
2
sin x ln 10 dx 2 sin x cos x
sin x ln 10 2 From Formula 6 we see one of the main reasons that natural logarithms (logarithms
with base e) are used in calculus: The differentiation formula is simplest when a e
because ln e 1.
Exponential Functions with Base a In Section 7.2 we showed that the derivative of the gen a x, a eral exponential function f x
fx 0, is a constant multiple of itself: f 0 ax where f0 lim hl0 ah 1
h We are now in a position to show that the value of the constant is f 0 d
ax
dx 7 Proof We use the fact that e ln a d
ax
dx a: d
e ln a
dx
e ln a a x ln a x x ln a d ln a x
e
dx
a x ln a e ln a x d
ln a x
dx ln a. 5E07(pp 442451) 446 ❙❙❙❙ 1/17/06 4:27 PM Page 446 CHAPTER 7 INVERSE FUNCTIONS In Example 6 in Section 3.4 we considered a population of bacteria cells that doubles
every hour and saw that the population after t hours is n n0 2 t, where n0 is the initial population. Formula 7 enables us to ﬁnd the growth rate:
dn
dt n0 2 t ln 2 EXAMPLE 13 Combining Formula 7 with the Chain Rule, we have d
(10 x 2 )
dx d
x2
dx 2 10 x ln 10 2 ln 10 x10 x 2 The integration formula that follows from Formula 7 is ya EXAMPLE 14 y 5 0 x ax
ln a dx 2x
ln 2 x 2 dx C 5 a 25
ln 2 0 1 20
ln 2 31
ln 2 Logarithmic Differentiation
The calculation of derivatives of complicated functions involving products, quotients, or
powers can often be simpliﬁed by taking logarithms. The method used in the following
example is called logarithmic differentiation.
x 3 4 sx 2 1
.
3x 2 5 EXAMPLE 15 Differentiate y SOLUTION We take logarithms of both sides of the equation and use properties of logarithms to simplify:
3
4 ln y 1
2 ln x ln x 2 1 5 ln 3x 2 Differentiating implicitly with respect to x gives
1 dy
y dx 3
4 1
x 1
2 2x
x2 5 1 3
3x 2 Solving for d y d x, we get
dy
dx y 3
4x x
x 2 15
3x 2 1 Because we have an explicit expression for y, we can substitute and write
 If we hadn’t used logarithmic differentiation in
Example 15, we would have had to use both the
Quotient Rule and the Product Rule. The resulting
calculation would have been horrendous. dy
dx x 3 4 sx 2 1
3x 2 5 3
4x x
x 2 1 15
3x 2 5E07(pp 442451) 1/17/06 4:27 PM Page 447 S ECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS ❙❙❙❙ 447 Steps in Logarithmic Differentiation
1. Take logarithms of both sides of an equation y f x and use properties of logarithms to simplify.
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y .
If f x
0 for some values of x, then ln f x is not deﬁned, but we can write
y
f x and use Equation 3. We illustrate this procedure by proving the general version of the Power Rule, as promised in Section 3.3.
x n, then The Power Rule If n is any real number and f x nx n fx  If x 0, we can show that f 0
0 for
n 1 directly from the deﬁnition of a derivative. Proof Let y x n and use logarithmic differentiation:
ln y ln x Therefore y
y
y n n n ln x x 0 n
x Hence  1 y
x n xn
x nx n 1 You should distinguish carefully between the Power Rule d dx x n n x n 1 , where
the base is variable and the exponent is constant, and the rule for differentiating exponential functions d dx a x a x ln a , where the base is constant and the exponent is variable.
In general there are four cases for exponents and bases:
db
a
0
(a and b are constants)
dx
d
2.
f x b b f x b 1f x
dx
d tx
a
a t x ln a t x
3.
dx
4. To ﬁnd d d x f x t x , logarithmic differentiation can be used, as in the next
example.
1.  Figure 5 illustrates Example 16 by showing
the graphs of f x
x sx and its derivative.
y EXAMPLE 16 Differentiate y f SOLUTION 1 Using logarithmic differentiation, we have fª ln y
1
0 FIGURE 5 1 x sx. x ln x sx y
y sx y y sx ln x 1
x ln x 1
sx ln x
2 sx 1
2 sx
x sx 2 ln x
2 sx 5E07(pp 442451) 448 ❙❙❙❙ 1/17/06 4:27 PM Page 448 CHAPTER 7 INVERSE FUNCTIONS SOLUTION 2 Another method is to write x sx d sx
(x )
dx d sx ln x
(e )
dx
2 x sx sx e ln x :
d
(s x ln x)
dx e sx ln x ln x
2 sx (as in Solution 1) The Number e a s a Limit
We have shown that if f x
ln x, then f x
1 x. Thus, f 1
fact to express the number e as a limit.
From the deﬁnition of a derivative as a limit, we have
h
h f1 x
x ln 1 xl0 lim ln 1 f1 x lim f1 hl0 lim ln 1 xl0 Because f 1 f1 lim xl0 x
x 1
ln 1
x lim xl0 1. We now use this f1 x 1x 1, we have
lim ln 1 x xl0 1x 1 Then, by Theorem 2.5.8 and the continuity of the exponential functions, we have
e e1 e lim x l 0 ln 1 e 8 x 1x lim e ln 1 x 1x xl0 lim 1 xl 0 x lim 1 xl0 x
3 2 y=(1+x)!?® 1
0 FIGURE 6 x 1x 1x Formula 8 is illustrated by the graph of the function y
table of values for small values of x.
y x 0.1
0.01
0.001
0.0001
0.00001
0.000001
0.0000001
0.00000001 1 x 1x (1 in Figure 6 and a x)1/x 2.59374246
2.70481383
2.71692393
2.71814593
2.71826824
2.71828047
2.71828169
2.71828181 5E07(pp 442451) 1/17/06 4:27 PM Page 449 SECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS If we put n
sion for e is 1 x in Formula 8, then n l e 9  7.4 25–28 ln x is used
much more frequently in calculus than the other logarithmic
functions y log a x.
 as x l 0 and so an alternative expres n 1
n 1 nl 449 Exercises 1. Explain why the natural logarithmic function y 2–24 lim ❙❙❙❙ 25. y Differentiate the function. 2. f x ln x 2 3. f log 2 1 7. f x 5
sln x 9. f x s x ln x 10 ■ 4. f x
6. f x log10 8. f x 3x cos ln x 5
ln sx 1
1 10. f t 11. F t 2t
ln
3t 1
1 12. h x ln( x sx 13. t x a
ln
a x ■ ■ x
ln x 1 x 2 ln 1 31. f x ln t
ln t ■ ■ ■ ■ x2 x
x tan x
■ ■ ■ ■ ■ ■ 1
ln x 1 32. f x
■ ■ 34. If f x 1) ■ ln ln ln x ■ ■ ■ ■ x
, ﬁnd f e .
ln x
x 2 ln x, ﬁnd f 1 . 4 2 ln sec x
■ 30. f x 1 33. If f x 3 ■ Differentiate f and ﬁnd the domain of f . 29. f x 1 28. y ■  ln x
x2 26. y log10 x
■ 29–32 x Find y and y .
x ln x 27. y ln cos 5. f x  35–36 t3
ln 2 x ey ln ln x, 36. y ln x 4 sin2x 16. y ln x 3 ■ 3t 17. h t y ln 1 35. y ln u
1 ln 2u 15. f u 19. y 14. F y  Find an equation of the tangent line to the curve at the
given point. 10 tan 18. y ■ 7, ■ 21. y ln e 23. y
■ ln 5
■ x ■ ■ ■ ■ ■ ■ ■ ■ ; 37–38  Find f x . Check that your answer is reasonable by comparing the graphs of f and f . 5x 2 3u
3u ■ x 22. y ■ x cos x ■ ■ ln 1 24. y xe sin x 38. f x 2
2 1x
■ 2, 0 ■ 37. f x
20. G u e, 0 3x ■ 2
■ ex ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 2 39–50  Use logarithmic differentiation to ﬁnd the derivative of
the function. 2 ■ ■ ln x ■ ■ 39. y 2x 1 5 x4 3 6 40. y 2 sx e x x 2 1 10 ■ 5E07(pp 442451) ❙❙❙❙ 450 1/17/06 4:27 PM Page 450 CHAPTER 7 INVERSE FUNCTIONS 41. y sin2x tan4x
x2 1 2 42. y 4 43. y xx 44. y x1 x 45. y x sin x 47. y ln x
xe 49. y
■ 46. y
x ■ ■ ■ 51. Find y if y ln x 2 52. Find y if x y ■ 73. ■ ■ ■ ■ ■ ; 55–56 x if f x ln x 1. ■ ■ 56. ln 4 ln x ■ ■ ■ dx ■ ■ ■ ■ ■ 2 x 2 , below the xaxis, and
4 and x
1. 1 ■ ■ ■ x ■ 2 1 from 0 to 1 about the xaxis.
80. Find the volume of the solid obtained by rotating the region under the curve x ■ sx ■ ■ ■ 1 y x2 57. Find the intervals of concavity and the inﬂection points of the function f x x2 dx ln sin x
C by (a) differentiating
the right side of the equation and (b) using the method of
Example 11. y Use a graph to estimate the roots of the equation. Then
use these estimates as the initial approximations in Newton’s
method to ﬁnd the roots correct to six decimal places.
4 ■ 1 under the curve  55. x ■ y x2 x 79. Find the volume of the solid obtained by rotating the region d
x 8 ln x .
dx 9 2 ■ above the hyperbola y
between the lines x 9 54. Find ■ ye 76. 10 t dt
■ ex 74. dx 78. Find, correct to three decimal places, the area of the region y x.
n 2 77. Show that x cot x dx y2 . 53. Find a formula for f 2 1 ■ cos x ln x ■ y ln x
x y x x ln x 50. y ■ 1
1 75. sin x 48. y x x2
x2 ln x s x. 1 from 0 to 3 about the yaxis. 58. Find the absolute minimum value of the function f x x ln x. 81. The work done by a gas when it expands from volume V1 to
V
volume V2 is W xV P d V , where P P V is the pressure as
a function of the volume V . (See Exercise 27 in Section 6.4.)
Boyle’s Law states that when a quantity of gas expands at constant temperature, PV C, where C is a constant. If the initial
volume is 600 cm3 and the initial pressure is 150 kPa, ﬁnd the
work done by the gas when it expands at constant temperature
to 1000 cm 3.
2 59–62 59. y ln 1 ■ C AS ■ ln tan2x 60. y ln sin x 61. y x2 ■ ln x 2 62. y ■ ■ ■ ■ ■ 3x ■ 2 ■ ■ 63. If f x ln 2 x x sin x , use the graphs of f , f , and f to
estimate the intervals of increase and the inﬂection points of f
on the interval 0, 15 .
ln x 2 c . What
happens to the inﬂection points and asymptotes as c changes?
Graph several members of the family to illustrate what you
discover. ; 64. Investigate the family of curves f x 65–76
4  y 67. y 69. y 71. y 6x 1 8 e x2 x
x 2 4 1 x2
dx
x3 dx y 68. 3t 1 2 66. dt 2 ■ 82. Find f if f x 2, x x 0, f 1 0, and f 2 83. If t is the inverse function of f x
84. If f x e x ln x and h x f 2x
1 y 70. y 72. y2 9 sx 4
6 e u
u3 1 x , ﬁnd h e . 85. For what values of m do the line y y m x and the curve
1 enclose a region? Find the area of the region. x x2 ln x near l.
(b) Illustrate part (a) by graphing f and its linearization.
(c) For what values of x is the linear approximation accurate to
within 0.1? 2 du
1
sx 2 dx 87. Use the deﬁnition of derivative to prove that lim dx
x ln x
cos x
dx
sin x 0. ln x, ﬁnd t 2 . ; 86. (a) Find the linear approximation to f x Evaluate the integral. 3
dx
x 65. 2 1 Discuss the curve under the guidelines of Section 4.5.  ln 1 88. Show that lim
nl 1 x
n x
x xl0 1 n e x for any x 0. 5E07(pp 442451) 1/17/06 4:27 PM Page 451 S ECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION  ❙❙❙❙ 451 7.2* The Natural Logarithmic Function  If your instructor has assigned Sections
7.2–7.4, you need not read Sections 7.2*, 7.3*,
and 7.4* (pp. 451–476). In this section we deﬁne the natural logarithm as an integral and then show that it obeys
the usual laws of logarithms. The Fundamental Theorem makes it easy to differentiate this
function.
1 Definition The natural logarithmic function is the function deﬁned by y area=ln x 0 y ln x y= 1
t 1 x t x 1 1
dt
t x 0 The existence of this function depends on the fact that the integral of a continuous function always exists. If x 1, then ln x can be interpreted geometrically as the area under
the hyperbola y 1 t from t 1 to t x. (See Figure 1.) For x 1, we have FIGURE 1 y ln 1 y For 0 x 1 1 1
dt
t 0 1, area=_ ln x y ln x y= 1
t x 1 1
dt
t y 1 x 1
dt
t 0 and so ln x is the negative of the area shaded in Figure 2.
0 x t 1 EXAMPLE 1
FIGURE 2 (a) By comparing areas, show that 1 ln 2 3 .
2
4
(b) Use the Midpoint Rule with n 10 to estimate the value of ln 2.
SOLUTION (a) We can interpret ln 2 as the area under the curve y 1 t from 1 to 2. From Figure 3
we see that this area is larger than the area of rectangle BCDE and smaller than the area
of trapezoid ABCD. Thus, we have y y= 1
t 1
2 ln 2 1
2 (1 1
2 ) 3
4 (b) If we use the Midpoint Rule with f t C
1 2 t ln 2
FIGURE 3 1 D E
B
0 ln 2 1
2 A 1 y 2 1 1
dt
t 0.1 1
1.05 0.1 f 1.05
1
1.15 1 t, n 10, and t f 1.15
1
1.95 0.1, we get
f 1.95 0.693 Notice that the integral that deﬁnes ln x is exactly the type of integral discussed in Part 1
of the Fundamental Theorem of Calculus (see Section 5.3). In fact, using that theorem,
we have
d x1
1
y dt x
dx 1 t 5E07(pp 452461) 452 ❙❙❙❙ 1/17/06 5:16 PM Page 452 CHAPTER 7 INVERSE FUNCTIONS and so
d
ln x
dx 2 1
x We now use this differentiation rule to prove the following properties of the logarithm
function. 3 Laws of Logarithms If x and y are positive numbers and r is a rational number, then
1. ln x y ln x 2. ln ln y x
y ln x 3. ln x r ln y r ln x Proof
1. Let f x ln ax , where a is a positive constant. Then, using Equation 2 and the
Chain Rule, we have
1d
ax
ax dx fx 1
ax 1
x a Therefore, f x and ln x have the same derivative and so they must differ by a constant:
ln ax
Putting x ln x 1 in this equation, we get ln a C ln 1 ln ax ln x C 0 C C. Thus ln a If we now replace the constant a by any number y, we have
ln x y
2. Using Law 1 with x ln and so ln x ln y 1 y, we have
1
y ln y ln 1
y ln 1
y y ln 1 0 ln y Using Law 1 again, we have
ln x
y ln x 1
y ln x ln The proof of Law 3 is left as an exercise.
EXAMPLE 2 Expand the expression ln x2 5 4 sin x
.
x
1
3 1
y ln x ln y 5E07(pp 452461) 1/17/06 5:16 PM Page 453 S ECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION ❙❙❙❙ 453 SOLUTION Using Laws 1, 2, and 3, we get ln x2 5 4 sin x
x3 1 ln x 2 5 4 ln x 2
1
2 EXAMPLE 3 Express ln a 4 ln x 3 ln sin x 5 ln x 3 ln sin x 1
1 ln b as a single logarithm. SOLUTION Using Laws 3 and 1 of logarithms, we have
1
2 ln a ln b ln b 1 2 ln a ln a ln sb ln(a sb )
In order to graph y ln x, we ﬁrst determine its limits:
(a) lim ln x 4 (b) lim ln x xl xl0 Proof
y y=ln x
0 x 1 (a) Using Law 3 with x 2 and r n (where n is any positive integer), we have
ln 2 n
n ln 2. Now ln 2 0, so this shows that ln 2 n l as n l . But ln x is an
increasing function since its derivative 1 x 0. Therefore, ln x l as x l .
(b) If we let t 1 x, then t l as x l 0 . Thus, using (a), we have
lim ln x xl0 F IGURE 4 If y ln x, x y 0 1 e x y=ln x
FIGURE 5 lim tl ln t 1
x 0 and d2y
dx 2 1
x2 0 which shows that ln x is increasing and concave downward on 0, . Putting this information together with (4), we draw the graph of y ln x in Figure 4.
Since ln 1 0 and ln x is an increasing continuous function that takes on arbitrarily
large values, the Intermediate Value Theorem shows that there is a number where ln x
takes on the value 1 (see Figure 5). This important number is denoted by e.
5 1.02 1
t 0, then
dy
dx 1 lim ln tl Definition e is the number such that ln e 1. y=ln x EXAMPLE 4 Use a graphing calculator or computer to estimate the value of e.
y=1 SOLUTION According to Deﬁnition 5, we estimate the value of e by graphing the curves
0.98
2.7 FIGURE 6 2.75 y ln x and y 1 and determining the xcoordinate of the point of intersection. By
zooming in repeatedly, as in Figure 6, we ﬁnd that
e 2.718 5E07(pp 452461) 454 ❙❙❙❙ 1/17/06 5:16 PM Page 454 CHAPTER 7 INVERSE FUNCTIONS With more sophisticated methods, it can be shown that the approximate value of e, to 20
decimal places, is
e 2.71828182845904523536 The decimal expansion of e is nonrepeating because e is an irrational number.
Now let’s use Equation 2 to differentiate functions that involve the natural logarithmic
function.
ln x 3 EXAMPLE 5 Differentiate y 1.
x3 SOLUTION To use the Chain Rule we let u dy
dx dy du
du dx 1 du
u dx 1. Then y
1
x 3 1 ln u, so
3x 2 3x 2 x 3 1 In general, if we combine Formula 2 with the Chain Rule as in Example 5, we get
d
ln u
dx 6 EXAMPLE 6 Find 1 du
u dx d
ln t x
dx or tx
tx d
ln sin x .
dx SOLUTION Using (6), we have d
ln sin x
dx 1d
sin x
sin x dx EXAMPLE 7 Differentiate f x 1
cos x
sin x cot x sln x. SOLUTION This time the logarithm is the inner function, so the Chain Rule gives fx EXAMPLE 8 Find 1
2 d
x
ln
dx
sx 12 ln x d
ln x
dx 1
2 sln x 1
x 1
2 x sln x 1
.
