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4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 476

476 Chapter 7: Transcendental Functions 7.2 Natural

For any positive number a, the value ƒsxd = ax is easy to define when x is an integer or rational number. When x is irrational, the meaning of ax is not so clear. x Similarly, the definition of the loga x, the of ƒsxd = a , is not completely obvious. In this section we use to define the natural loga- rithm function, for which the number a is a particularly important value. This function al- lows us to define and analyze general exponential and logarithmic functions, y = ax and y = loga x. Logarithms originally played important roles in arithmetic computations. Historically, considerable labor went into producing long tables of logarithms, correct to five, eight, or even more, decimal places of accuracy. Prior to the modern age of electronic calculators and computers, every engineer owned slide rules marked with logarithmic scales. Calcula- tions with logarithms made possible the great seventeenth-century advances in offshore navigation and celestial mechanics. Today we know such calculations are done using calculators or computers, but the properties and numerous applications of logarithms are as important as ever.

Definition of the Natural Logarithm Function One solid approach to defining and understanding logarithms begins with a study of the natural logarithm function defined as an integral through the Fundamental Theorem of Calculus. While this approach may seem indirect, it enables us to derive quickly the fa- miliar properties of logarithmic and exponential functions. The functions we have studied so far were analyzed using the techniques of calculus, but here we do something more fundamental. We use calculus for the very definition of the logarithmic and exponential functions. The natural logarithm of a positive number x, written as ln x, is the value of an integral.

DEFINITION The Natural Logarithm Function

x = 1 7 ln x t dt, x 0 L1

If x 7 1, then ln x is the area under the curve y = 1 t from t = 1 to t = x > (Figure 7.9). For 0 6 x 6 1, ln x gives the negative of the area under the curve from x to 1. The function is not defined for x … 0. From the Zero Width Interval Rule for definite , we also have 1 1 ln 1 = dt = 0. L1 t 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 477

7.2 Natural Logarithms 477

y x 1 1 1 If 0 x 1, then ln x t dt t dt LL1 x gives the negative of this area. x If 1, then ln 1 x x t dt L1 gives this area. y ln x 1 1 y x x 0 xx1

1 If 1, then ln 1 0. x x t dt L1

y ln x

FIGURE 7.9 The graph of y = ln x and its relation to the function y = 1 x, x 7 0. The > graph of the logarithm rises above the x-axis as x moves from 1 to the right, and it falls below the TABLE 7.1 Typical 2-place axis as x moves from 1 to the left. values of ln x x ln x Notice that we show the graph of y = 1 x in Figure 7.9 but use y = 1 t in the inte- 0 undefined > > gral. Using x for everything would have us writing 0.05 -3.00 x 0.5 -0.69 = 1 ln x x dx, 10 L1 2 0.69 with x meaning two different things. So we change the variable of integration to t. 3 1.10 By using rectangles to obtain finite approximations of the area under the graph of y = 1 t and over the interval between t = 1 and t = x, as in Section 5.1, we can 4 1.39 > approximate the values of the function ln x. Several values are given in Table 7.1. There is 10 2.30 an important number whose natural logarithm equals 1.

DEFINITION The Number e The number e is that number in the domain of the natural logarithm satisfying

ln (e) = 1

Geometrically, the number e corresponds to the point on the x-axis for which the area under the graph of y = 1 t and above the interval [1, e] is the exact area of the unit square. > The area of the region shaded blue in Figure 7.9 is 1 sq unit when x = e. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 478

478 Chapter 7: Transcendental Functions

The of y = ln x By the first part of the Fundamental Theorem of Calculus (Section 5.4), x d = d 1 = 1 ln x dt x . dx dx L1 t For every positive value of x, we have

d 1 ln x = . dx x

Therefore, the function y = ln x is a solution to the initial value problem dy dx = 1 x, > > x 7 0, with y s1d = 0. Notice that the derivative is always positive so the natural loga- rithm is an increasing function, hence it is one-to-one and invertible. Its inverse is studied in Section 7.3. If u is a differentiable function of x whose values are positive, so that ln u is defined, then applying the dy dy du = dx du dx

to the function y = ln u gives d d du 1 du ln u = ln u # = . dx du dx u dx

d 1 du ln u = , u 7 0 (1) dx u dx

EXAMPLE 1 of Natural Logarithms d 1 d 1 1 (a) ln 2x = s2xd = s2d = dx 2x dx 2x x (b) Equation (1) with u = x2 + 3 gives d 1 d 1 2x ln sx2 + 3d = # sx2 + 3d = # 2x = . dx x2 + 3 dx x2 + 3 x2 + 3 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 479