2 SOLUTION 1 d
x
ln
dx
sx 1
2 1 dx
dx s x x
sx 1
2 sx
x 2 sx
1 x 2
x 1
2 1x 1
2 21 x x
x 1
2 1
2
x 2x (1) x
2 5
1x 2 2 12 5E07(pp 452461) 1/17/06 5:16 PM Page 455 S ECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION  Figure 7 shows the graph of the function f
of Example 8 together with the graph of its derivative. It gives a visual check on our calculation.
Notice that f x is large negative when f is
0 when f has a
rapidly decreasing and f x
minimum. ❙❙❙❙ 455 SOLUTION 2 If we ﬁrst simplify the given function using the laws of logarithms, then the
differentiation becomes easier: d
x
ln
dx
sx 1
2 d
[ln x
dx
1 y x f 1
2 1
1
2 1 ln x 2 1
x 2 (This answer can be left as written, but if we used a common denominator we would see
that it gives the same answer as in Solution 1.) 1
0 x EXAMPLE 9 Discuss the curve y fª x 2 using the guidelines of Section 4.5. ln 4 SOLUTION
A. The domain is
FIGURE 7 x4 x2 x x2 0 B. The yintercept is f 0 4 x x 2 2, 2 ln 4. To ﬁnd the xintercept we set
y x2 ln 4 0 We know that ln 1 0, so we have 4 x 2 1 ? x 2 3 and therefore the
xintercepts are s3.
C. Since f
x
f x , f is even and the curve is symmetric about the yaxis.
D. We look for vertical asymptotes at the endpoints of the domain. Since 4
x2 l 0
as x l 2 and also as x l 2 , we have
lim ln 4 x l2 x2 by (4). Thus, the lines x and
2 and x E. lim ln 4 xl 2 x2 2 are vertical asymptotes.
2x
4 x2 fx y
(0, ln 4) x=_2 x=2
0
{_ œ„ , 0}
3 x 0 when 0 x 2, f is increasing
Since f x
0 when 2 x 0 and f x
on 2, 0 and decreasing on 0, 2 .
F. The only critical number is x
0. Since f changes from positive to negative at 0,
f0
ln 4 is a local maximum by the First Derivative Test. {œ„ , 0}
3 G. FIGURE 8
y=ln(4 ≈) fx 4 x2 2
4 2x
x2 2x 8
4 2 0 for all x, the curve is concave downward on
Since f x
inﬂection point.
H. Using this information, we sketch the curve in Figure 8.
EXAMPLE 10 Find f x if f x ln x . SOLUTION Since fx ln x
ln x if x
if x 0
0 2x 2
x2 2
2, 2 and has no 5E07(pp 452461) 456 ❙❙❙❙ 1/17/06 5:16 PM Page 456 CHAPTER 7 INVERSE FUNCTIONS it follows that 1
x
1
x fx Thus, f x 1 x for all x if x
if x 1
x 1 0
0 0. The result of Example 10 is worth remembering:
d
(ln x
dx 7 1
x ) The corresponding integration formula is y 8 1
dx
x ln x C Notice that this ﬁlls the gap in the rule for integrating power functions: yx
The missing case n n xn 1
n1 dx C if n 1 1 is supplied by Formula 8. EXAMPLE 11 Evaluate x y x2 1 d x. x 2 1 because the differential du
1
occurs (except for the constant factor 2). Thus, x dx 2 du and
SOLUTION We make the substitution u x  Since the function f x
ln x x in
Example 12 is positive for x 1, the integral
represents the area of the shaded region in
Figure 9. x 2 du
u 1 1
2 dx y= FIGURE 9 ln u 1 C
1
2 C ln x 2 1 1
1 C
0 for all x. We could C but this isn’t necessary. ln x
x SOLUTION We let u
1 ln x 2 ln s x 2 EXAMPLE 12 Calculate 0 1
2 Notice that we removed the absolute value signs because x 2
use the properties of logarithms to write the answer as y
0.5 y 1
2 y 2 x dx e x x 1, u ln 1 y e 1 ln x
d x.
x ln x because its differential du d x x occurs in the integral. When
0; when x e, u ln e 1. Thus y e 1 ln x
dx
x y 1 0 u du u2
2 1 0 1
2 5E07(pp 452461) 1/17/06 5:16 PM Page 457 S ECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION ❙❙❙❙ 457 y tan x dx. EXAMPLE 13 Calculate SOLUTION First we write tangent in terms of sine and cosine: sin x
dx
cos x y tan x dx y
This suggests that we should substitute u
sin x dx
du: cos x since then du sin x
dx
cos x y tan x dx y ln u
Since ln cos x
written as ln cos x C ln sec x , the result of Example 13 can also be ln 1 cos x y tan x dx 9 du
u y
C sin x dx and so ln sec x C Logarithmic Differentiation
The calculation of derivatives of complicated functions involving products, quotients, or
powers can often be simpliﬁed by taking logarithms. The method used in the following
example is called logarithmic differentiation.
EXAMPLE 14 Differentiate x 3 4 sx 2 1
3x 2 5 y SOLUTION We take logarithms of both sides of the equation and use the Laws of
Logarithms to simplify:
3
4 ln y 1
2 ln x ln x 2 1 5 ln 3x 2 Differentiating implicitly with respect to x gives
1 dy
y dx 3
4 1
x 1
2 2x
x 2 5 1 3
3x 2 Solving for d y d x, we get
dy
dx
 If we hadn’t used logarithmic differentiation in
Example 14, we would have had to use both the
Quotient Rule and the Product Rule. The resulting
calculation would have been horrendous. y 3
4x x
x2 15
3x 2 1 Because we have an explicit expression for y, we can substitute and write
dy
dx x 3 4 sx 2 1
3x 2 5 3
4x x
x2 1 15
3x 2 5E07(pp 452461) ❙❙❙❙ 458 1/17/06 5:17 PM Page 458 CHAPTER 7 INVERSE FUNCTIONS Steps in Logarithmic Differentiation
1. Take logarithms of both sides of an equation y f x and use the Laws of Logarithms to simplify.
2. Differentiate implicitly with respect to x.
3. Solve the resulting equation for y . y  7.2*
1–4 If f x
0 for some values of x, then ln f x is not deﬁned, but we can write
f x and use Equation 7. Exercises Use the Laws of Logarithms to expand the quantity.  27. y 3 xy
1. ln 2
z
3. ln uv
■ ■ 5–8 2. ln sa b 2 ■ 1
2 ln x 8. ln x
■ 9–12 29. y ■ ■ ■ ■ ■ ■ 5
■ ■ ■ 1
3 6. ln 3 ln 2
5 ln x 2 ■ ln 8 ■ ■ ■ ■ ■ ■ ■ 11. y
■ 13–30 10. y ln x
■  ■ ■ ■ ■ 1
■ ln x
■ ■ ■ Differentiate the function. 13. f x s x ln x 14. f x ln x 2 15. f ln cos 16. f x cos ln x 17. f x 5
sln x 18. f x 5
ln sx 20. h x ln( x 19. t x ln a
a x
x 21. f u ` 23. F t 25. y ln ln 2 2t
3t
x 1
1 ■ 22. f t 10 5x2  ■ 1
1 24. y 26. G u ln tan 2 x
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ln x
x2 32. y
■ ■ ■ x
ln x ■ ■ ■ 1 1
1 ln x 34. f x 1 s1
■ 36. f x ln x
■ ■ ■ 37. If f x t ln 4 ln ln ln x ■ ■ ■ ■ x
, ﬁnd f e .
ln x
3t , ﬁnd f 1. ; 39–40  Find f x . Check that your answer is reasonable by comparing the graphs of f and f . ■ ■ sin x
■ 40. f x ln x
■ ■ ■ ln x 2 ■ ■ ■ x
■ 1
■ ■ 41–42  Find an equation of the tangent line to the curve at the
given point. 1) sx 2
ln t
ln t 41. y 3u
3u sin 2 ln x ,
■ ■ 43. Find y if y ■ 46. Find ■ ■ ■ y2 . y sin x. 45. Find a formula for f 2
2 ■ ln x 2 ln x 3 42. y 1, 0 ■ 44. Find y if ln x y ln x 4 sin2x ln ■ Differentiate f and ﬁnd the domain of f . ■ 39. f x 3
4 ■ 35. f x ■ ln u
1 ln 2u ■ 38. If f t 2
■ ■ ln ln x ln x 12. y 3 ■ 30. y b
■ 2 ■ Make a rough sketch of the graph of each function. Do not
use a calculator. Just use the graph given in Figure 4 and, if necessary, the transformations of Section 1.3.
ln x ln tan x Find y and y . 33. f x  9. y  33–36 ■ ■ 31. y 1 ■ 28. y ■ b ln z a ln y 35 1
1 tan ln ax
■ 31–32 ■ Express the quantity as a single logarithm.  5. 2 ln 4
7. c2 3x 2
4. ln
x1 10 x
x ln d9
x 8 ln x .
dx 9 n x if f x ln x 1. 7,
■ 2, 0
■ ■ 5E07(pp 452461) 1/17/06 5:17 PM Page 459 S ECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION
 Use a graph to estimate the roots of the equation correct to one decimal place. Then use these estimates as the initial
approximations in Newton’s method to ﬁnd the roots correct to
six decimal places. ■ 2 4
■ ■ ■ x2 48. ln 4 ln x
■ ■ ■ ■ under the curve y 1 from 0 to 1 about the xaxis. 1 sx 72. Find the volume of the solid obtained by rotating the region 1 x2 under the curve y x ■ 459 71. Find the volume of the solid obtained by rotating the region ; 47–48 47. x ❙❙❙❙ 1 from 0 to 3 about the yaxis. 73. The work done by a gas when it expands from volume V1 to ■ ■ ■ V
volume V2 is W xV P d V , where P P V is the pressure as
a function of the volume V . (See Exercise 27 in Section 6.4.)
Boyle’s Law states that when a quantity of gas expands at constant temperature, PV C, where C is a constant. If the initial
volume is 600 cm3 and the initial pressure is 150 kPa, ﬁnd the
work done by the gas when it expands at constant temperature
to 1000 cm 3.
2 1 49–52 Discuss the curve under the guidelines of Section 4.5.  49. y
51. y ln 1 ■ C AS 50. y ■ x2 ■ ■ ■ ■ ln tan2x 52. y ln sin x ln x 2 ■ ■ 3x ■ 2 ■ ■ ■ 74. Find f if f 53. If f x ln 2 x x sin x , use the graphs of f , f , and f to
estimate the intervals of increase and the inﬂection points of f
on the interval 0, 15 . 57. y
■ ■ 4 58. y ■ ■ ■ ■ ■ 1
3 ln x, ﬁnd t 2 . 2x ln 1.5 5
12 (b) Use the Midpoint Rule with n 10 to estimate ln 1.5. 78. Refer to Example 1. (a) Find an equation of the tangent line to the curve y
that is parallel to the secant line AD.
(b) Use part (a) to show that ln 2 0.66. 1
1 ■ 0. 77. (a) By comparing areas, show that Use logarithmic differentiation to ﬁnd the derivative of
the function.
x 3 1 4 sin2x
55. y
2x 1 5 x 4 3 6
56. y
3
sx
x2
x2 0, and f 2 ln x near l.
(b) Illustrate part (a) by graphing f and its linearization.
(c) For what values of x is the linear approximation accurate to
within 0.1?  sin2x tan4x
x2 1 2 0, f 1 ; 76. (a) Find the linear approximation to f x ln x 2 c . What
happens to the inﬂection points and asymptotes as c changes?
Graph several members of the family to illustrate what you
discover. ■ x 2, x 75. If t is the inverse function of f x ; 54. Investigate the family of curves f x 55–58 x ■ ■ 1t ■ 79. By comparing areas, show that
59–68 59.
61.
63. y  3
dx
x 4 2 y Evaluate the integral. dt 2 1 8 e x2 x
x 62. 3t y 1 1 u3 1 y ln x
x 9 6 66. y2 y 68. dx y ■ ■ ■ ■ ■ (1
■ 1
n ln n 1
2 1 1
3 1
n 1 [Hint: Start by showing
that both sides of the equation have the same derivative.] dx 81. For what values of m do the line y y cos x
dx
sin x ■ m x and the curve
1 enclose a region? Find the area of the region.
ln 2 x ln 1 x. 83. Use the deﬁnition of derivative to prove that sx )4
dx
sx
■ x x2 82. Find lim x l lim xl0 ■ 69. Show that x cot x dx ln sin x
C by (a) differentiating
the right side of the equation and (b) using the method of
Example 13. 70. Find, correct to three decimal places, the area of the region above the hyperbola y
between the lines x 1
3 80. Prove the third law of logarithms.
2 dx
x ln x y e 1
2 du
1
sx sx 4 2 67. u2 4 64. dx x
dx
x3 y 6x ■ y 2 2 2 65. ■ 60. 2 x 2 , below the xaxis, and
4 and x
1. ■ ln 1 x
x 1 x 0.1 and t x
ln x
by graphing both f and t in several viewing rectangles.
When does the graph of f ﬁnally surpass the graph of t ?
(b) Graph the function h x
ln x x 0.1 in a viewing rectangle
that displays the behavior of the function as x l .
(c) Find a number N such that ; 84. (a) Compare the rates of growth of f x ln x
x 0.1 0.1 whenever x N 5E07(pp 452461) 460 ❙❙❙❙ 1/17/06 5:17 PM Page 460 CHAPTER 7 INVERSE FUNCTIONS  7.3* The Natural Exponential Function
Since ln is an increasing function, it is onetoone and therefore has an inverse function,
which we denote by exp. Thus, according to the deﬁnition of an inverse function, 1 f x y&
? fy 1 x exp x y &? ln y x and the cancellation equations are
f 1 ff fx x 1 x x 2 exp ln x x and ln exp x x In particular, we have
exp 0 e 1 y=x ln e r 1 x Therefore, by (1), r ln e exp r y=ln x
0 since ln e We obtain the graph of y exp x by reﬂecting the graph of y ln x about the line y x.
(See Figure 1.) The domain of exp is the range of ln, that is,
, ; the range of exp is
the domain of ln, that is, 0, .
If r is any rational number, then the third law of logarithms gives y=exp x 1 0 exp 1
y 1 since ln 1 er r Thus, exp x
e x whenever x is a rational number. This leads us to deﬁne e x, even for
irrational values of x, by the equation
FIGURE 1 ex exp x In other words, for the reasons given, we deﬁne e x to be the inverse of the function ln x. In
this notation (1) becomes
3 ex y &? ln y x and the cancellation equations (2) become
4 5 e ln x ln e x x x x 0 for all x 5E07(pp 452461) 1/17/06 5:17 PM Page 461 SECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION E XAMPLE 1 Find x if ln x ❙❙❙❙ 461 5. SOLUTION 1 From (3) we see that ln x
Therefore, x 5 e5 means x e 5. SOLUTION 2 Start with the equation ln x 5 and apply the exponential function to both sides of the equation:
e ln x
But (4) says that e ln x e 5. x. Therefore, x EXAMPLE 2 Solve the equation e 5 3x e5 10. SOLUTION We take natural logarithms of both sides of the equation and use (5): ln e 5 ln 10 3x ln 10 3x 5 3x 5
1
3 x ln 10
5 ln 10 Since the natural logarithm is found on scientiﬁc calculators, we can approximate the
solution to four decimal places: x 0.8991.
y The exponential function f x
e x is one of the most frequently occurring functions
in calculus and its applications, so it is important to be familiar with its graph (Figure 2)
and its properties (which follow from the fact that it is the inverse of the natural logarithmic function). y=´ 6 1
0 1 x Properties of the Natural Exponential Function The exponential function f x an increasing continuous function with domain
for all x. Also
lim e x lim e x 0 xl and range 0, xl FIGURE 2 The natural exponential function e x. So the xaxis is a horizontal asymptote of f x EXAMPLE 3 Find lim
xl e 2x
e 2x 1 . SOLUTION We divide numerator and denominator by e 2 x : lim xl e 2x
e 2x 1 lim xl 1 1
1 0 1
e
1 2x 1 1
lim e xl 2x . Thus, e x e x is
0 5E07(pp 462471) 462 ❙❙❙❙ 1/17/06 4:49 PM Page 462 CHAPTER 7 INVERSE FUNCTIONS We have used the fact that t 2x l as x l
2x lim e lim e t xl 0 tl e x has the properties expected of an exponential function. We now verify that f x
7 and so Laws of Exponents If x and y are real numbers and r is rational, then 1. e x y e xe y 2. e x y ex
ey 3. e x r e rx Proof of Law 1 Using the ﬁrst law of logarithms and Equation 5, we have ln e xe y ln e x ln e y x y ln e x y Since ln is a onetoone function, it follows that e xe y e x y.
Laws 2 and 3 are proved similarly (see Exercises 85 and 86). As we will see in the
next section, Law 3 actually holds when r is any real number. Differentiation
The natural exponential function has the remarkable property that it is its own derivative.
d
ex
dx 8 ex e x is differentiable because it is the inverse function of y ln x,
which we know is differentiable with nonzero derivative. To ﬁnd its derivative, we use
the inverse function method. Let y e x. Then ln y x and, differentiating this latter
equation implicitly with respect to x, we get Proof The function y
y slope=e®
{x, e ®} 1 dy
y dx
1 1 dy
dx y slope=1 y=e®
0 F IGURE 3 x ex The geometric interpretation of Formula 8 is that the slope of a tangent line to the curve
y e x at any point is equal to the ycoordinate of the point (see Figure 3). This property
implies that the exponential curve y e x grows very rapidly (see Exercise 90).
EXAMPLE 4 Differentiate the function y
SOLUTION To use the Chain Rule, we let u dy
dx dy du
du dx e tan x.
tan x. Then we have y
eu du
dx e tan x sec2x e u, so 5E07(pp 462471) 1/17/06 4:49 PM Page 463 S ECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION ❙❙❙❙ 463 In general if we combine Formula 8 with the Chain Rule, as in Example 4, we get
d
eu
dx 9 EXAMPLE 5 Find y if y e 4x eu du
dx sin 5x. SOLUTION Using Formula 9 and the Product Rule, we have y e 4x cos 5x 5 sin 5x e 4x 4 4x e 5 cos 5x EXAMPLE 6 Find the absolute maximum value of the function f x 4 sin 5x
xe x. SOLUTION We differentiate to ﬁnd any critical numbers: fx xe x 1 x e 1 e x 1 x 0 when 1 x 0,
Since exponential functions are always positive, we see that f x
0 when x 1. By the First Derivative Test for
that is, when x 1. Similarly, f x
Absolute Extreme Values, f has an absolute maximum value when x 1 and the value is
f1 1e 1
e 1 0.37
e 1 x, together with asymptotes, EXAMPLE 7 Use the ﬁrst and second derivatives of f x to sketch its graph.
0 , so we check for vertical asymptotes
SOLUTION Notice that the domain of f is x x
by computing the left and right limits as x l 0. As x l 0 , we know that t 1 x l ,
so
lim e 1 x lim e t
xl0 and this shows that x
so tl 0 is a vertical asymptote. As x l 0 , we have t
lim e 1 x xl0 As x l lim e t 1 xl , 0 tl , we have 1 x l 0 and so
lim e 1 x xl e0 1 This shows that y 1 is a horizontal asymptote.
Now let’s compute the derivative. The Chain Rule gives
fx e1 x
x2 0 for all x 0. Thus, f is
Since e 1 x 0 and x 2 0 for all x 0, we have f x
, 0 and on 0, . There is no critical number, so the function has no
decreasing on
maximum or minimum. The second derivative is
fx x 2e 1 x 1 x2
x4 e 1 x 2x e 1 x 2x
x4 1 5E07(pp 462471) 464 ❙❙❙❙ 1/17/06 4:49 PM Page 464 CHAPTER 7 INVERSE FUNCTIONS
1
Since e 1 x 0 and x 4 0, we have f x
0 when x
0 and f x
0
2x
1
when x
, 1 ) and concave upward on
2 . So the curve is concave downward on (
2
( 1 , 0) and on 0, . The inﬂection point is ( 1 , e 2).
2
2
To sketch the graph of f we ﬁrst draw the horizontal asymptote y 1 (as a dashed
line), together with the parts of the curve near the asymptotes in a preliminary sketch
[Figure 4(a)]. These parts reﬂect the information concerning limits and the fact that f is
decreasing on both
, 0 and 0, . Notice that we have indicated that f x l 0 as
x l 0 even though f 0 does not exist. In Figure 4(b) we ﬁnish the sketch by incorporating the information concerning concavity and the inﬂection point. In Figure 4(c) we
check our work with a graphing device. y y
4 y=‰ inﬂection
point
y=1
0 FIGURE 4 y=1
0 x (a) Preliminary sketch _3 x 3
0 (b) Finished sketch (c) Computer confirmation Integration
e x has a simple derivative, its integral is also simple: Because the exponential function y ye 10 EXAMPLE 8 Evaluate 2 x3 yx e x 2 x3 yx e C dx.
x 3. Then du SOLUTION We substitute u ex dx 1
3 dx ye u 1
3 3x 2 dx, so x 2 dx
1
3 du EXAMPLE 9 Find the area under the curve y e eu
3x 1
3 C ex du and 3 C from 0 to l. SOLUTION The area is A y 1 0 e 3x dx 1
3 e 3x 1
0 1
3 1 e 3 5E07(pp 462471) 1/17/06 4:50 PM Page 465 S ECTION 7.3* THE NATURAL EXPONENTIAL FUNCTION  7.3* (c) reﬂecting about the xaxis
(d) reﬂecting about the yaxis
(e) reﬂecting about the xaxis and then about the yaxis (b) What is an approximate value for e ?
(c) Sketch, by hand, the graph of the function f x
e x with
particular attention to how the graph crosses the yaxis.
What fact allows you to do this? ln 6 2. (a) e 3. (a) ln e
4. (a) ln e sin x N N 5–12 (b) e
N N N 6. (a) e 2x 3 7 x N N N N 27. lim
xl x 5
2x 10. ln 2x 2 4 Ce bx, where a 12. 7e x e 2x
N N N N N N N N N N 13. e 15. ln e x
N 14. ln(1 100
2 N N 16. e 1 3
N N N N sx ) x4 17. (a) e x 10 18. (a) 2 N N N N N N N N 21. y N N N N N 3x
N N 20. y N N N 1
2 N 51 N N 3 1 23. Starting with the graph of y graph that results from
(a) shifting 2 units downward
(b) shifting 2 units to the right N N N x 34. y e u cos u e t sin 2 t 38. y
40. y cos e x 42. y s1 xe 44. y ex
ex e
e cu e k tan sx s1
ee ae x
ce x 2e3x x b
d N N N N N N e 2 x cos x, N N N N N N e
N e x x, 46. y 0, 1 2x x
x N N N N 47. Find y if e x 2 y N N x tial equation y N N N 2y Ae x
y 0. Bxe 49. For what values of r does the function y tion y 6y e , write the equation of the N N N 8y x satisﬁes the differene rx satisfy the equa 0? N y y 51. If f x N y. 48. Show that the function y x
N 1, e N 50. Find the values of
x N ex 32. y 37. F t 45. y 2e x 22. y ex
N N  Find an equation of the tangent line to the curve at the
given point. 4 N x 3 N N Make a rough sketch of the graph of each function. Do
not use a calculator. Just use the graph given in Figure 2 and, if
necessary, the transformations of Section 1.3.
e x 2e x 43. y  19. y N e x ln x N 19–22 N 36. y 2 1 (b) e 2 9
N N 45–46 (b) ln x ln x N Differentiate the function. 41. y Solve each inequality for x.  N e 1/ u N 17–18 N 39. y 7 3x
3x 2x 35. f u places.
2 5x e
e xl2 e ax 33. y b Find the solution of the equation correct to four decimal  xl 30. lim e 3 N  e 3x
e 3x 28. lim 3x 2x 31. f x 12
N 2 ln x 11. e ax N N 31–44 1 xl 3x e
e xl2 10
ln 3 x e 3x
e 3x 29. lim e 3 3 N ln 2 lim e tan x 26. N 1 9. 2 ln x x3 1 xl (b) ln 5 0 4
2 Find the limit.  25. lim e ln x N (b) e 1 7. ln ln x 13–16 N x Solve each equation for x. 5. (a) 2 ln x N 25–30 (b) e 3 ln 2 N  8. e e graph that results from
(a) reﬂecting about the line y
(b) reﬂecting about the line x (b) ln se
s2 e x, ﬁnd the equation of the 24. Starting with the graph of y Simplify each expression.  465 Exercises 1. (a) How is the number e deﬁned? 2–4 ❙❙❙❙ for which y e x satisﬁes the equation y.
e 2 x, ﬁnd a formula for f n 52. Find the thousandth derivative of f x x.
xe x. 5E07(pp 462471) 466 ❙❙❙❙ 1/17/06 4:51 PM Page 466 CHAPTER 7 INVERSE FUNCTIONS 53. (a) Use the Intermediate Value Theorem to show that there is a (d) Use the graph to estimate the time after which the displacement is no more than 2 cm from equilibrium. root of the equation e x x 0.
(b) Use Newton’s method to ﬁnd the root of the equation in
part (a) correct to six decimal places. 59. Find the absolute maximum value of the function fx ; 54. Use a graph to ﬁnd an initial approximation (to one decimal 60. Find the absolute minimum value of the function t x place) to the root of the equation
e x2 x 3 63–65 e3x N (a) Find the mass remaining after 40 years.
(b) At what rate does the mass decay after 40 years?
(c) How long does it take for the mass to be reduced to 5 mg? 1 N 0.5 t ; N N e cos x N N N N ex 67. f x
N N N N 1
s2 fx N N N N N 3 x N N N N e 2 x 2 2 e x2 2 2 (a) Find the asymptote, maximum value, and inﬂection points
of f .
(b) What role does play in the shape of the curve?
(c) Illustrate by graphing four members of this family on the
same screen. ; 69–76 ; 58. An object is attached to the end of a vibrating spring and
its displacement from its equilibrium position is
y 8e t 2 sin 4 t, where t is measured in seconds and y is
measured in centimeters.
(a) Graph the displacement function together with the
functions y 8e t 2 and y
8e t 2. How are these
graphs related? Can you explain why?
(b) Use the graph to estimate the maximum value of the
displacement. Does it occur when the graph touches the
graph of y 8e t 2 ?
(c) What is the velocity of the object when it ﬁrst returns to its
equilibrium position? N occurs in probability and statistics, where it is called the
normal density function. The constant is called the mean and
the positive constant is called the standard deviation. For
simplicity, let’s scale the function so as to remove the factor
1 ( s2 ) and let’s analyze the special case where
0. So
we study the function kt where p t is the proportion of the population that knows the
rumor at time t and a and k are positive constants. [In Section 10.5 we will see that this is a reasonable model for p t .]
(a) Find lim t l p t .
(b) Find the rate of spread of the rumor.
(c) Graph p for the case a 10, k 0.5 with t measured in
hours. Use the graph to estimate how long it will take for
80% of the population to hear the rumor. N y equation
1
ae N 68. The family of bellshaped curves 57. Under certain circumstances a rumor spreads according to the 1 ex 2x e N 66. f x where the time t is measured in years since midyear 1980, so
0 t 20. Use a graph to estimate the time at which the
number of VCRs was increasing most rapidly. Then use derivatives to give a more accurate estimate. pt N e2 x 64. y  Draw a graph of f that shows all the important aspects of
the curve. Estimate the local maximum and minimum values and
then use calculus to ﬁnd these values exactly. Use a graph of f to
estimate the inﬂection points. holds in the United States with at least one VCR has been
modeled by the function
85
53 e increasing? ; 66–67 ; 56. For the period from 1980 to 2000, the percentage of house Vt 1x1 e 65. y ln 2 t 25 x Discuss the curve using the guidelines of Section 4.5.  63. y after t years is
24 e x 2e 62. On what interval is the function f x 55. If the initial mass of a sample of 90Sr is 24 mg, then the mass mt x e 3 x concave upward? 61. On what interval is the curve y Then use Newton’s method to ﬁnd the root correct to six decimal places. e x x, 0. x x3 e x. x  5 Evaluate the integral.
3x 69. y 71. y e s1 73. y 75. y N 0 e ex 72. y sec x e 1
x dx 74. y 76. ye esx
dx
sx
N x2 y e x dx x e 1 70. dx N N N N N N 0 xe
2 dx
tan x dx e1 x
dx
x2
x N sin e x d x
N N N N 5E07(pp 462471) 1/17/06 4:51 PM Page 467 S ECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS 78. Find f x if f x x 3e e x, y e 3x, and x 5 sin x, f 0 x 1. 1, and f 0 xaxis the region bounded by the curves y
and x 1. e,y ex 0, x 0, e x 2 N 83. If f x N N ,y 0,
ex 3 x e sin x
84. Evaluate lim
xl
x 1
1 82. f x
N N N e x, ﬁnd f
1 N 1 N N ex
ex
N N x2
2! x lim 0 and any xl N xn
n!
2.7. ex
xk for any positive integer k. ; 90. This exercise illustrates Exercise 89(c) for the case k . 85. Prove the second law of exponents [see (7)].
86. Prove the third law of exponents [see (7)].
87. (a) Show that e x  7.4* 1 (b) Use part (a) to show that e
(c) Use part (a) to show that 4. 1 x if x 0.
[Hint: Show that f x
1
ex
for x 0.]
2
(b) Deduce that 4 x01 e x dx e.
3 x2 positive integer n,  Find the inverse function of f . Check your answer by
graphing both f and f 1 on the same screen. N 1
2 2 ; 81–82 3 x 89. (a) Use mathematical induction to prove that for x yaxis the region bounded by the curves y
x 0, and x 1. ln x 1 (b) Use part (a) to improve the estimate of x01 e x dx given in
Exercise 87(b). 80. Find the volume of the solid obtained by rotating about the 81. f x 0, 2. 79. Find the volume of the solid obtained by rotating about the
x 467 88. (a) Use the inequality of Exercise 87(a) to show that, for 77. Find, correct to three decimal places, the area of the region bounded by the curves y ❙❙❙❙ x is increasing 10.
(a) Compare the rates of growth of f x
x 10 and t x
ex
by graphing both f and t in several viewing rectangles.
When does the graph of t ﬁnally surpass the graph of f ?
(b) Find a viewing rectangle that shows how the function
hx
e x x 10 behaves for large x.
(c) Find a number N such that
ex
x 10 10 10 whenever x N General Logarithmic and Exponential Functions
In this section we use the natural exponential and logarithmic functions to study exponential and logarithmic functions with base a 0. General Exponential Functions
If a 0 and r is any rational number, then by (4) and (7) in Section 7.3*,
ar e ln a r e r ln a Therefore, even for irrational numbers x, we deﬁne
ax 1 e x ln a Thus, for instance,
2 s3 e s3 ln 2 e 1.20 3.32 The function f x
a x is called the exponential function with base a. Notice that a x is
positive for all x because e x is positive for all x. 5E07(pp 462471) 468 ❙❙❙❙ 1/17/06 4:52 PM Page 468 CHAPTER 7 INVERSE FUNCTIONS Deﬁnition 1 allows us to extend one of the laws of logarithms. We know that
ln a r
r ln a when r is rational. But if we now let r be any real number we have, from
Deﬁnition 1,
ln a r ln e r ln a
r ln a
Thus
ln a r 2 r ln a for any real number r The general laws of exponents follow from Deﬁnition 1 together with the laws of exponents for e x.
3 1. a Laws of Exponents If x and y are real numbers and a, b
xy xy aa 2. a xy x aa y 3. a xy a 0, then xy 4. ab x a xb x Proof
1. Using Deﬁnition 1 and the laws of exponents for e x, we have ax y e x y ln a e x ln a e x ln ae y ln a y ln a a xa y 3. Using Equation 2 we obtain ax y e y ln a x e xy ln a e yx ln a
a xy The remaining proofs are left as exercises.
The differentiation formula for exponential functions is also a consequence of Deﬁnition 1:
d
ax
dx 4 Proof d
ax
dx a x ln a d
e x ln a
dx e x ln a d
x ln a
dx a x ln a
Notice that if a e, then ln e 1 and Formula 4 simpliﬁes to a formula that we
already know: d dx e x e x. In fact, the reason that the natural exponential function is
used more often than other exponential functions is that its differentiation formula is
simpler.
EXAMPLE 1 In Example 6 in Section 3.4 we considered a population of bacteria cells in a
homogeneous nutrient medium. We showed that if the population doubles every hour,
then the population after t hours is
n n0 2 t where n0 is the initial population. Now we can use (4) to compute the growth rate:
dn
dt n0 2 t ln 2 5E07(pp 462471) 1/17/06 4:52 PM Page 469 S ECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS For instance, if the initial population is n0
hours is
dn
dt ❙❙❙❙ 469 1000 cells, then the growth rate after two 1000 2 t ln 2 t2 t2 4000 ln 2 2773 cells h EXAMPLE 2 Combining Formula 4 with the Chain Rule, we have d
(10 x 2 )
dx 2 10 x ln 10 d
x2
dx 2 ln 10 x10 x 2 Exponential Graphs
If a 1, then ln a 0, so d dx a x a x ln a 0, which shows that y a x is increasing (see Figure 1). If 0 a 1, then ln a 0 and so y a x is decreasing (see Figure 2).
y y 1 1 0
x 0 x lim a ®=0, lim a ®=`
_` x ` x FIGURE 1 y=a®, a>1 x lim a ®=`, lim a ®=0
_` x` FIGURE 2 y=a®, 0<a<1 Notice from Figure 3 that as the base a gets larger, the exponential function grows more
rapidly (for x 0).
Figure 4 shows how the exponential function y 2 x compares with the power function
y x 2. The graphs intersect three times, but ultimately the exponential curve y 2 x
grows far more rapidly than the parabola y x 2. (See also Figure 5.)
” ’®
2
1 ” ’®
4
1 y 10® 4® e® y 2® y y=2® 1.5® y=2®
200 y=≈ 1® 100 y=≈
10 0 FIGURE 3 1 x 0 FIGURE 4 2 4 x 0 FIGURE 5 2 4 6 x 5E07(pp 462471) 470 ❙❙❙❙ 1/17/06 4:52 PM Page 470 CHAPTER 7 INVERSE FUNCTIONS In Chapter 10 we will show why exponential functions occur in the description of population growth and radioactive decay. Let’s look at human population growth. Table 1
shows data for the population of the world in the 20th century and Figure 6 shows the corresponding scatter plot.
TABLE 1 P Year Population
(millions) 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 6x10 ' 1900 1920 1940 1960 1980 2000 t FIGURE 6 Scatter plot for world population growth The pattern of the data points in Figure 6 suggests exponential growth, so we use a
graphing calculator with exponential regression capability to apply the method of least
squares and obtain the exponential model
P 0.008079266 1.013731 t Figure 7 shows the graph of this exponential function together with the original data
points. We see that the exponential curve ﬁts the data reasonably well. The period of
relatively slow population growth is explained by the two world wars and the Great
Depression of the 1930s.
P
6x10 ' FIGURE 7 Exponential model for
population growth 1900 1920 1940 1960 1980 Exponential Integrals
The integration formula that follows from Formula 4 is ya x dx ax
ln a C a 1 2000 t 5E07(pp 462471) 1/17/06 4:52 PM Page 471 S ECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS y EXAMPLE 3 5 0 5 2x
ln 2 x 2 dx 25
ln 2 0 20
ln 2 ❙❙❙❙ 471 31
ln 2 The Power Rule Versus the Exponential Rule
Now that we have deﬁned arbitrary powers of numbers, we are in a position to prove the
general version of the Power Rule, as promised in Section 3.3.
x n, then The Power Rule If n is any real number and f x nx n fx  If x 0, we can show that f 0
0 for
n 1 directly from the deﬁnition of a derivative. Proof Let y x n and use logarithmic differentiation:
ln x n Therefore y
y n
x Hence y n ln y  db
a
dx 2. d
fx
dx 3. d tx
a
dx 0
b bfx n xn
x nx n 0 1 b1 fx a t x ln a t x
tx , logarithmic differentiation can be used, as in the next example.
EXAMPLE 4 Differentiate y f y
x x (a and b are constants) 4. To ﬁnd d d x f x y n ln x You should distinguish carefully between the Power Rule d dx x n n x n 1 , where
the base is variable and the exponent is constant, and the rule for differentiating exponential functions d dx a x a x ln a , where the base is constant and the exponent is variable.