7.2 Natural Logarithms 479

Notice the remarkable occurrence in Example 1a. The function y = ln 2x has the same derivative as the function y = ln x. This is true of y = ln ax for any positive number a: d 1 d 1 1 ln ax = # saxd = sad = . (2) dx ax dx ax x

Since they have the same derivative, the functions y = ln ax and y = ln x differ by a constant.

HISTORICAL BIOGRAPHY Properties of Logarithms John Napier Logarithms were invented by John Napier and were the single most important improve- (1550–1617) ment in arithmetic calculation before the modern electronic computer. What made them so useful is that the properties of logarithms enable multiplication of positive numbers by addition of their logarithms, division of positive numbers by subtraction of their loga- rithms, and of a number by multiplying its logarithm by the exponent. We summarize these properties as a of rules in Theorem 2. For the moment, we restrict the exponent r in Rule 4 to be a rational number; you will see why when we prove the rule.

THEOREM 2 Properties of Logarithms For any numbers a 7 0 and x 7 0, the natural logarithm satisfies the following rules:

1. : ln ax = ln a + ln x

a = - 2. : ln x ln a ln x

1 =- = 3. Reciprocal Rule: ln x ln x Rule 2 with a 1

4. : ln xr = r ln x r rational

We illustrate how these rules apply.

EXAMPLE 2 Interpreting the Properties of Logarithms

(a) ln 6 = ln s2 # 3d = ln 2 + ln 3 Product 4 (b) ln 4 - ln 5 = ln = ln 0.8 Quotient 5 1 () ln =-ln 8 Reciprocal 8 =-ln 23 =-3 ln 2 Power

EXAMPLE 3 Applying the Properties to Function Formulas

(a) ln 4 + ln sin x = ln s4 sin xd Product x + 1 (b) ln = ln sx + 1d - ln s2x - 3d Quotient 2x - 3 4100 AWL/Thomas_ch07p466-552 08/25/04 1:13 PM Page 480

480 Chapter 7: Transcendental Functions

= 1 =- (c) ln sec x ln cos x ln cos x Reciprocal

3 1 3 1 (d) ln2x + 1 = ln sx + 1d > = ln sx + 1d Power 3 We now give the proof of Theorem 2. The steps in the proof are similar to those used in solving problems involving logarithms.

Proof that ln ax = ln a + ln x The argument is unusual—and elegant. It starts by ob- serving that ln ax and ln x have the same derivative (Equation 2). According to Corollary 2 of the , then, the functions must differ by a constant, which means that

ln ax = ln x + C for some C. Since this last equation holds for all positive values of x, it must hold for x = 1. Hence,

ln sa # 1d = ln 1 + C

ln a = 0 + C ln 1 = 0

C = ln a. By substituting we conclude,

ln ax = ln a + ln x.

Proof that ln x r = r ln x (assuming r rational) We use the same-derivative argument again. For all positive values of x, d 1 d ln xr = sxrd Eq. (1) with u = x r dx xr dx

1 - Here is where we need r to be rational, = rxr 1 xr at least for now. We have proved the Power Rule only for rational 1 d exponents. = r # = sr ln xd. x dx

Since ln xr and r ln x have the same derivative,

ln xr = r ln x + C

for some constant C. Taking x to be 1 identifies C as zero, and we’re done. You are asked to prove Rule 2 in Exercise 84. Rule 3 is a special case of Rule 2, obtained by setting a = 1 and noting that ln 1 = 0. So we have established all cases of Theorem 2.

We have not yet proved Rule 4 for r irrational; we will return to this case in Section 7.3. The rule does hold for all r, rational or irrational.