In general there are four cases for exponents and bases:
1.  Figure 8 illustrates Example 4 by showing
the graphs of f x
x sx and its derivative. 1 x sx. SOLUTION 1 Using logarithmic differentiation, we have
fª ln y 1
0 FIGURE 8 1 x ln x sx y
y sx y y s x ln x 1
x ln x 1
sx ln x
2 sx 1
2 sx
x sx 2 ln x
2 sx 5E07(pp 472481) 472 ❙❙❙❙ 1/17/06 5:14 PM Page 472 CHAPTER 7 INVERSE FUNCTIONS SOLUTION 2 Another method is to write x sx d sx
(x )
dx e ln x sx : d sx ln x
(e )
dx
e sx ln x x sx d
(s x ln x)
dx
2 ln x
2 sx (as in Solution 1) General Logarithmic Functions
If a 0 and a 1, then f x
a x is a onetoone function. Its inverse function is called
the logarithmic function with base a and is denoted by log a . Thus log a x 5 y &? a y x In particular, we see that
log e x ln x The cancellation equations for the inverse functions log a x and a x are
a log a x x and log a a x x Figure 9 shows the case where a 1. (The most important logarithmic functions have
base a 1.) The fact that y a x is a very rapidly increasing function for x 0 is
reﬂected in the fact that y log a x is a very slowly increasing function for x 1.
y 1 y y=log™ x
y=log£ x
y=log∞ x
y=log¡¸ x y=x y=a®, a>1
0
0 1 x x y=log a x, a>1
FIGURE 9 FIGURE 10 Figure 10 shows the graphs of y log a x with various values of the base a. Since
log a 1 0, the graphs of all logarithmic functions pass through the point 1, 0 .
The laws of logarithms are similar to those for the natural logarithm and can be deduced
from the laws of exponents (see Exercise 61).
The following formula shows that logarithms with any base can be expressed in terms
of the natural logarithm. 5E07(pp 472481) 1/17/06 5:14 PM Page 473 S ECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS 6 Change of Base Formula For any positive number a a ❙❙❙❙ 473 1 , we have ln x
ln a log a x log a x. Then, from (1), we have a y x. Taking natural logarithms of both
sides of this equation, we get y ln a ln x. Therefore Proof Let y ln x
ln a y Scientiﬁc calculators have a key for natural logarithms, so Formula 6 enables us to use
a calculator to compute a logarithm with any base (as shown in the next example). Similarly, Formula 6 allows us to graph any logarithmic function on a graphing calculator or
computer (see Exercises 14–16).
EXAMPLE 5 Evaluate log 8 5 correct to six decimal places.
SOLUTION Formula 6 gives ln 5
ln 8 log 8 5 0.773976 Formula 6 enables us to differentiate any logarithmic function. Since ln a is a constant,
we can differentiate as follows:
d
log a x
dx d
dx ln x
ln a 1d
ln x
ln a dx d
log a x
dx 7  NOTATION FOR LOGARITHMS
Most textbooks in calculus and the sciences, as
well as calculators, use the notation ln x for the
natural logarithm and log x for the “common
logarithm,” log 10 x. In the more advanced mathematical and scientiﬁc literature and in computer
languages, however, the notation log x usually
denotes the natural logarithm. 1
x ln a 1
x ln a EXAMPLE 6 d
log10 2
dx 2 1
d
2
sin x ln 10 dx 2 sin x cos x
sin x ln 10 sin x From Formula 7 we see one of the main reasons that natural logarithms (logarithms
with base e) are used in calculus: The differentiation formula is simplest when a e
because ln e 1. 5E07(pp 472481) 474 ❙❙❙❙ 1/17/06 5:15 PM Page 474 CHAPTER 7 INVERSE FUNCTIONS The Number e a s a Limit
ln x, then f x
1 x. Thus, f 1
We have shown that if f x
fact to express the number e as a limit.
From the deﬁnition of a derivative as a limit, we have
h
h f1 x
x ln 1 xl0 lim ln 1 f1 x lim f1 hl0 lim ln 1 xl0 Because f 1 lim f1 x
x xl0 1
ln 1
x lim xl0 1. We now use this f1 x 1x 1, we have
lim ln 1 x xl0 1x 1 Then, by Theorem 2.5.8 and the continuity of the exponential functions, we have
e e1 e lim x l 0 ln 1 e 8 x 1x lim e ln 1 x 1x xl0 lim 1 xl0 x lim 1 y x 2 0.1
0.01
0.001
0.0001
0.00001
0.000001
0.0000001
0.00000001 y=(1+x)!?® 1
0 x FIGURE 11 If we put n
sion for e is 9 1 x in Formula 8, then n l e lim nl 1 1x 1x Formula 8 is illustrated by the graph of the function y
table of values for small values of x. 3 x xl0 1 x 1x (1 in Figure 11 and a x)1/x 2.59374246
2.70481383
2.71692393
2.71814593
2.71826824
2.71828047
2.71828169
2.71828181 as x l 0 and so an alternative expres 1
n n 5E07(pp 472481) 1/17/06 5:15 PM Page 475 S ECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS  7.4* 475 Exercises 1. (a) Write an equation that deﬁnes a x when a is a positive num 17–18 ber and x is a real number.
(b) What is the domain of the function f x
a x?
(c) If a 1, what is the range of this function?
(d) Sketch the general shape of the graph of the exponential
function for each of the following cases.
(i) a 1
(ii) a 1
(iii) 0 a 1 Ca x whose graph is given. Find the exponential function f x  17. 18. y y (3, 24)
2 (1, 6) 2. (a) If a is a positive number and a 1, how is log a x deﬁned?
(b) What is the domain of the function f x
log a x ?
(c) What is the range of this function?
(d) If a 1, sketch the general shapes of the graphs of
y log a x and y a x with a common set of axes. 2 ”2, 9 ’ 0
N x N N N N N N N N 3. 5 s 7 4. 10 x
x 5. cos x
N N 7–10  2 6. x cos x N N N N N N N N N N N Evaluate the expression.
1 7. (a) log 10 1000 (b) log 2 16 8. (a) log 10 0.1 (b) log 8 320 9. (a) log 12 3
10. (a) log a 1
a (b) 10 N N N N N N N log10 4 log10 7 N 21–22 N N N Graph the given functions on a common screen. How are
;
these graphs related? t2 22. lim log10 x 2 5x xl3 N N N N N N N N N N 6
N N  11. y x 2, y e, 12. y 3 x, y 10 x, N N Find the limit.  tl N N 11–12 N x x 5 and
tx
5 x by graphing both functions in several viewing rectangles. Find all points of intersection of the graphs correct to
one decimal place. 21. lim 2
N N ; 20. Compare the rates of growth of the functions f x
log 8 5 (b) log 5 5 s 2 log 12 48 N 2 x and t x
2 are
drawn on a coordinate grid where the unit of measurement
is 1 inch, then at a distance 2 ft to the right of the origin the
height of the graph of f is 48 ft but the height of the graph
of t is about 265 mi.
(b) Suppose that the graph of y log 2 x is drawn on a coordinate grid where the unit of measurement is an inch. How
many miles to the right of the origin do we have to move
before the height of the curve reaches 3 ft? Write the expression as a power of e.  x 0 19. (a) Show that if the graphs of f x
3–6 N x N N x y 5, N y N 23. h t
25. y N N decimal places.
(a) log 12 e
(b) log 6 13.54 ; 14–16 N N N N (c) log 2 15. y log 1.5 x, 16. y ln x,
N N y log 4 x,
y y log 6 x, y log 50 x N y log 10 x, y log 10 x, y ln x, y e x, 10 x N N N y
N N log 3 x 2 28. y u 10 4 23 30. f x x
sin x 34. y ln x
x
N x N x N N N N x cos x ln x
N N 39. Find an equation of the tangent line to the curve y
N N N N point 1, 10 . 1 x ln x 38. y
N x
x 1x sin x 36. y ex x2 log10 32. y x 37. y
N 2 10 tan x 35. y x44x 24. t x
26. y 2u 27. f u 33. y log 8 x 3t 1x 5 31. y Use Formula 6 to graph the given functions on a common screen. How are these graphs related?
log 2 x, t3 29. f x  14. y Differentiate the function.  1
(10 ) x 20 ( 1 ) x,
3 y N y 23–38 x 13. Use Formula 6 to evaluate each logarithm correct to six N ❙❙❙❙ N N N 10 x at the 5E07(pp 472481) ❙❙❙❙ 476 1/17/06 5:15 PM Page 476 CHAPTER 7 INVERSE FUNCTIONS (b) If I0 8 and a 0.38, ﬁnd the rate of change of intensity
with respect to depth at a depth of 20 m.
(c) Using the values from part (b), ﬁnd the average light intensity between the surface and a depth of 20 m. x cos x, ﬁnd f x . Check that your answer is reasonable by comparing the graphs of f and f . ; 40. If f x
41–46
2 1 10 t dt 42. y log10 x
dx
x 44. y 46. y2 41. y 43. y 45. y3 1 N Evaluate the integral.  N sin cos d
N N N N N N 0 2u 4
x5 capacitor and releasing it suddenly when the ﬂash is set off.
The following data describe the charge remaining on the capacitor (measured in microcoulombs, C) at time t (measured in
seconds). 5 x dx 2x
x ; 57. The ﬂash unit on a camera operates by storing charge on a du 1 N dx N t
N 47. Find the area of the region bounded by the curves y y 5 x, x 1, and x N 2 x, 1. 10 x from x 0 to x 1 is
rotated about the xaxis. Find the volume of the resulting solid. 48. The region under the curve y x
; 49. Use a graph to ﬁnd the root of the equation 2 1 3 x correct to one decimal place. Then use this estimate as the initial
approximation in Newton’s method to ﬁnd the root correct to
six decimal places. 50. Find y if x y Q 100.00
81.87
67.03 0.06
0.08
0.10 54.88
44.93
36.76 (a) Use a graphing calculator or computer to ﬁnd an exponential model for the charge. (See Section 1.2.)
(b) The derivative Q t represents the electric current
(measured in microamperes, A) ﬂowing from the capacitor to the ﬂash bulb. Use part (a) to estimate the current
when t 0.04 s. Compare with the result of Example 2 in
Section 2.1. ; 58. The table gives the U.S. population from 1790 to 1860. y x. 51. Find the inverse function of f x
52. Calculate lim x l x t 0.00
0.02
0.04 N Q ln x 10 x 10 x 1. . 53. The geologist C. F. Richter deﬁned the magnitude of an earthquake to be log10 I S , where I is the intensity of the
quake (measured by the amplitude of a seismograph 100 km
from the epicenter) and S is the intensity of a “standard”
earthquake (where the amplitude is only 1 micron 10 4 cm).
The 1989 Loma Prieta earthquake that shook San Francisco
had a magnitude of 7.1 on the Richter scale. The 1906 San
Francisco earthquake was 16 times as intense. What was its
magnitude on the Richter scale?
54. A sound so faint that it can just be heard has intensity I0 10 12 watt m2 at a frequency of 1000 hertz (Hz). The
loudness, in decibels (dB), of a sound with intensity I is then
deﬁned to be L 10 log10 I I0 . Ampliﬁed rock music is measured at 120 dB, whereas the noise from a motordriven lawn
mower is measured at 106 dB. Find the ratio of the intensity of
the rock music to that of the mower.
55. Referring to Exercise 54, ﬁnd the rate of change of the loud ness with respect to the intensity when the sound is measured
at 50 dB (the level of ordinary conversation).
56. According to the BeerLambert Law, the light intensity at a depth of x meters below the surface of the ocean is I x
I0 a x,
where I0 is the light intensity at the surface and a is a constant
such that 0 a 1.
(a) Express the rate of change of I x with respect to x in terms
of I x . Year Population Year Population 1790
1800
1810
1820 3,929,000
5,308,000
7,240,000
9,639,000 1830
1840
1850
1860 12,861,000
17,063,000
23,192,000
31,443,000 (a) Use a graphing calculator or computer to ﬁt an exponential
function to the data. Graph the data points and the exponential model. How good is the ﬁt?
(b) Estimate the rates of population growth in 1800 and 1850
by averaging slopes of secant lines.
(c) Use the exponential model in part (a) to estimate the rates
of growth in 1800 and 1850. Compare these estimates with
the ones in part (b).
(d) Use the exponential model to predict the population in
1870. Compare with the actual population of 38,558,000.
Can you explain the discrepancy?
59. Prove the second law of exponents [see (3)].
60. Prove the fourth law of exponents [see (3)].
61. Deduce the following laws of logarithms from (3): (a) log a x y
(b) log a x y
(c) log a x y
62. Show that lim
nl log a x log a y
log a x log a y
y log a x
1 x
n n e x for any x 0. 5E07(pp 472481) 1/17/06 5:15 PM Page 477 SECTION 7.5 INVERSE TRIGONOMETRIC FUNCTIONS  7.5 ❙❙❙❙ 477 Inverse Trigonometric Functions
In this section we apply the ideas of Section 7.1 to ﬁnd the derivatives of the socalled
inverse trigonometric functions. We have a slight difﬁculty in this task: Because the
trigonometric functions are not onetoone, they do not have inverse functions. The difﬁculty is overcome by restricting the domains of these functions so that they become onetoone.
You can see from Figure 1 that the sine function y sin x is not onetoone (use the
Horizontal Line Test). But the function f x
sin x,
2x
2 (see Figure 2), is
onetoone. The inverse function of this restricted sine function f exists and is denoted by
sin 1 or arcsin. It is called the inverse sine function or the arcsine function.
y y y=sin x
_π
2
0 _π π
2 0 x π FIGURE 1 FIGURE 2 π
2 π 1 x y &? fy x we have sin 1x 1  sin 1x 1
sin x Thus, if 1 x y &? sin y x and 1, sin 1x is the number between EXAMPLE 1 Evaluate (a) sin 2 2 and y 2 2 whose sine is x. ( ) and (b) tan(arcsin 1 ).
3 11
2 SOLUTION (a) We have
sin 3
1
¨
2 œ2
„
FIGURE 3 () 11
2 6 1
because sin 6
6 lies between
2 and 2.
2 and
1
(b) Let
. Then we can draw a right triangle with angle
arcsin 1 , so sin
3
3
as in Figure 3 and deduce from the Pythagorean Theorem that the third side has length
s9 1 2 s2. This enables us to read from the triangle that tan(arcsin 3 )
1 tan 1
2 s2 π y=sin x, _ 2 ¯x¯ 2 Since the deﬁnition of an inverse function says that
f x 5E07(pp 472481) 478 ❙❙❙❙ 1/17/06 5:15 PM Page 478 CHAPTER 7 INVERSE FUNCTIONS The cancellation equations for inverse functions become, in this case, sin 2 1 sin x x
x 1 sin sin x  for
for x 2
1 x 2
1 We must be careful when using the ﬁrst cancellation equation because it is valid only
2, 2 . The following example shows how to proceed
when x lies in the interval
when x lies outside this interval.
EXAMPLE 2 Evaluate: (a) sin sin 1 0.6 1 (b) sin sin (c) sin 12 1 sin 2
3 SOLUTION 1 and 1, the second cancellation equation in (2) gives (a) Since 0.6 lies between sin sin 1 0.6
(b) Since 12 lies between 2 and
sin 1 0.6 2, the ﬁrst cancellation equation gives
sin 12 12 2, 2 , we can’t use the cancellation
(c) Since 2 3 does not lie in the interval
3 because 3
equation. Instead we note that sin 2 3
s3 2 and sin 1(s3 2)
2 and 2. Therefore
lies between
sin
y
π
2 _1 0 1 x _π
2 1 sin 2
3 1 sin s3
2 3 2, 2 , and its
The inverse sine function, sin 1, has domain 1, 1 and range
graph, shown in Figure 4, is obtained from that of the restricted sine function (Figure 2)
by reﬂection about the line y x.
We know that the sine function f is continuous, so the inverse sine function is also continuous. We also know from Section 3.5 that the sine function is differentiable, so the
inverse sine function is also differentiable. We could calculate the derivative of sin 1 by
the formula in Theorem 7.1.7, but since we know that sin 1 is differentiable, we can just
as easily calculate it by implicit differentiation as follows.
2y
2. Differentiating sin y x
Let y sin 1x. Then sin y x and
implicitly with respect to x, we obtain FIGURE 4 y=sin–! x=arcsin x and
Now cos y 0 since 2 y
cos y dy
dx 1 dy
dx cos y 1
cos y 2, so
s1 sin 2 y s1 x2 5E07(pp 472481) 1/17/06 5:15 PM Page 479 SECTION 7.5 INVERSE TRIGONOMETRIC FUNCTIONS dy
dx Therefore d
sin 1x
dx 3 4 1
cos y EXAMPLE 3 If f x sin 1 x2 ❙❙❙❙ 479 1
x2 s1
1 1 x2 s1 x 1 1 , ﬁnd (a) the domain of f , (b) f x , and (c) the domain of f . fª SOLUTION
_2 2 1, 1 , the domain of f is (a) Since the domain of the inverse sine function is f x x2 1 1 1 x2 x0 {x _4 2
s2 } x [ s2, s2 ] (b) Combining Formula 3 with the Chain Rule, we have FIGURE 5 s1 fx 1
x2 s1  The graphs of the function f of Example 3
and its derivative are shown in Figure 5. Notice
that f is not differentiable at 0 and this is consistent with the fact that the graph of f makes
a sudden jump at x 0. x4 1 2 d2
x
dx 1
2x 2 1
2x
s2 x 2 x 4 2x 1 (c) The domain of f is
x 1 x2 1 1 {x y
1
0 π
2 π x x2 x0
0 2
s2 } x ( s2, 0) (0, s2 ) The inverse cosine function is handled similarly. The restricted cosine function
fx
cos x, 0 x
, is onetoone (see Figure 6) and so it has an inverse function
denoted by cos 1 or arccos.
cos 1x cos y x cos x x for 0 cos cos 1x 4 y &? x for and 0 y The cancellation equations are FIGURE 6 y=cos x, 0¯x¯π
5 y cos 1 x
1 x 1 π The inverse cosine function, cos 1, has domain 1, 1 and range 0,
and is a continuous function whose graph is shown in Figure 7. Its derivative is given by π
2 _1 0 1 x 6 d
cos 1x
dx 1
s1 x2 1 x 1 FIGURE 7 y=cos –! x=arccos x Formula 6 can be proved by the same method as for Formula 3 and is left as Exercise 17. 5E07(pp 472481) 480 ❙❙❙❙ 1/17/06 5:15 PM Page 480 CHAPTER 7 INVERSE FUNCTIONS y _π
2 The tangent function can be made onetoone by restricting it to the interval
2, 2 . Thus, the inverse tangent function is deﬁned as the inverse of the function
fx
tan x,
2x
2. (See Figure 8.) It is denoted by tan 1 or arctan.
π
2 0 x tan 1x 7 y &? tan y x and y 2 2 EXAMPLE 4 Simplify the expression cos tan 1x . tan 1x. Then tan y x and
2
but, since tan y is known, it is easier to ﬁnd sec y ﬁrst: FIGURE 8 y=tan x, SOLUTION 1 Let y π
π
_ 2 <x< 2 sec2 y 1 tan2 y sec y s1 x2 since sec y 1
sec y cos y œ„„„„„
1+≈
x 2. We want to ﬁnd cos y x2 1 cos tan 1x Thus y 0 for 2 y 2 1
s1 x2 SOLUTION 2 Instead of using trigonometric identities as in Solution 1, it is perhaps easier
to use a diagram. If y tan 1x, then tan y x, and we can read from Figure 9 (which
illustrates the case y 0) that
1
cos tan 1x
cos y
s1 x 2 y
1
FIGURE 9
y The inverse tangent function, tan
Its graph is shown in Figure 10.
We know that π
2 1 lim tan x 0 xl arctan, has domain and 2 xl lim 2 and range 2, 2. tan x x _π
2 and so the lines x
2 are vertical asymptotes of the graph of tan. Since the graph of
tan 1 is obtained by reﬂecting the graph of the restricted tangent function about the line
y x, it follows that the lines y
2 and y
2 are horizontal asymptotes of the
graph of tan 1. This fact is expressed by the following limits: F IGURE 10 y=tan–! x=arctan x
8 lim tan 1x 1 EXAMPLE 5 Evaluate lim arctan x x l2 lim tan 1x 2 xl 2 xl . SOLUTION Since 1
x 2 as x l 2 l the ﬁrst equation in (8) gives
lim arctan x l2 1
x 2 2 2 5E07(pp 472481) 1/17/06 5:15 PM Page 481 ❙❙❙❙ SECTION 7.5 INVERSE TRIGONOMETRIC FUNCTIONS 481 Since tan is differentiable, tan 1 is also differentiable. To ﬁnd its derivative, let
y tan 1x. Then tan y x. Differentiating this latter equation implicitly with respect to
x, we have
dy
sec2 y
1
dx
dy
dx and so 1
sec2y 1
tan2y 1 d
tan 1x
dx 9 1
x2 1 1
x2 1 The remaining inverse trigonometric functions are not used as frequently and are summarized here.
10 y 0 FIGURE 11 π 2π x 1 &? csc y x and y 0, 2 ,3 2 sec 1x x 1 &? sec y x and y 0, 2 ,3 2 y _1 x y y csc 1x cot 1x x &? cot y x and y 0, The choice of intervals for y in the deﬁnitions of csc 1 and sec 1 is not universally
agreed upon. For instance, some authors use y
in the deﬁnition of
0, 2
2,
sec 1. [You can see from the graph of the secant function in Figure 11 that both this choice
and the one in (10) will work.] The reason for the choice in (10) is that the differentiation
formulas are simpler (see Exercise 79).
We collect in Table 11 the differentiation formulas for all of the inverse trigonometric
functions. The proofs of the formulas for the derivatives of csc 1, sec 1, and cot 1 are left
as Exercises 19–21. y=sec x
11 Table of Derivatives of Inverse Trigonometric Functions d
sin 1x
dx 1 d
cos 1x
dx
d
tan 1x
dx d
csc 1x
dx x2 s1 1 d
sec 1x
dx 1
1 x 2 1
x sx 2 d
cot 1x
dx x2 s1 1
x sx 2 1 1
1 1
x2 Each of these formulas can be combined with the Chain Rule. For instance, if u is a differentiable function of x, then
d
sin 1u
dx 1
s1 du
u 2 dx and d
tan 1u
dx 1
1 du
u 2 dx 5E07(pp 482493) 482 ❙❙❙❙ 1/17/06 4:41 PM Page 482 CHAPTER 7 INVERSE FUNCTIONS 1
and (b) f x
sin 1x EXAMPLE 6 Differentiate (a) y x tan 1s x. SOLUTION dy
dx (a) d
sin 1x
dx 1 sin 1x 1
sin x s1
1 2 x2
1
1
2
(sx ) 2 x tan 1sx x tan 1sx fx (b) d
sin 1x
dx 2 sx
21 x EXAMPLE 7 Prove the identity tan 1x 1 cot 1x 12 2. SOLUTION Although calculus is not needed to prove this identity, the proof using calculus
tan 1x cot 1x, then
is quite simple. If f x 1 fx 1 for all values of x. Therefore, f x
x 1. Then
C
Thus, tan 1x f1 cot 1x 1
x 2 0 x2 1 C, a constant. To determine the value of C, we put
tan 1 1 1 cot 1 4 4 2 2. Each of the formulas in Table 11 gives rise to an integration formula. The two most useful of these are the following: 12 y s1 13 yx EXAMPLE 8 Find y 1 14 0 4x 2 s1 1
x2 1
2 1 sin 1x dx tan 1x dx C C dx. SOLUTION If we write y 1 14 0 s1 4x 2 dx y 1 14 0 s1 2x 2 then the integral resembles Equation 12 and the substitution u dx
2 x is suggested. This 5E07(pp 482493) 1/17/06 4:42 PM Page 483 ❙❙❙❙ SECTION 7.5 INVERSE TRIGONOMETRIC FUNCTIONS gives du 2 dx, so dx y du 2. When x
1 14 1
2 dx 4x 2 s1 0 1
2 1 yx EXAMPLE 9 Evaluate 2 a2 y 0, u
du 12 [sin ( )
11
2 sin 1 1
2 ,u . So 1
2 u2 s1 0 1
4 0; when x 483 sin 1u 12
0 1
2 0 6 12 d x. SOLUTION To make the given integral more like Equation 13 we write dx y x2 a2 This suggests that we substitute u y dx
x2 1
a2 a2 dx
x2
a2 y a2 1
a2 1 x a. Then du yu a du
2 1
a 1 y y dx
x
a 2 1 d x a, dx
du u2 a du, and 1
tan 1u
a 1 C Thus, we have the formula
 One of the main uses of inverse trigonometric functions is that they often arise when
we integrate rational functions. yx 14 EXAMPLE 10 Find y x
x 4 9 a2  x
4 9 dx (s3 2) 1 2. (a) arctan (b) cos 1 (b) csc 1 1
2 y 7. tan(sin 2 x dx and we can use Equation 14 du
2 u 1 9 1
2 x2
3 1
tan
3 1 u
3 C C 1 ■ 2 (b) arcsin( 1 s2 ) 4. (a) sec 1s2 (b) arcsin 1 5. (a) arccos cos 2 (b) tan tan 1 tan 3 4 ■ 8. csc (arccos 5 )
3 5 (b) cos(arcsin 2 )
1 ■ ■  1 10. cos tan
■ 11. Prove that cos sin 1x
12–14 1 ( )) 12
3 9. sin( 2 tan 1s 2 ) 1 3. (a) tan 1s3 6. (a) tan C Exercises Find the exact value of each expression. 1. (a) sin x
a d x. 1
tan
6 1–10 1 3: yx  7.5 1
tan
a dx x 2 because then du SOLUTION We substitute u with a 1
2 ■ s1 ■ ■ x 2 for 2 ■ 1 tan
■ x 1 3 ■ ■ ■ ■ 1. Simplify the expression. 12. tan sin 1x
13. sin tan 1x
■ ■ ■ 14. csc arctan 2 x
■ ■ ■ ■ ■ ■ ■ 5E07(pp 482493) ❙❙❙❙ 484 ; 15–16 1/17/06 4:42 PM Page 484 CHAPTER 7 INVERSE FUNCTIONS 43–46 Graph the given functions on the same screen. How are
these graphs related?
 15. y sin x, 16. y
■ 2
2 tan x,
■ x ■ x ■ 2; ■ sin 1x; y 2; tan 1x; y ■ y ■ ■ 43. x y
■ lim sin 1x 17. Prove Formula 6 for the derivative of cos xl xl 1 x 46. lim tan xl ■ by the same ■ ■ ■ ■ 1 18. (a) Prove that sin x cos x
2.
(b) Use part (a) to prove Formula 6.
d
cot 1x
dx 1 d
csc 1x
dx 22–35 x2 1 d
sec 1x
dx 21. Prove that stan 23. y tan 1 1 25. y sin 1 2x x sx 2 1 29. y cos 1 31. y
33. h t e tan 35. y arccos
■ cot
ln b
a ■ e 1 ¨ x cos 1x
tan (x 1 s1 x
x ■ x2)
49. A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away from the base of the wall at a speed
of 2 ft s, how fast is the angle between the ladder and the wall
changing when the bottom of the ladder is 6 ft from the base of
the wall? a
a
0 ■ x ,a ■ ■ b 50. A lighthouse is located on a small island, 3 km away from the 0 ■ ■ ■ 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 ? ■ 36–37  Find the derivative of the function. Find the domains of
the function and its derivative. arcsin e x 36. f x
■ ■ ■ 37. t x ■ 1 38. Find y if tan ■ xy ■ ■ cos
■ 1 ■ 3 51–54 2x
■ ■ x sin 1 x4 C AS . ; 41–42  Find f x . Check that your answer is reasonable by comparing the graphs of f and f . ■ ■ e
■ x 42. f x arctan x
■ ■ ■ 52. y tan x
■ ■ ■ ■ 1 54. y 1
1 tan
tan 1 ■ ■ ■ x
x 1
1 ln x
■ ■ ■ x 2, ﬁnd t 2 . s16 3 arccos x 2 at the point 1, 41. f x x x
■ x 1 sin 53. y 40. Find an equation of the tangent line to the curve y Sketch the curve using the guidelines of Section 4.5. ■ ■ 39. If t x  51. y x 2 y. 1 d x2 s1 1t a cos x
,
b cos x
■ h sec 1 t 32. y t ■ lower edge is a distance d above the eye of an observer (as in
the ﬁgure). How far from the wall should the observer stand to
get the best view? (In other words, where should the observer
stand so as to maximize the angle subtended at his eye by the
painting?) x 2 arcsin x 30. y x
a 1 34. y ■ arctan x 2x 1 cot ■ 5 x ln arctan x 28. h t arctan cos x s1 26. f x 1 1 ■ 48. A painting in an art gallery has height h and is hung so that its 24. h x 27. H x ■ 2 A . x 2 ■ . 1 sx 1 ■ 3 possible.
22. y ■ ¨ P . 1
x sx 2 B Find the derivative of the function. Simplify where  ln x as to maximize the angle ? 1 20. Prove that x2
2x 2 47. Where should the point P be chosen on the line segment AB so method as for Formula 3. 19. Prove that 1 xl0 1
1 ■
■ 1 44. lim arccos 45. lim arctan e x ■ Find the limit.  ■ x2 x arcsin 1 ■ ■ ■ ■ ■ 55. If f x arctan cos 3 arcsin x , use the graphs of f , f , and
f to estimate the xcoordinates of the maximum and minimum
points and inﬂection points of f .
x c sin 1x.
What happens to the number of maxima and minima as c
changes? Graph several members of the family to illustrate
what you discover. ; 56. Investigate the family of curves given by f x 5E07(pp 482493) 1/17/06 4:43 PM Page 485 A PPLIED PROJECT WHERE TO SIT AT THE MOVIES 57. Find the most general antiderivative of the function fx 2x 58. Find f x if f x
59–70 59. y 31 6 s3 2 and f dx
1 16x 2 62. y s1 sin 1x
dx
s1 x 2 64. y 9
dx
9 66.
68. y x sx 63. y 65. yx 67. y s1 0 x
2 dt t2 t2
t6 dt dx y sx 1
■ ■ ■ ■ t2 0 1 1
2 x2 y
xy 2. sin x
dx
cos2x tan 1x
dx
x2 y s1 ■ ■ dx sin 1 2 4 dx 2 sin 1x 2x e 4x
■ x
a sin sin 1x .
sin 1 sin x , , and ﬁnd its 77. Use the method of Example 7 to prove the identity 1 e 4 (b) Sketch the graph of the function t x
x
.
cos x
(c) Show that t x
.
cos x
(d) Sketch the graph of h x
cos 1 sin x , x
derivative. 4t 2 1 4 76. (a) Sketch the graph of the function f x cos 1 2x 2 1 x 0 78. Prove the identity dx
■ 71. Use the method of Example 9 to show that, if a y sa x
1 2 and (a) arctan 1 arctan 1
2
3
(b) 2 arctan 1 arctan 1
3
7 dt dt 2 0 70. x
■ arctan if the left side lies between y1 s3 4 y 0 0. arctan y 75. Use the result of Exercise 74 to prove the following: 4 1 y s1 12 4 485 1, arctan x 60. 61. ■ 1 x2 74. Prove that, for xy 12 Evaluate the integral.  12 69. 4 x 51 2 ❙❙❙❙ ■ arcsin ■ 0, 1
1 2 arctan s x 2 ? sec y x and
sec 1x &
y
0, 2
2, . Show that with this deﬁnition,
we have (instead of the formula given in Exercise 20) 79. Some authors deﬁne y C d
sec 1x
dx 72. The region under the curve y 1 s x 2 4 from x 0 to
x 2 is rotated about the xaxis. Find the volume of the resulting solid. x
x 1
x sx 2 80. Let f x x arctan 1 x if x
(a) Is f continuous at 0?
(b) Is f differentiable at 0? 1
73. Evaluate x0 sin 1x dx by interpreting it as an area and inte grating with respect to y instead of x. 1 0 and f 0 x 1
0. APPLIED PROJECT
C AS Where to Sit at the Movies
A movie theater has a screen that is positioned 10 ft off the ﬂoor and is 25 ft high. The ﬁrst row
of seats is placed 9 ft from the screen and the rows are set 3 ft apart. The ﬂoor of the seating area
is inclined at an angle of
20 above the horizontal and the distance up the incline that you sit
is x. The theater has 21 rows of seats, so 0 x 60. Suppose you decide that the best place to
sit is in the row where the angle subtended by the screen at your eyes is a maximum. Let’s also
suppose that your eyes are 4 ft above the ﬂoor, as shown in the ﬁgure. (In Exercise 48 in Section 7.5 we looked at a simpler version of this problem, where the ﬂoor is horizontal, but this
project involves a more complicated situation and requires technology.)
1. Show that 25 ft
¨ arccos 4 ft a2 b 2 625
2ab x where 10 ft a2 9 x cos 2 31 and b2 9 x cos 2 x sin 2 x sin å
9 ft 6 2 5E07(pp 482493) 486 ❙❙❙❙ 1/17/06 4:43 PM Page 486 CHAPTER 7 INVERSE FUNCTIONS 2. Use a graph of as a function of x to estimate the value of x that maximizes . In which row
should you sit? What is the viewing angle in this row? 3. Use your computer algebra system to differentiate of the equation d d x and ﬁnd a numerical value for the root
0. Does this value conﬁrm your result in Problem 2? 4. Use the graph of to estimate the average value of on the interval 0 x 60. Then use
your CAS to compute the average value. Compare with the maximum and minimum values
of .  7.6 Hyperbolic Functions
Certain combinations of the exponential functions e x and e x arise so frequently in mathematics and its applications that they deserve to be given special names. In many ways they
are analogous to the trigonometric functions, and they have the same relationship to the
hyperbola that the trigonometric functions have to the circle. For this reason they are collectively called hyperbolic functions and individually called hyperbolic sine, hyperbolic
cosine, and so on.
Definition of the Hyperbolic Functions sinh x cosh x tanh x ex e x e x csch x ex
2 sinh x
cosh x sech x 1
cosh x coth x 2 1
sinh x cosh x
sinh x The graphs of hyperbolic sine and cosine can be sketched using graphical addition as
in Figures 1 and 2.
y y y=cosh x
1
y= ´
2 y y=1 y=sinh x
0 1 x
1 y=_ e –®
2 0 1 1
y= e–®
2 y= ´
2 x y=_1
0 x FIGURE 1 FIGURE 2 FIGURE 3 1
1
y=sinh x= 2 ´ 2 e–® 1
1
y=cosh x= 2 ´+ 2 e–® y=tanh x 5E07(pp 482493) 1/17/06 4:44 PM Page 487 SECTION 7.6 HYPERBOLIC FUNCTIONS y 0 FIGURE 4 A catenary y=c+a cosh (x/a) x ❙❙❙❙ 487 Note that sinh has domain and range , while cosh has domain and range 1, .
The graph of tanh is shown in Figure 3. It has the horizontal asymptotes y
1. (See
Exercise 23.)
Some of the mathematical uses of hyperbolic functions will be seen in Chapter 8.
Applications to science and engineering occur whenever an entity such as light, velocity,
electricity, or radioactivity is gradually absorbed or extinguished, for the decay can be represented by hyperbolic functions. The most famous application is the use of hyperbolic
cosine to describe the shape of a hanging wire. It can be proved that if a heavy ﬂexible
cable (such as a telephone or power line) is suspended between two points at the same
height, then it takes the shape of a curve with equation y c a cosh x a called a catenary (see Figure 4). (The Latin word catena means “chain.”)
The hyperbolic functions satisfy a number of identities that are similar to wellknown
trigonometric identities. We list some of them here and leave most of the proofs to the
exercises.
Hyperbolic Identities sinh x sinh x cosh x cosh x cosh2x sinh2x sinh x y sinh x cosh y cosh x sinh y cosh x y cosh x cosh y sinh x sinh y EXAMPLE 1 Prove (a) cosh2x 1 tanh2x 1 sinh2x sech2x tanh2x 1 and (b) 1 sech2x. SOLUTION (a) cosh2x ex sinh2x e x 2 ex 2
e 2x
4
4 e x 2 2 2
4 e 2x e 2x 2
4 e 2x 1 (b) We start with the identity proved in part (a):
cosh2x sinh2x 1 If we divide both sides by cosh2x, we get
1
or 1 sinh2x
cosh2x 1
cosh2x tanh2x sech2x The identity proved in Example 1(a) gives a clue to the reason for the name “hyperbolic” functions:
If t is any real number, then the point P cos t, sin t lies on the unit circle x 2 y 2 1
because cos2t sin2t 1. In fact, t can be interpreted as the radian measure of POQ 5E07(pp 482493) 488 ❙❙❙❙ 1/17/06 4:44 PM Page 488 CHAPTER 7 INVERSE FUNCTIONS y
P(cos t, sin t) O Q x ≈ +¥=1 in Figure 5. For this reason the trigonometric functions are sometimes called circular
functions.
Likewise, if t is any real number, then the point P cosh t, sinh t lies on the right branch
of the hyperbola x 2 y 2 1 because cosh2t sinh2t 1 and cosh t 1. This time, t
does not represent the measure of an angle. However, it turns out that t represents twice
the area of the shaded hyperbolic sector in Figure 6, just as in the trigonometric case t represents twice the area of the shaded circular sector in Figure 5.
The derivatives of the hyperbolic functions are easily computed. For example, FIGURE 5
y d
sinh x
dx ex d
dx e x ex 2 e x cosh x 2 P(cosh t, sinh t) We list the differentiation formulas for the hyperbolic functions as Table 1. The remaining
proofs are left as exercises. Note the analogy with the differentiation formulas for trigonometric functions, but beware that the signs are different in some cases.