The Graph and Range of ln x

The derivative dsln xd dx = 1 x is positive for x 7 0, so ln x is an increasing function of > > x. The second derivative, -1 x2 , is negative, so the graph of ln x is concave down. > 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 481

7.2 Natural Logarithms 481

y We can estimate the value of ln 2 by considering the area under the graph of y = 1 x > 1 and above the interval [1, 2]. In Figure 7.10 a rectangle of height 1 2 over the interval [1, 2] y x > fits under the graph. Therefore the area under the graph, which is ln 2, is greater than the area, 1 2, of the rectangle. So ln 2 7 1 2. Knowing this we have, 1 > >

n 1 n 1 ln 2 = n ln 2 7 n = 2 a2b 2 and x 0 12 - 1 n ln 2 n =-n ln 2 6-n =- . FIGURE 7.10 The rectangle of height a2b 2 y = 1 2 fits beneath the graph of y = 1 x > > It follows that for the interval 1 … x … 2. lim ln x = q and lim ln x =-q . : q : + x x 0

We defined ln x for x 7 0, so the domain of ln x is the set of positive real numbers. The above discussion and the Intermediate Value Theorem show that its range is the entire real line giving the graph of y = ln x shown in Figure 7.9.

The Integral 1s1/ud du Equation (1) leads to the integral formula

1 du = ln u + C (3) L u when u is a positive differentiable function, but what if u is negative? If u is negative, then -u is positive and 1 1 du = ds -ud Eq. (3) with u replaced by -u L u L s -ud = ln s -ud + C. (4) We can combine Equations (3) and (4) into a single formula by noticing that in each case the expression on the right is ln ƒ u ƒ + C. In Equation (3), ln u = ln ƒ u ƒ because u 7 0; in Equation (4), ln s -ud = ln ƒ u ƒ because u 6 0. Whether u is positive or nega- tive, the integral of (1 u) du is ln ƒ u ƒ + C. >

If u is a differentiable function that is never zero,

1 du = ln ƒ u ƒ + C. (5) L u

Equation (5) applies anywhere on the domain of 1 u, the points where u Z 0. > We know that un+1 un du = + C, n Z-1 and rational n + 1 L 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 482

482 Chapter 7: Transcendental Functions

Equation (5) explains what to do when n equals -1. Equation (5) says integrals of a certain form lead to logarithms. If u = ƒsxd, then du = ƒ¿sxd dx and 1 ƒ¿sxd du = dx. L u L ƒsxd So Equation (5) gives ƒ¿sxd dx = ln ƒ ƒsxd ƒ + C L ƒsxd whenever ƒ(x) is a differentiable function that maintains a constant sign on the domain given for it.

EXAMPLE 4 Applying Equation (5) - - u = x2 - 5, du = 2x dx, 2 2x 1 du 1 = = ƒ ƒ us0d =-5, us2d =-1 (a) 2 dx u ln u L0 x - 5 L-5 d -5 = ln ƒ -1 ƒ - ln ƒ -5 ƒ = ln 1 - ln 5 =-ln 5

p 2 5 u = 3 + 2 sin u, du = 2 cos u du, > u 4 cos = 2 - = = (b) + du u du us p 2d 1, usp 2d 5 L-p 2 3 2 sin u L1 > > > 5 = 2 ln ƒ u ƒ d 1 = 2 ln ƒ 5 ƒ - 2 ln ƒ 1 ƒ = 2 ln 5

Note that u = 3 + 2 sin u is always positive on[-p 2, p 2], so Equation (5) applies. > >

The Integrals of tan x and cot x Equation (5) tells us at last how to integrate the and cotangent functions. For the tangent function,

sin x -du u = cos x 7 0 on s -p 2, p 2d, tan x dx = dx = > > L L cos x L u du =-sin x dx du =- =-ln ƒ u ƒ + C L u 1 =-ln ƒ cos x ƒ + C = ln + C Reciprocal Rule ƒ cos x ƒ

= ln ƒ sec x ƒ + C. For the cotangent,

= cos x dx = du u = sin x, cot x dx u L L sin x L du = cos x dx

= ln ƒ u ƒ + C = ln ƒ sin x ƒ + C =-ln ƒ csc x ƒ + C. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 483

7.2 Natural Logarithms 483

tan u du =-ln ƒ cos u ƒ + C = ln ƒ sec u ƒ + C L

cot u du = ln ƒ sin u ƒ + C =-ln ƒ csc x ƒ + C L

EXAMPLE 5 = p 6 p 3 p 3 Substitute u 2x, > > du 1 > tan 2x dx = tan u # = tan u du dx = du 2, 2 2 > L0 L0 L0 us0d = 0, p 3 usp 6d = p 3 1 > 1 1 > > = ln ƒ sec u ƒ = sln 2 - ln 1d = ln 2 2 d 0 2 2

Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the laws of logarithms to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the next example.