0 x 1 Derivatives of Hyperbolic Functions d
sinh x
dx d
csch x
dx csch x coth x sinh x d
sech x
dx sech x tanh x d
tanh x
dx FIGURE 6 cosh x d
cosh x
dx ≈¥=1 sech2 x d
coth x
dx csch2 x EXAMPLE 2 Any of the differentiation rules in Table 1 can be combined with the Chain Rule. For instance,
d
(cosh s x )
dx sinh s x d
sx
dx sinh s x
2 sx Inverse Hyperbolic Functions
You can see from Figures 1 and 3 that sinh and tanh are onetoone functions and so they
have inverse functions denoted by sinh 1 and tanh 1. Figure 2 shows that cosh is not onetoone, but when restricted to the domain 0, it becomes onetoone. The inverse hyperbolic cosine function is deﬁned as the inverse of this restricted function. y sinh 1x y cosh 1x &? cosh y x and y y 2 tanh 1x x &? sinh y &? tanh y x
0 The remaining inverse hyperbolic functions are deﬁned similarly (see Exercise 28). 5E07(pp 482493) 1/17/06 4:44 PM Page 489 S ECTION 7.6 HYPERBOLIC FUNCTIONS We can sketch the graphs of sinh 1, cosh 1, and tanh
Figures 1, 2, and 3. 1 ❙❙❙❙ 489 in Figures 7, 8, and 9 by using
y y
y 0
0 x _1
0 FIGURE 7 y=sinh –! x
domain=R range=R 1 x x 1 FIGURE 8 y=cosh –! x
domain=[1, `} range=[0, `} FIGURE 9 y=tanh –! x
domain=(_1, 1) range=R Since the hyperbolic functions are deﬁned in terms of exponential functions, it’s not
surprising to learn that the inverse hyperbolic functions can be expressed in terms of logarithms. In particular, we have:
3 ln( x sx 2 1) x 4 cosh 1x ln( x sx 2 1) x 5  Formula 3 is proved in Example 3. The
proofs of Formulas 4 and 5 are requested in
Exercises 26 and 27. sinh 1x tanh 1x 1
2 ln( x EXAMPLE 3 Show that sinh 1x
SOLUTION Let y 1
1 ln x
x 1 1
x 1 1 ). sx 2 sinh 1x. Then
x ey sinh y y 2 ey 2x e 2y so e 2 xe y e y 0 or, multiplying by e y,
1 0 This is really a quadratic equation in e y:
ey 2 2x e y 1 0 x sx 2 1 0 (because x sx 2 1 ). Thus, the minus sign Solving by the quadratic formula, we get
ey
Note that e y 0, but x s x 2
is inadmissible and we have s4 x 2
2 2x 1
ey x 4 sx 2 1 5E07(pp 482493) 490 ❙❙❙❙ 1/17/06 4:45 PM Page 490 CHAPTER 7 INVERSE FUNCTIONS ln( x ln e y y Therefore 1) sx 2 (See Exercise 25 for another method.)
6 Derivatives of Inverse Hyperbolic Functions d
sinh 1x
dx  Notice that the formulas for the derivatives
of tanh 1x and coth 1x appear to be identical.
But the domains of these functions have no num1,
bers in common: tanh 1x is deﬁned for x
1.
whereas coth 1x is deﬁned for x 1
x2 d
cosh 1x
dx d
csch 1x
dx 1 s1 d
sech 1x
dx 1
x s1 d
coth 1x
dx 1 1
sx d
tanh 1x
dx 2 1
1 x 2 1
x sx 2 1 x2 x2 1 The inverse hyperbolic functions are all differentiable because the hyperbolic functions
are differentiable. The formulas in Table 6 can be proved either by the method for inverse
functions or by differentiating Formulas 3, 4, and 5.
EXAMPLE 4 Prove that d
sinh 1x
dx 1
x2 s1 sinh 1x. Then sinh y
with respect to x, we get
SOLUTION 1 Let y x. If we differentiate this equation implicitly cosh y
Since cosh2 y sinh2 y 1 and cosh y
dy
dx 1
cosh y . dy
dx 1 0, we have cosh y
1
sinh 2 y s1 sinh2 y, so s1
1 s1 x2 SOLUTION 2 From Equation 3 (proved in Example 3), we have d
sinh 1x
dx d
ln( x
dx 1) sx 2 x 1
sx 2 d
(x
1 dx x 1
sx 2 1 (x
1
sx 2 sx 2
sx 2
1 1 1x
1 ) sx 2 1) sx 2
x
sx 2
1 1 5E07(pp 482493) 1/17/06 4:45 PM Page 491 ❙❙❙❙ SECTION 7.6 HYPERBOLIC FUNCTIONS EXAMPLE 5 Find d
tanh
dx 1 491 sin x . SOLUTION Using Table 6 and the Chain Rule, we have d
tanh
dx 1 sin x 1
sin x 1 y dx 1 x2 s1 0 d
sin x
dx 1
cos x
sin2x 1
EXAMPLE 6 Evaluate 2 cos x
cos2x sec x . SOLUTION Using Table 6 (or Example 4) we know that an antiderivative of 1 s1 x 2 is 1 sinh x. Therefore y dx 1 0 1 sinh
ln(1  7.6
1–6 1
s2 ) (from Equation 3) Exercises
13. coth2x 1 14. tanh x Find the numerical value of each expression.  1
0 sinh 1x x2 s1 csch2x y 1. (a) sinh 0 (b) cosh 0 2. (a) tanh 0 (b) tanh 1 3. (a) sinh ln 2 (b) sinh 2 15. sinh 2 x 2 sinh x cosh x 4. (a) cosh 3 (b) cosh ln 3 16. cosh 2 x cosh2x 5. (a) sech 0 (b) cosh 1 1 17. tanh ln x x2
x2 1 6. (a) sinh 1
■ ■ 7–19  (b) sinh 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ 18. ■ Prove the identity. ■ x
cosh x
(This shows that cosh is an even function.) sinh x e 11. sinh x y ■ 21. If tanh x ■ ■ ■ sinh n x
■ ■ ■ ■ ■ 3
4 , ﬁnd the values of the other hyperbolic functions 4
5 , ﬁnd the values of the other hyperbolic functions at x.
x sinh x cosh y 22. (a) Use the graphs of sinh, cosh, and tanh in Figures 1–3 to cosh x sinh y ;
y e 2x at x. ex 10. cosh x ■ 20. If sinh x 8. cosh sinh x tanh x
tanh x 1
1 sinh x n cosh n x
(n any real number) x
sinh x
(This shows that sinh is an odd function.) 9. cosh x 1
1 sinh2x 19. cosh x 7. sinh 12. cosh x tanh x tanh y
1 tanh x tanh y cosh x cosh y sinh x sinh y draw the graphs of csch, sech, and coth.
(b) Check the graphs that you sketched in part (a) by using a
graphing device to produce them. ■ 5E07(pp 482493) ❙❙❙❙ 492 1/17/06 4:46 PM Page 492 CHAPTER 7 INVERSE FUNCTIONS 23. Use the deﬁnitions of the hyperbolic functions to ﬁnd each of the following limits.
(a) lim tanh x (b) lim tanh x (c) lim sinh x (d) lim sinh x (e) lim sech x (f) lim coth x (g) lim coth x family of functions y
change as a varies? (h) lim coth x xl 49. A telephone line hangs between two poles 14 m apart in the xl xl shape of the catenary y 20 cosh x 20
15, where x and y
are measured in meters.
(a) Find the slope of this curve where it meets the right pole.
(b) Find the angle between the line and the pole. xl xl xl xl0 a cosh x a . How does the graph xl0 y (i) lim csch x
xl ¨ 5 24. Prove the formulas given in Table 1 for the derivatives of the functions (a) cosh, (b) tanh, (c) csch, (d) sech, and (e) coth.
25. Give an alternative solution to Example 3 by letting 0 _7 y sinh 1x and then using Exercise 9 and Example 1(a)
with x replaced by y. 50. Using principles from physics it can be shown that when a 26. Prove Equation 4. cable is hung between two poles, it takes the shape of a curve
y f x that satisﬁes the differential equation 27. Prove Equation 5 using (a) the method of Example 3 and (b) Exercise 18 with x replaced by y. d2y
dx 2 28. For each of the following functions (i) give a deﬁnition like those in (2), (ii) sketch the graph, and (iii) ﬁnd a formula similar to Equation 3.
(a) csch 1
(b) sech 1
(c) coth 1
following functions.
(a) cosh 1
(b) tanh
(d) sech 1
(e) coth
 30. f x 1 y Find the derivative. sinh x 35. G x 1
1 37. h t 2 32. t x sinh x y 34. F x sinh x tanh x cosh x
cosh x 36. f t e t sech t t2 38. f t ln sinh t tanh e t 39. H t 40. y e 43. y x tanh 1x 45. y x sinh 46. y sech 1s1 1 xl 53. At what point of the curve y x3 ■ ■ ■ 55–63 x tan , show that sec cosh x. Evaluate the integral.
2 55. 0 y sinh x cosh x dx 57. y 59. y cosh x 1
■  ln sec x2 s9
x 2, cosh x does the tangent have slope 1?
54. If x coth 1s x 2 47. y
■ 1 2x x2 ln s1 B cosh m x sinh x
.
ex 52. Evaluate lim tanh 1sx 44. y x 2 sinh A sinh m x satisﬁes the differential equation y
m 2 y.
(b) Find y y x such that y
9y, y 0
4,
and y 0
6. sinh cosh x 42. y cosh 3x 41. y tx
T 51. (a) Show that any function of the form 2 coth s1 T
cosh
t fx is a solution of this differential equation. x cosh x 33. h x 2 dy
dx 1 1 (c) csch 1 tanh 4 x 31. f x t
T where is the linear density of the cable, t is the acceleration
due to gravity, and T is the tension in the cable at its lowest
point, and the coordinate system is chosen appropriately. Verify
that the function 29. Prove the formulas given in Table 6 for the derivatives of the 30–47 7x ■ ■ ■ ■ ■ ■ ■ y sinh 1 58. sinh sx
dx
sx 56. y tanh x dx 60. y2 ; 48. A ﬂexible cable always hangs in the shape of a catenary
y c a cosh x a , where c and a are constants and a 0
(see Figure 4 and Exercise 50). Graph several members of the cosh x
2 1 dx 4 x dx sech2x
dx
tanh x 5E07(pp 482493) 1/17/06 4:47 PM Page 493 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE 61.
63.
■ y 1
dt
st 2 9 6 4 y1 e 62. y dt
s16 t 2 1 0 1 ■ ■ is A t dx 1
2 t. [Hint: First show that ■ ■ ■ ■ ■ ■ ■ sinh c x between x 0 and x 1 is equal to 1. and then verify that A t 1
2 cosh t 1 sx 2 1 dx . 67. Show that if a 0 and b 0, then there exist numbers
and such that ae x be x equals either sinh x
or
. In other words, almost every function of the
cosh x
form f x
ae x be x is a shifted and stretched hyperbolic
sine or cosine function. ; 65. (a) Use Newton’s method or a graphing device to ﬁnd approximate solutions of the equation cosh 2 x 1 sinh x.
(b) Estimate the area of the region bounded by the curves
y cosh 2 x and y 1 sinh x.  7.7 y sinh t cosh t ■ ; 64. Estimate the value of the number c such that the area under the
curve y 1
2 At
■ 493 66. Show that the area of the shaded hyperbolic sector in Figure 6 dt x e 2x ❙❙❙❙ Indeterminate Forms and L’ Hospital ’s Rule
Suppose we are trying to analyze the behavior of the function
ln x
x1 Fx Although F is not deﬁned when x 1, we need to know how F behaves near 1. In particular, we would like to know the value of the limit
lim 1 x l1 ln x
x1 In computing this limit we can’t apply Law 5 of limits (the limit of a quotient is the quotient of the limits, see Section 2.3) because the limit of the denominator is 0. In fact,
although the limit in (1) exists, its value is not obvious because both numerator and denominator approach 0 and 0 is not deﬁned.
0
In general, if we have a limit of the form
lim xla fx
tx where both f x l 0 and t x l 0 as x l a, then this limit may or may not exist and is
called an indeterminate form of type 0 . We met some limits of this type in Chapter 2. For
0
rational functions, we can cancel common factors:
lim
x l1 x2
x2 x
1 lim
x l1 xx 1
x 1x 1 lim
x l1 x
x 1 1
2 We used a geometric argument to show that
lim xl0 sin x
x 1 But these methods do not work for limits such as (1), so in this section we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms. 5E07(pp 494503) 494 ❙❙❙❙ 1/17/06 5:01 PM Page 494 CHAPTER 7 INVERSE FUNCTIONS Another situation in which a limit is not obvious occurs when we look for a horizontal
asymptote of F and need to evaluate the limit
ln x
x1 lim 2 xl It isn’t obvious how to evaluate this limit because both numerator and denominator
become large as x l . There is a struggle between numerator and denominator. If the
numerator wins, the limit will be ; if the denominator wins, the answer will be 0. Or there
may be some compromise, in which case the answer may be some ﬁnite positive number.
In general, if we have a limit of the form
lim xla where both f x l (or
) and t x l (or
), then the limit may or may not exist
and is called an indeterminate form of type
. We saw in Section 4.4 that this type of
limit can be evaluated for certain functions, including rational functions, by dividing
numerator and denominator by the highest power of x that occurs in the denominator. For
instance,
1
1
2
x
1
x2
10
1
lim
lim
2
xl
xl
2x
1
1
20
2
2
2
x y f This method does not work for limits such as (2), but l’Hospital’s Rule also applies to this
type of indeterminate form. g 0 a x y L’Hospital’s Rule Suppose f and t are differentiable and t x
possibly at a). Suppose that y=m¡(xa) lim f x xla or that
y=m™(xa)
0 a x 0 lim f x and lim a
a m1
m2 which is the ratio of their derivatives. This suggests that
lim xla fx
tx lim xla fx
tx lim t x xla fx
tx if the limit on the right side exists (or is
NOTE 1 lim t x ■ .) Then fx
tx
). ■ NOTE 3 or 0 ■ NOTE 2 lim xla 0 near a (except xla (In other words, we have an indeterminate form of type 0 or
0
xla  Figure 1 suggests visually why l’Hospital’s
Rule might be true. The ﬁrst graph shows two
differentiable functions f and t, each of which
approaches 0 as x l a. If we were to zoom in
toward the point a, 0 , the graphs would start
to look almost linear. But if the functions actually
were linear, as in the second graph, then their
ratio would be and xla FIGURE 1 m1 x
m2 x fx
tx L’Hospital’s Rule says that the limit of a quotient of functions is equal to the
limit of the quotient of their derivatives, provided that the given conditions are satisﬁed. It
is especially important to verify the conditions regarding the limits of f and t before using
l’Hospital’s Rule.
L’Hospital’s Rule is also valid for onesided limits and for limits at inﬁnity or
negative inﬁnity; that is, “ x l a ” can be replaced by any of the symbols x l a , x l a ,
x l , or x l
.
ta For the special case in which f a
ta
0, f and t are continuous, and
0, it is easy to see why l’Hospital’s Rule is true. In fact, using the alternative form 5E07(pp 494503) 1/17/06 5:02 PM Page 495 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE  L’Hospital’s Rule is named after a
French nobleman, the Marquis de l’Hospital
(1661–1704), but was discovered by a Swiss
mathematician, John Bernoulli (1667–1748).
See Exercise 78 for the example that the
Marquis used to illustrate his rule. See the
project on page 504 for further historical details. ❙ ❙❙❙ 495 of the deﬁnition of a derivative, we have lim xla fx
tx fx
x
tx
lim
xla
x lim fa
ta xla fx
x
lim
xla t x
x
lim xla fa
a
ta
a fa
a
ta
a
fx
tx lim xla fa
ta fx
tx The general version of l’Hospital’s Rule for the indeterminate form 0 is somewhat more
0
difﬁcult and its proof is deferred to the end of this section. The proof for the indeterminate
form
can be found in more advanced books.
EXAMPLE 1 Find lim
x l1 ln x
.
x1 SOLUTION Since lim ln x ln 1 x l1 0 and lim x
x l1 1 0 we can apply l’Hospital’s Rule:  Notice that when using l’Hospital’s Rule we
differentiate the numerator and denominator
separately. We do not use the Quotient Rule. ln x
x1 lim
x l1 d
ln x
dx
lim
x l1 d
x1
dx
lim
x l1 EXAMPLE 2 Calculate lim
 The graph of the function of Example 2 is
shown in Figure 2. We have noticed previously
that exponential functions grow far more rapidly
than power functions, so the result of Example 2
is not unexpected. See also Exercise 87. xl SOLUTION We have lim x l e x lim xl FIGURE 2 1x
1 1 and lim x l x 2 e
x2 , so l’Hospital’s Rule gives dx
e
dx
lim
xl
d2
x
dx lim xl ex
2x Since e x l and 2 x l as x l , the limit on the right side is also indeterminate,
but a second application of l’Hospital’s Rule gives y= ´
≈
0 x l1 ex
.
x2 x 20 1
x lim 10 lim xl ex
x2 lim xl ex
2x lim xl ex
2 5E07(pp 494503) 496 ❙❙❙❙ 1/17/06 5:03 PM Page 496 CHAPTER 7 INVERSE FUNCTIONS  The graph of the function of Example 3 is
shown in Figure 3. We have discussed previously
the slow growth of logarithms, so it isn’t surprising that this ratio approaches 0 as x l . See
also Exercise 88. E XAMPLE 3 Calculate lim
xl SOLUTION Since ln x l ln x
.
3
sx
3
and sx l 2 as x l , l’Hospital’s Rule applies: lim xl y= ln x
Œ„
x ln x
3
sx lim xl 1
3 1x
x 23
0 0 Notice that the limit on the right side is now indeterminate of type 0 . But instead of
applying l’Hospital’s Rule a second time as we did in Example 2, we simplify the
expression and see that a second application is unnecessary: 10,000 lim _1 xl ln x
3
sx lim 1
3 xl 1x
x 23 lim xl 3
sx 0 3 FIGURE 3 EXAMPLE 4 Find lim xl0 tan x
x3 x . [See Exercise 36(d) in Section 2.2.]
x l 0 and x 3 l 0 as x l 0, we use l’Hospital’s SOLUTION Noting that both tan x Rule:
lim xl0 tan x
x3 x lim xl0 sec2x 1
3x 2
0 Since the limit on the right side is still indeterminate of type 0 , we apply l’Hospital’s
Rule again:
sec2x 1
2 sec2x tan x
lim
lim
xl0
xl0
3x 2
6x  The graph in Figure 4 gives visual conﬁrmation of the result of Example 4. If we were to
zoom in too far, however, we would get an
inaccurate graph because tan x is close to x
when x is small. See Exercise 36(d) in
Section 2.2. Because lim x l 0 sec2 x
lim xl0 1 0 2 sec2x tan x
6x 1
tan x
lim sec2 x lim
xl0
3 xl0
x 1
tan x
lim
3 xl0 x We can evaluate this last limit either by using l’Hospital’s Rule a third time or by
writing tan x as sin x cos x and making use of our knowledge of trigonometric limits.
Putting together all the steps, we get
y= _1 1, we simplify the calculation by writing tan x x
˛ lim xl0 tan x
x3 x lim xl0 sec 2 x 1
3x2 lim xl0 2 sec 2 x tan x
6x 1 1
tan x
lim
3 xl0 x FIGURE 4 EXAMPLE 5 Find lim
xl 1
sec 2 x
lim
3 xl0 1 1
3 sin x
.
1 cos x SOLUTION If we blindly attempted to use l’Hospital’s Rule, we would get  lim xl sin x
1 cos x lim xl cos x
sin x This is wrong! Although the numerator sin x l 0 as x l
, notice that the denominator 1 cos x does not approach 0, so l’Hospital’s Rule can’t be applied here. 5E07(pp 494503) 1/17/06 5:03 PM Page 497 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ❙ ❙❙❙ 497 The required limit is, in fact, easy to ﬁnd because the function is continuous and the
denominator is nonzero at :
lim xl 1 sin x
cos x sin
cos 1 0
1 0 1 Example 5 shows what can go wrong if you use l’Hospital’s Rule without thinking.
Other limits can be found using l’Hospital’s Rule but are more easily found by other methods. (See Examples 3 and 5 in Section 2.3, Example 3 in Section 2.6, and the discussion
at the beginning of this section.) So when evaluating any limit, you should consider other
methods before using l’Hospital’s Rule. Indeterminate Products
If lim x l a f x
0 and lim x l a t x
(or
), then it isn’t clear what the value of
lim x l a f x t x , if any, will be. There is a struggle between f and t. If f wins, the answer
will be 0; if t wins, the answer will be (or
). Or there may be a compromise where
the answer is a ﬁnite nonzero number. This kind of limit is called an indeterminate form
of type 0
. We can deal with it by writing the product f t as a quotient:
f
1t ft or ft t
1f This converts the given limit into an indeterminate form of type 0 or
0
use l’Hospital’s Rule.
 Figure 5 shows the graph of the function in
Example 6. Notice that the function is undeﬁned
at x 0; the graph approaches the origin but
never quite reaches it.
y EXAMPLE 6 Evaluate lim x ln x.
xl0 SOLUTION The given limit is indeterminate because, as x l 0 , the ﬁrst factor (x)
approaches 0 while the second factor ln x approaches
. Writing x 1 1 x , we
have 1 x l as x l 0 , so l’Hospital’s Rule gives y=x ln x lim x ln x xl0 NOTE
0 FIGURE 5 1 so that we can ■ ln x
1x lim xl0 1x
1 x2 lim xl0 lim x xl0 0 In solving Example 6 another possible option would have been to write
lim x ln x x lim xl0 xl0 x
1 ln x This gives an indeterminate form of the type 0 0, but if we apply l’Hospital’s Rule we get
a more complicated expression than the one we started with. In general, when we rewrite
an indeterminate product, we try to choose the option that leads to the simpler limit.
x e x. EXAMPLE 7 Use l’Hospital’s Rule to help sketch the graph of f x
SOLUTION Because both x and e x become large as x l , we have lim x l x e x
. As
xl
, however, e l 0 and so we have an indeterminate product that requires the
use of l’Hospital’s Rule:
x lim x e x xl lim xl x
ex lim xl Thus, the xaxis is a horizontal asymptote. 1
e x lim xl ex 0 5E07(pp 494503) 498 ❙❙❙❙ 1/17/06 5:04 PM Page 498 CHAPTER 7 INVERSE FUNCTIONS We use the methods of Chapter 4 to gather other information concerning the graph.
The derivative is
xe x fx
y y=x´ 1
_2 ex 1 ex x Since e x is always positive, we see that f x
0 when x 1 0, and f x
0
when x 1 0. So f is increasing on 1,
and decreasing on
, 1 . Because
f
1
0 and f changes from negative to positive at x
1, f 1
e 1 is a local
(and absolute) minimum. The second derivative is _1 fx 1 ex x ex 2 ex x x
(_1, _1/e) FIGURE 6 Since f x
0 if x
2 and f x
0 if x
2, f is concave upward on
and concave downward on
, 2 . The inﬂection point is 2, 2e 2 .
We use this information to sketch the curve in Figure 6. 2, Indeterminate Differences
If lim x l a f x and lim x l a t x , then the limit
lim f x tx xla is called an indeterminate form of type
. Again there is a contest between f and
t. Will the answer be ( f wins) or will it be
( t wins) or will they compromise on a
ﬁnite number? To ﬁnd out, we try to convert the difference into a quotient (for instance,
by using a common denominator, or rationalization, or factoring out a common factor) so
that we have an indeterminate form of type 0 or
.
0
EXAMPLE 8 Compute lim xl sec x 2 tan x . SOLUTION First notice that sec x l and tan x l
terminate. Here we use a common denominator: lim xl sec x 2 tan x 1
cos x lim xl 2 lim xl as x l 1 2 sin x
cos x sin x
cos x lim xl Note that the use of l’Hospital’s Rule is justiﬁed because 1
as x l
2. Indeterminate Powers
Several indeterminate forms arise from the limit
lim f x tx xla 1. lim f x
xla 0 2. lim f x and xla 3. lim f x
xla and 1 and lim t x 0 type 0 0 lim t x 0 type xla xla lim t x xla type 1 2 , so the limit is inde 0 2 cos x
sin x 0 sin x l 0 and cos x l 0 5E07(pp 494503) 1/17/06 5:05 PM Page 499 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ❙ ❙❙❙ 499 Each of these three cases can be treated either by taking the natural logarithm:
let y tx fx , then ln y t x ln f x or by writing the function as an exponential:
tx fx et x ln f x (Recall that both of these methods were used in differentiating such functions.) In either
method we are led to the indeterminate product t x ln f x , which is of type 0
.
EXAMPLE 9 Calculate lim 1 sin 4 x xl0 cot x . SOLUTION First notice that as x l 0 , we have 1
given limit is indeterminate. Let y
Then ln y ln 1 1 sin 4 x sin 4 x l 1 and cot x l , so the sin 4 x
cot x cot x cot x ln 1 sin 4 x so l’Hospital’s Rule gives
lim ln y ln 1 lim xl0 xl0 1 lim xl0 sin 4 x
tan x 4 cos 4 x
sin 4 x
sec2x 4 So far we have computed the limit of ln y, but what we want is the limit of y. To ﬁnd this
we use the fact that y e ln y :
lim 1 xl0  The graph of the function y x x, x 0, is
shown in Figure 7. Notice that although 0 0 is not
deﬁned, the values of the function approach 1 as
x l 0 . This conﬁrms the result of Example 10.
2 sin 4 x cot x lim e ln y lim y xl0 xl0 e4 EXAMPLE 10 Find lim x x.
xl0 SOLUTION Notice that this limit is indeterminate since 0 x for any x
exponential: 0 for any x 0 but x 0
0. We could proceed as in Example 9 or by writing the function as an
xx e ln x x 1 e x ln x In Example 6 we used l’Hospital’s Rule to show that
lim x ln x _1 0 xl0 2 FIGURE 7 Therefore
lim x x xl0  See the biographical sketch of Cauchy on
page 97. 0 lim e x ln x xl0 e0 1 In order to give the promised proof of l’Hospital’s Rule we ﬁrst need a generalization
of the Mean Value Theorem. The following theorem is named after another French mathematician, AugustinLouis Cauchy (1789–1857). 5E07(pp 494503) 500 ❙❙❙❙ 1/17/06 5:06 PM Page 500 CHAPTER 7 INVERSE FUNCTIONS 3 Cauchy’s Mean Value Theorem Suppose that the functions f and t are continuous
on a, b and differentiable on a, b , and t x
0 for all x in a, b . Then there is
a number c in a, b such that fc
tc fb
tb fa
ta Notice that if we take the special case in which t x
x, then t c
1 and Theorem 3
is just the ordinary Mean Value Theorem. Furthermore, Theorem 3 can be proved in a similar manner. You can verify that all we have to do is change the function h given by
Equation 4.2.4 to the function
hx fx fb
tb fa fa
ta tx ta and apply Rolle’s Theorem as before.
Proof of L’Hospital’s Rule We are assuming that lim x l a f x L
We must show that lim x l a f x t x
fx
0 Fx lim xla 0 and lim x l a t x 0. Let fx
tx L. Deﬁne if x
if x a
a Gx Then F is continuous on I since f is continuous on x
lim F x lim f x xla 0 xla tx
0
Ix if x
if x a
a a and Fa Likewise, G is continuous on I . Let x I and x a. Then F and G are continuous on
a, x and differentiable on a, x and G
0 there (since F
f and G
t ). Therefore, by Cauchy’s Mean Value Theorem there is a number y such that a y x and
Fy
Gy Fx
Gx Fa
Ga Here we have used the fact that, by deﬁnition, F a
x l a , then y l a (since a y x), so
lim xla fx
tx lim xla Fx
Gx lim yla Fx
Gx
0 and G a Fy
Gy lim yla 0. Now, if we let fy
ty A similar argument shows that the lefthand limit is also L. Therefore
lim xla fx
tx L This proves l’Hospital’s Rule for the case where a is ﬁnite. L 5E07(pp 494503) 1/17/06 5:08 PM Page 501 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE If a is inﬁnite, we let t
fx
tx lim f1t
t1t
f 1t
t 1t 1 t2
1 t2 lim xl f 1t
t 1t lim tl0 tl0 tl0  7.7
1–4 9. lim f x 0 x la xl lim t x 0 x la lim h x lim xl 1 x la 11. lim
lim p x (c) lim xla et xla fx
tx xla hx
px xla t3 ln x
x 16. lim ex
x 17. lim ln x
x 18. lim ln ln x
x 20. lim ln x
sin x xl xl0 19. lim xla 5t 3t
t xl1 ex 1
x2 ■ ■ (b) lim p x sin x
x3 cos x
x2 28. lim ln x
x (b) lim f x px (e) lim p x qx (c) lim h x sin x
cos x 30. lim xl 1 ■ ■ px qx (f) lim sp x 29. lim xla ■ ■ xl0 ■ ■ xl0 1
1 6. lim 1
1 xa
8. lim b
xl1 x xl 2 x
x2 xl 33. lim xl1 2
3x 2 1
1 x
x xl0 xl x 2 35. lim ln 1 xa ax a
x 12 xl0 x ln x
cos x 37. lim s x ln x
xl0 32. lim 2e x 1
1 cos m x 34. lim
xl 1 36. lim xl0 x
tan 1 sx 2
s2x2
1 4x
2
1 e 2x
sec x 38. lim x 2e x
xl cos n x
x2 xl0 x ■ xl1 x9
7. lim 5
x l1 x x2 2 xla xla ■ 1 xl0 Find the limit. Use l’Hospital’s Rule where appropriate. If
there is a more elementary method, consider using it. If l’Hospital’s
Rule doesn’t apply, explain why.
x2
x x
x3 xl0 26. lim xl0  5. lim 1 sin 1x
x 31. lim
5–62 ex 25. lim qx xla xla ■ 22. lim sin x
sinh x 27. lim ■ x 24. lim qx fx xl sin
csc e
x3 xl px xla xl 1
2 23. lim xla (d) lim p x lim l x (c) lim p x q x tx 1
t (b) lim h x p x xla xla e3t tl0 xl0 2. (a) lim f x p x 4. (a) lim f x 12. lim 14. 21. lim xla tan x
sin x 15. lim px
qx (c) lim p x x xl0 tan px
tan qx xl0 px
fx (d) lim xla 10. lim 13. lim fx
px (b) lim 3. (a) lim f x cos x
sin x
1 tl0 (e) lim fx
tx x la which of the following limits are indeterminate forms? For those
that are not an indeterminate form, evaluate the limit where
possible.
xla 1 2 tl0 lim q x x la 1. (a) lim (by l’Hospital’s Rule for ﬁnite a) Exercises Given that  501 1 x. Then t l 0 as x l , so we have lim lim ❙❙❙❙ 5E07(pp 494503) ❙❙❙❙ 502 1/17/06 5:09 PM CHAPTER 7 INVERSE FUNCTIONS 39. lim cot 2 x sin 6 x 40. lim sin x ln x xl0 41. lim x 3e x 42. 43. lim ln x tan 1
x 45. lim xl0 46. lim csc x xl 49. lim x
51. lim x x 50. lim xe 1 1x ln 2 ln x 60. lim cos 3x 1 5x xl0 2 62. lim xl0 xl ■ ■ ■ ■ ■ ■ 2x 1 3
5 ■ ■ ■ ■ Use a graph to estimate the value of the limit. Then use
l’Hospital’s Rule to ﬁnd the exact value.
5 xl 64. lim tan x
xl ■ 65–66 ln x ■ ■ ■ ■ ■ ■ ■ ■ ■  65. f x ex 66. f x 2 x sin x,
■ 1, tx
tx ■ ■ x3 4x sec x
■ 1
■ ■ ■ ■ ■ ■ A0 1 xe 71. y x 68. y x ln x x2 70. y ex x 72. y ex x ■ ■ ln 1
■ x
■ ■ ■ ■ ■ 2 ■ x ■ 4x
■ nt If we let n l , we refer to the continuous compounding of
interest. Use l’Hospital’s Rule to show that if interest is compounded continuously, then the amount after n years is v 3e i
n A0 e it 80. If an object with mass m is dropped from rest, one model for
its speed v after t seconds, taking air resistance into account, is guidelines of Section 4.5. 69. y ■ compounded n times a year, the value of the investment after
t years is A xe ■ x 3
s2a 3x x 4 a saax
4
a sax 3 A 67–72 Use l’Hospital’s Rule to help sketch the curve. Use the 67. y n 79. If an initial amount A0 of money is invested at an interest rate i
■ ■ ■ as x approaches a, where a 0. (At that time it was common
to write aa instead of a 2.) Solve this problem. tan 2 x Illustrate l’Hospital’s Rule by graphing both f x t x
;
and f x t x near x 0 to see that these ratios have the same
limit as x l 0. Also calculate the exact value of the limit. ■ ■ book Analyse des Inﬁniment Petits published by the Marquis
de l’Hospital in 1696. This was the ﬁrst calculus textbook ever
published and the example that the Marquis used in that book
to illustrate his rule was to ﬁnd the limit of the function
y 4 ■ ■ 78. The ﬁrst appearance in print of l’Hospital’s Rule was in the 2x
2x  63. lim x ln x ■ x e c x, where
.
c is a real number. Start by computing the limits as x l
Identify any transitional values of c where the basic shape
changes. What happens to the maximum or minimum points
and inﬂection points as c changes? Illustrate by graphing several members of the family. x 1x ■ ; 77. Investigate the family of curves given by f x 1x x xl ■ bx 1 58. lim e x x ■ x e , where n
is a positive integer. What features do these curves have in
common? How do they differ from one another? In particular,
what happens to the maximum and minimum points and inﬂection points as n increases? Illustrate by graphing several members of the family. x xl x 61. lim cos x ; 63–64 1 56. lim x xl xl ■ x 5
x2 x ■ sin x x a
x 54. lim sin x x1 x ■ 74. f x ; 76. Investigate the family of curves given by f x xl 3
x 1 x x 75. f x 1 xl0 xl0 59. lim x  73. f x ■ 52. lim tan 2 x 2x 57. lim x 1 x xl 2 53. lim 1 xl 1 xl1 xl0 55. lim 1
ln x 48. lim ln x xl cot x xl0 x) x tan x sec x 4 xl csc x 47. lim (s x 2 lim 1 xl 73–75 (a) Graph the function.
(b) Use l’Hospital’s Rule to explain the behavior as x l 0 or
as x l .
(c) Estimate the maximum and minimum values and then use
calculus to ﬁnd the exact values.
(d) Use a graph of f to estimate the xcoordinates of the inﬂection points. 44. lim x tan 1 x x2 xl1 C AS xl0 2 xl ■ Page 502 ■ mt
1
c e ct m where t is the acceleration due to gravity and c is a positive
constant. (In Chapter 10 we will be able to deduce this 5E07(pp 494503) 1/17/06 5:10 PM Page 503 S ECTION 7.7 INDETERMINATE FORMS AND L’HOSPITAL’S RULE equation from the assumption that the air resistance is proportional to the speed of the object.)
(a) Calculate lim t l v. What is the meaning of this limit?
(b) For ﬁxed t, use l’Hospital’s Rule to calculate lim m l v.
What can you conclude about the speed of a very heavy
falling object?
81. In Section 5.3 we investigated the Fresnel function Sx
x0x sin( 1 t 2) dt, which arises in the study of the dif2
fraction of light waves. Evaluate
lim xl0 Sx
x3 89. Prove that lim x l 0 x ln x
90. Evaluate lim xl0 1
x3 y x 0 0 for any y 0. sin t 2 dt. 91. The ﬁgure shows a sector of a circle with central angle . Let A be the area of the segment between the chord PR and
the arc PR. Let B be the area of the triangle PQR. Find
lim l 0 A
B.
P
A(¨ ) the xaxis is initially C 2a if x
a and 0 if x
a. It can
be shown that if the heat diffusivity of the rod is k, then the
temperature of the rod at the point x at time t is
C
a s4 k t 503 for any number p 0. This shows that the logarithmic function approaches more slowly than any power of x. 82. Suppose that the temperature in a long thin rod placed along T x, t ❙❙❙❙ a x u 2 4kt e 0 B(¨ )
¨
O du To ﬁnd the temperature distribution that results from an initial
hot spot concentrated at the origin, we need to compute
lim T x, t R Q 92. The ﬁgure shows two regions in the ﬁrst quadrant: A t is the area under the curve y sin x 2 from 0 to t , and B t is
the area of the triangle with vertices O, P, and t, 0 . Find
lim t l 0 A t B t . al0 y Use l’Hospital’s Rule to ﬁnd this limit.
83. If f is continuous, f 2 lim 0, and f 2 f2 3x f2 sin 2 x
x3 lim xl0 a b
x2 O lim fx h fx h fx 2h Explain the meaning of this equation with the aid of a diagram.