EXAMPLE 6 Using Logarithmic Differentiation Find dy dx if > 2 1 2 sx + 1dsx + 3d > y = , x 7 1. x - 1

Solution We take the natural logarithm of both sides and simplify the result with the properties of logarithms: 2 1 2 sx + 1dsx + 3d > ln y = ln x - 1 2 1 2 = ln ssx + 1dsx + 3d > d - ln sx - 1d Rule 2

2 1 2 = ln sx + 1d + ln sx + 3d > - ln sx - 1d Rule 1 1 = ln sx2 + 1d + ln sx + 3d - ln sx - 1d. Rule 3 2 We then take derivatives of both sides with respect to x, using Equation (1) on the left:

1 dy 1 1 1 1 = # 2x + # - . y dx x2 + 1 2 x + 3 x - 1

Next we solve for dy dx: > dy 2x 1 1 = y + - . dx ax2 + 1 2x + 6 x - 1 b 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 484

484 Chapter 7: Transcendental Functions

Finally, we substitute for y: 2 1 2 dy sx + 1dsx + 3d > 2x 1 1 = + - . dx x - 1 ax2 + 1 2x + 6 x - 1 b A direct computation in Example 6, using the Quotient and Product Rules, would be much longer. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 484

484 Chapter 7: Transcendental Functions

EXERCISES 7.2

Using the Properties of Logarithms 25. y = ussin sln ud + cos sln udd 26. y = ln ssec u + tan ud 1. Express the following logarithms in terms of ln 2 and ln 3. 1 1 1 + x a. ln 0.75 b. ln (4 9) c. ln (1 2) 27. y = ln 28. y = ln > > 2 + 2 1 - x d. ln23 9 e. ln 322 f. ln 213.5 x x 1 1 + ln t 2. Express the following logarithms in terms of ln 5 and ln 7. 29. y = 30. y = 2ln 1t 1 - ln t a. ln (1 125) b. ln 9.8 c. ln 727 > 2sin u cos u d. ln 1225 e. ln 0.056 31. y = ln ssec sln udd 32. y = ln a 1 + 2 ln u b f. sln 35 + ln s1 7dd sln 25d > > sx2 + 1d5 sx + 1d5 = = Use the properties of logarithms to simplify the expressions in 33. y ln 34. y ln 20 a 21 - x b Csx + 2d Exercises 3 and 4. 3 x2 2x - sin u 2 - + 1 35. y = ln 2t dt 36. y = ln t dt 3. a. ln sin u ln b. ln s3x 9xd ln Lx2 2 L2x a 5 b a3x b > 1 c. ln s4t4d - ln 2 2 Integration Evaluate the integrals in Exercises 37–54. 4. a. ln sec u + ln cos u b. ln s8x + 4d - 2 ln 2 -2 0 c. 3 ln23 t2 - 1 - ln st + 1d dx 3 dx 37. x 38. - L-3 L-1 3x 2 Derivatives of Logarithms 2y dy 8r dr 39. 40. In Exercises 5–36, find the derivative of y with respect to x, t, or u, as L y2 - 25 L 4r2 - 5 p p 3 appropriate. sin t > 4 sin u 41. - dt 42. - du 5. y = ln 3x 6. y = ln kx, k constant L0 2 cos t L0 1 4 cos u = 2 = 3 2 2 4 7. y ln st d 8. y ln st > d 2 ln x dx 3 10 43. x dx 44. = = L1 L2 x ln x 9. y ln x 10. y ln x 4 dx 16 dx = su + d = s u + d 11. y ln 1 12. y ln 2 2 45. 2 46. L2 xsln xd L2 2x2ln x 13. y = ln x3 14. y = sln xd3 3 sec2 t sec y tan y 15. y = tsln td2 16. y = t2ln t 47. dt 48. dy L 6 + 3 tan t L 2 + sec y 4 4 3 3 x x x x p 2 p 2 17. y = ln x - 18. y = ln x - > x > 4 16 3 9 49. tan dx 50. cot t dt + L0 2 Lp 4 = ln t = 1 ln t > 19. y 20. y p p 12 t t u > 51. 2 cot du 52. 6 tan 3x dx ln x x ln x 3 21. y = 22. y = Lp 2 L0 1 + ln x 1 + ln x > dx sec x dx 23. y = ln sln xd 24. y = ln sln sln xdd 53. 54. L 21x + 2x L 2ln ssec x + tan xd 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 485