86. If f is continuous, show that lim hl0 fx h 2f x
h2 fx h fx 87. Prove that for any positive integer n. This shows that the exponential function approaches inﬁnity faster than any power of x.
88. Prove that lim ln x
xp fx 0 x O t x e
0 1 x2 if x
if x 0
0 (a) Use the deﬁnition of derivative to compute f 0 .
(b) Show that f has derivatives of all orders that are deﬁned
on . [Hint: First show by induction that there is a polynomial pn x and a nonnegative integer k n such that
pn x f x x k n for x 0.]
fn x ; 94. Let
fx ex
lim n
xl x xl t 93. Let 0 85. If f is continuous, use l’Hospital’s Rule to show that hl0 B(t) y=sin{≈} 5x 84. For what values of a and b is the following equation true? P{t, sin ( t @ )} A(t) 7, evaluate x xl0 y P{t, sin ( t @ )} x
1 x if x
if x 0
0 (a) Show that f is continuous at 0.
(b) Investigate graphically whether f is differentiable at 0 by
zooming in several times toward the point 0, 1 on the
graph of f .
(c) Show that f is not differentiable at 0. How can you reconcile this fact with the appearance of the graphs in part (b)? 5E07(pp 504507) 504 ❙❙❙❙ 1/17/06 4:54 PM Page 504 CHAPTER 7 INVERSE FUNCTIONS WRITING PROJECT
The Origins of L’Hospital’s Rule
L’Hospital’s Rule was ﬁrst published in 1696 in the Marquis de l’Hospital’s calculus textbook
Analyse des Inﬁniment Petits, but the rule was discovered in 1694 by the Swiss mathematician
John (Johann) Bernoulli. The explanation is that these two mathematicians had entered into a
curious business arrangement whereby the Marquis de l’Hospital bought the rights to Bernoulli’s
mathematical discoveries. The details, including a translation of l’Hospital’s letter to Bernoulli
proposing the arrangement, can be found in the book by Eves [1].
Write a report on the historical and mathematical origins of l’Hospital’s Rule. Start by providing brief biographical details of both men (the dictionary edited by Gillispie [2] is a good
source) and outline the business deal between them. Then give l’Hospital’s statement of his rule,
which is found in Struik’s sourcebook [4] and more brieﬂy in the book of Katz [3]. Notice that
l’Hospital and Bernoulli formulated the rule geometrically and gave the answer in terms of differentials. Compare their statement with the version of l’Hospital’s Rule given in Section 7.7 and
show that the two statements are essentially the same.
1. Howard Eves, In Mathematical Circles (Volume 2: Quadrants III and IV) (Boston: Prindle, Weber and Schmidt, 1969), pp. 20–22.
2. C. C. Gillispie, ed., Dictionary of Scientiﬁc Biography (New York: Scribner’s, 1974). See the
 The Internet is another source of information for this project. See the web site www.stewartcalculus.com
and click on History of Mathematics.  7 Review article on Johann Bernoulli by E. A. Fellmann and J. O. Fleckenstein in Volume II and the
article on the Marquis de l’Hospital by Abraham Robinson in Volume VIII.
3. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993),
p. 484.
4. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, NJ: Princeton University Press, 1969), pp. 315–316. ■ CONCEPT CHECK 1. (a) What is a onetoone function? How can you tell if a function is onetoone by looking at its graph?
(b) If f is a onetoone function, how is its inverse function
f 1 deﬁned? How do you obtain the graph of f 1 from
the graph of f ?
(c) Suppose f is a onetoone function and t f 1. If
f ta
0, write a formula for t a .
2. (a) What are the domain and range of the natural exponential function f x
ex?
(b) What are the domain and range of the natural logarithmic
function f x
ln x ?
(c) How are the graphs of these functions related? Sketch these
graphs, by hand, using the same axes.
(d) If a is a positive number, a 1, write an equation that
expresses log a x in terms of ln x.
sin 1x deﬁned?
What are its domain and range?
(b) How is the inverse cosine function f x
cos 1x deﬁned?
What are its domain and range?
(c) How is the inverse tangent function f x
tan 1x
deﬁned? What are its domain and range? Sketch its
graph. 3. (a) How is the inverse sine function f x ■ 4. Write the deﬁnitions of the hyperbolic functions sinh x, cosh x, and tanh x.
5. State the derivative of each function. (a)
(d)
(g)
( j)
(m) y
y
y
y
y ex
log a x
tan 1x
tanh x
tanh 1x (b)
(e)
(h)
(k) y
y
y
y ax
sin 1x
sinh x
sinh 1x (c)
(f)
(i)
(l) y
y
y
y ln x
cos 1x
cosh x
cosh 1x 6. (a) How is the number e deﬁned? (b) Express e as a limit.
(c) Why is the natural exponential function y e x used more
often in calculus than the other exponential functions y a x ?
(d) Why is the natural logarithmic function y ln x used more
often in calculus than the other logarithmic functions y log a x ?
7. (a) What does l’Hospital’s Rule say? (b) How can you use l’Hospital’s Rule if you have a product
f x t x where f x l 0 and t x l as x l a ?
(c) How can you use l’Hospital’s Rule if you have a difference
fx
t x where f x l and t x l as x l a ?
(d) How can you use l’Hospital’s Rule if you have a power
f x t x where f x l 0 and t x l 0 as x l a ? 5E07(pp 504507) 1/17/06 4:55 PM Page 505 C HAPTER 7 REVIEW ■ TRUEFALSE QUIZ 1. If f is onetoone, with domain 1 , then f f6 2. If f is onetoone and differentiable,with domain
1 f 3. The function f x
1 4. tan 1 5. If 0 a s5 6. 3 d
ln 10
dx , then 1 f 6. 6 d
10 x
dx 11. 6. x10 x 1 1
10 2 x 2, is onetoone. 1
cos x 15. cosh x ln b. 1 for all x 16. ln 17. 7. You can always divide by e x. 0 and b 9. If x 0, then ln a 0, then ln x 6 b y dx
x 16 2 ln b. ln a xl ■ EXERCISES 1. The graph of f is shown. Is f onetoone? Explain. y 10 1 dx
x lim xl sec2x
sin x ■ 7. y ■ 8. y ln x 9. y y 1
10 3 ln 2 tan x
1 cos x 18. lim 6 ln x. sin 1x
cos 1x 14. tan 1x e s5 ln 8. If a ln x. 13. cos 1x 4 b, then ln a 1
3 e 3x is y 12. The inverse function of y cos x, 505 ■ 10. 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. ❙ ❙❙❙ ln x 1 2 arctan x
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1. For large values of x, which of the functions y x a,
y a , and y log a x has the largest values and which has the
smallest values? 10. Let a x 0 x 11–12 Find the exact value of each expression.  11. (a) e
2. The graph of t is given. (a)
(b)
(c)
(d) (b) log 10 25 ■ ■ 13–20 ■ 15. e e x 1 ln x 1 ■ (a) f 1 3 and (b) f 1 ■ ■ 4. Find the inverse function of f x
5–9 8. Find x
2x 1
.
1 ■ ■ Sketch a rough graph of the function without using a
calculator.
 5. y 5x 1 6. y e x x ■ ■ ■ 3 0.3
■ ■ Differentiate.  22. t t 23. h e tan 2
ln sec 5x et 24. h u 29. y ■ d ■ t 2 ln t 27. y x e 20. sin x
■ 25. y 3, and f 7 3. ■ 1
3 18. log 5 c 21. f t
3. Suppose f is onetoone, f 7 ) ■ 1 1 ■ 21–47 ■ 16. ln 1 17 19. tan x
01 ■ 14. e x 1 ■ ■ 1
3 x 17. ln x
1 ■ 1
2 Solve the equation for x.  13. ln x
g log 10 4 (b) tan(arcsin 12. (a) ln e Why is t onetoone?
Estimate the value of t 1 2 .
Estimate the domain of t 1.
Sketch the graph of t 1.
y 2 ln 3 e cx c sin x
2 ln sec x 31. y xe 33. y 2 1x
t2 tan 5x
cos x et 1 10 su 26. y e t t2 28. y 1 sin 2t
e x 2x 30. y ln x e 32. y x re sx 34. y e cos x cos e x 2 5E07(pp 504507) ❙❙❙❙ 506 1/17/06 Page 506 CHAPTER 7 INVERSE FUNCTIONS v tan 1v 35. H v
37. y x sinh x 39. y
41. y
43. y 1
2 sin2x 42. xe y 2 e sin 47. f x
■ ln x ■ y 1
x2
1 3 1 5 ■ 49–52  tan 1 x ■ ■ 1
1
tan
ln 1 xl ■ ■ ■ ■ xl0 ■ 1
4 x
x2 ln 12
1 1
x1 1 tx e 51. f x ln t x te 52. f x ■ 53–54 ■  Find f n ■ ■ ■ ■ ■ ■ ■ then f x x 56. Find y if y
57–58  ■ x ■ ■ ■ ■ ■ x xe , x ne. ■ x e x, ■ ■ ■ ■ 58. y
■ 59. At what point on the curve y e4x 1
x2 xl 4x 76. lim x 2 ln x
xl0 1
ln x 1 78. lim xl ■ ■ ■ cos x tan x 2 1 80. y 1x ■ ■ x e
■ 3x e ■ ■ ■ ■ ■ ■ 1x 2x x 2 ln x 2 84. y ■ 1 sin 82. y x ln x
e ■ Sketch the curve using the guidelines of Section 4.5. tan ■ ■ 1 ■ ■ ■ ■ 2 ■ x ln x,
■ ln x ■ 4 2 e, e
■ ■ ■ is the tangent x e sin x, ﬁnd f x . Graph f and f on the same screen
and comment. ; 60. If f x 61. (a) Find an equation of the tangent to the curve y parallel to the line x 4y 1.
(b) Find an equation of the tangent to the curve y
passes through the origin. x e that is
e x that K e a t e b t , where a, b, and K are positive constants and b a, is used to model the concentration at
time t of a drug injected into the bloodstream.
(a) Show that lim t l C t
0.
(b) Find C t , the rate at which the drug is cleared from
circulation.
(c) When is this rate equal to 0? 2 c xe c x . What happens to the maximum and minimum points and the inﬂection
points as c changes? Illustrate your conclusions by graphing
several members of the family. ; 86. Investigate the family of functions f x horizontal? 62. The function C t cos x
x x2 e 1 x in a viewing rectangle that shows all the
main aspects of this function. Estimate the inﬂection points.
Then use calculus to ﬁnd them exactly. arctan y. 0, 2 74. lim x ; 85. Graph f x Find an equation of the tangent to the curve at the given
2 4x ■  83. y
■ point.
57. y 79–84 81. y ln 2 x ■ 55. Use mathematical induction to show that if f x
n ■ ■ 79. y ■ 1 xl0 1
x2 x xl1 ■ 54. f x ■ 72. lim x 4
x t ln x ■ sin x 1 xl x 77. lim x. 2x 53. f x
■ ■ 70. lim xl ■ ■ e4x 75. lim x 3e x2 x 50. f x x xl x
x xl0 Find f in terms of t . 49. f x 68. lim e 2x
2x 69. lim x tanh 1s x 46. y x xl 71. lim 2x 73. lim 1
2 66. lim arctan x 3 4 48. Show that d
dx x3 67. lim ln sinh x 1
3x x2 64. lim ln 100
x l 10 65. lim e 2 1 ■ 3x xl xl0 sinh x 3 63. lim e arctan(arcsin sx ) 44. y 1 Evaluate the limit.  xl3 1
ln x cosh 63–78 x cos x 40. y ln cosh 3x 45. y z2 log10 1 38. y 1
x ln 36. F z 2 ln sin x ■ 4:56 PM repAe ct cos t
resents damped oscillation of an object. Find the velocity and
acceleration of the object. 87. An equation of motion of the form s 88. (a) Show that there is exactly one root of the equation ln x 3 x and that it lies between 2 and e.
(b) Find the root of the equation in part (a) correct to four decimal places.
89. The biologist G. F. Gause conducted an experiment in the 1930s with the protozoan Paramecium and used the population
function
64
Pt
1 31e 0.7944t
to model his data, where t was measured in days. Use this
model to determine when the population was increasing most
rapidly. 5E07(pp 504507) 1/17/06 4:57 PM Page 507 C HAPTER 7 REVIEW 90–103 90. y 1 4 16 0 92. y 94. 98. 91. dt dr 5 y 96. t 2 1 2 2 0 y1 97. 2x e
x 99. dx y sinh au du 104. y 0 x4 ■  1 x2 x
x2 2 with two sides on the axes and the third side tangent to the
curve y e x ? dy 1
116. Evaluate x0 e x dx without using the Fundamental Theorem of dx Calculus. [Hint: Use the deﬁnition of a deﬁnite integral with
right endpoints, sum a geometric series, and then use
l’Hospital’s Rule.] e sx
dx
sx yx x
2 xab t x dt, where a, b 117. If F x 1
dx
2x 101. ■ ■ ■ y2 103. dx y ■ sec 2 d sec tan
1 sec ■ e 2 x dx e 106. y 1 0 y 1 0 ■ e x cos x dx ■ ■ ■ x ■ ■ 120. (a) Show that ln x  ■ ■ ■ ■ ■ ■ y ■ sx 1 ■ ■ y 108. f x
■ ■ ■ ■ 2x ln x ■ 109. Find the average value of the function f x t2 e
■ ■ ■ 1 x on the e x, x 2, and x yaxis the region under the curve y
to x 1. 1
b x2 e x and t x 113. If t is the inverse function of f x t 1. ln x ln b
b ln a
a 1
a y 11 x 4 from x 0
0 x 0, x 1. 111. Find the volume of the solid obtained by rotating about the 112. If f x e t f t dt if b a 0.
(d) Give a geometric proof of Napier’s Inequality by comparing the slopes of the three lines shown in the ﬁgure. 110. Find the area of the region bounded by the curves e x, y x 0 (c) Deduce Napier’s Inequality: dt interval 1, 4 .
y 1
x es
ds
s
■ 1
2 ■ Find f x . 107. f x y 1 for x
0, x 1, x x
107–108 1. for all x, ﬁnd an explicit formula for f x . 4 ■ x2
x2 xe 2 x f t dt (b) Show that, for x ■ 1 ln a cos arctan sin arccot x y 1 x sin 1x dx x 119. If f is a continuous function such that 1 e ln b 1 118. Show that Use properties of integrals to prove the inequality. s1 ax
1 Use l’Hospital’s Rule to show that F is continuous at d ■ 1 x 1 0 105. bx y tan x ln cos x dx
tan 0, then, by the Fundamental Theorem, F 102. 1 y 2y2 Fx y s1 104–106 y 4 ye x e ■ 1 0 95. cos x
dx
1 sin2x 100. ■ y 93. 2r cos ln x
dx
x y 507 115. What is the area of the largest triangle in the ﬁrst quadrant Evaluate the integral.  ❙❙❙❙ f 1 x , ﬁnd t 1 . ln x tan 1x, ﬁnd a b x y=ln x 4. 114. What is the area of the largest rectangle in the ﬁrst quadrant with two sides on the axes and one vertex on the curve
y e x? (e) Give another proof of Napier’s Inequality by applying
Property 8 of integrals (see Section 5.2) to xab 1 x d x. 5E07(pp 508509) 1/17/06 4:59 PM Page 508 PROBLEMS
PLUS Cover up the solution to the example and try it yourself.
c x 2 have exactly one EXAMPLE For what values of c does the equation ln x
solution? y 3≈ ≈ 1 ≈
2 0.3≈
0.1≈ x 0 y=ln x SOLUTION One of the most important principles of problem solving is to draw a diagram,
even if the problem as stated doesn’t explicitly mention a geometric situation. Our present problem can be reformulated geometrically as follows: For what values of c does
the curve y ln x intersect the curve y c x 2 in exactly one point?
Let’s start by graphing y ln x and y c x 2 for various values of c. We know that,
for c 0, y c x 2 is a parabola that opens upward if c 0 and downward if c 0.
Figure 1 shows the parabolas y c x 2 for several positive values of c. Most of them
don’t intersect y ln x at all and one intersects twice. We have the feeling that there
must be a value of c (somewhere between 0.1 and 0.3) for which the curves intersect
exactly once, as in Figure 2.
To ﬁnd that particular value of c, we let a be the xcoordinate of the single point of
intersection. In other words, ln a ca 2, so a is the unique solution of the given equation. We see from Figure 2 that the curves just touch, so they have a common tangent
line when x a. That means the curves y ln x and y c x 2 have the same slope when
x a. Therefore 1
a FIGURE 1 2ca y Solving the equations ln a y=c≈
c=? ca 2 and 1 a
ca 2 ln a
0 a x Thus, a 2ca, we get
c 1
2c 1
2 e 1 2 and y=ln x c
FIGURE 2 ln a
a2 ln e 1 2
e 1
2e For negative values of c we have the situation illustrated in Figure 3: All parabolas
y c x 2 with negative values of c intersect y ln x exactly once. And let’s not forget
about c 0: The curve y 0 x 2 0 is just the xaxis, which intersects y ln x exactly
once.
y y=ln x
0
x FIGURE 3 To summarize, the required values of c are c 508 1 2e and c 0. 5E07(pp 508509) 1/17/06 P RO B L E M S 4:59 PM Page 509 2 e x , show that the
rectangle has the largest possible area when the two vertices are at the points of inﬂection of
the curve. 1. If a rectangle has its base on the xaxis and two vertices on the curve y 2. Prove that log 2 5 is an irrational number.
3. Show that dn
e ax sin bx
dx n
where a and b are positive numbers, r 2
4. Show that sin 1 tanh x 5. Show that, for x tan 1 r ne ax sin bx
a2 b 2, and 1 ba. 0,
x 6. Suppose f is continuous, f 0 the integral x f tan sinh x . 1
1
0 n 1 0, f 1 x
0, and x01 f x d x 1, f x 1
3 . Find the value of y d y. x1x s1 7. Show that f x tan 1x x2 t 3 dt is onetoone and ﬁnd f 1 0. 8. If y show that y a x
sa 2 1 2
sa 2 1 arctan sin x
sa 2 1 a cos x 1
.
cos x 9. For what value of a is the following equation true? x
x lim xl 10. Sketch the set of all points x, y such that x
11. Prove that cosh sinh x x a
a e
e x. y sinh cosh x for all x. 12. Show that, for all positive values of x and y, ex y
xy
13. For what value of k does the equation e 2 x e2
k sx have exactly one solution? 14. For which positive numbers a is it true that a x
15. For which positive numbers a does the curve y x for all x ? 1
x a intersect the line y x? 509 ...
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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.
 Fall '09
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