7.2 Natural Logarithms 485

Logarithmic Differentiation x = 2 is In Exercises 55–68, use logarithmic differentiation to find the deriva- 2 = + 1 L 1 2 dx. tive of y with respect to the given independent variable. L1 A x 55. y = 2xsx + 1d 56. y = 2sx2 + 1dsx - 1d2 T 79. a. Find the centroid of the region between the curve y = 1 x = = > t 1 and the x-axis from x 1 to x 2. Give the coordinates to 57. y = 58. y = At + 1 Atst + 1d two decimal places. b. Sketch the region and show the centroid in your sketch. 59. y = 2u + 3 sin u 60. y = stan ud22u + 1 80. a. Find the center of mass of a thin plate of constant density 1 61. y = tst + 1dst + 2d 62. y = covering the region between the curvey = 1 1x and the x- tst + 1dst + 2d > axis from x = 1 to x = 16. u + 5 u sin u 63. y = 64. y = b. Find the center of mass if, instead of being constant, the u cos u 2 sec u density function is dsxd = 4 1x. > x2x2 + 1 sx + 1d10 65. y = 66. y = Solve the initial value problems in Exercises 81 and 82. 2 3 5 sx + 1d > Cs2x + 1d dy = + 1 = xsx - 2d xsx + 1dsx - 2d 81. 1 x, ys1d 3 = 3 = 3 dx 67. y 2 68. y 2 B x + 1 Bsx + 1ds2x + 3d d 2y 82. = sec2 x, ys0d = 0 and y¿s0d = 1 dx2 Theory and Applications T 83. The linearization of ln s1 + xd at x = 0 Instead of approxi- = + = 69. Locate and identify the absolute extreme values of mating ln x near x 1, we approximate ln s1 xd near x 0. We get a simpler formula this way. a. ln (cos x) on [-p 4, p 3], > > a. Derive the linearization ln s1 + xd L x at x = 0. b. cos (ln x) on [1 2, 2]. > b. Estimate to five decimal places the error involved in replacing s d = - 7 70. a. Prove that ƒ x x ln x is increasing for x 1. ln s1 + xd by x on the interval [0, 0.1]. 6 7 b. Using part (a), show that ln x x if x 1. c. Graph ln s1 + xd and x together for 0 … x … 0.5. Use 71. Find the area between the curves y = ln x and y = ln 2x from different colors, if available. At what points does the x = 1 to x = 5. approximation of ln s1 + xd seem best? Least good? By 72. Find the area between the curve y = tan x and the x-axis from reading coordinates from the graphs, find as good an upper x =-p 4 to x = p 3. bound for the error as your grapher will allow. > > 73. The region in the first quadrant bounded by the coordinate axes, 84. Use the same-derivative argument, as was done to prove Rules 1 the line y = 3, and the curve x = 2 2y + 1 is revolved about and 4 of Theorem 2, to prove the Quotient Rule property of loga- N the y-axis to generate a solid. Find the volume of the solid. rithms. 74. The region between the curve y = 2cot x and the x-axis from x = p 6 to x = p 2 is revolved about the x-axis to generate a Grapher Explorations > > solid. Find the volume of the solid. 85. Graph ln x, ln 2x, ln 4x, ln 8x, and ln 16x (as many as you can) to- 75. The region between the curve y = 1 x2 and the x-axis from gether for 0 6 x … 10. What is going on? Explain. > x = 1 2 to x = 2 is revolved about the y-axis to generate a solid. = ƒ ƒ … … - … … > 86. Graph y ln sin x in the window 0 x 22, 2 y 0. Find the volume of the solid. Explain what you see. How could you change the formula to turn 76. In Section 6.2, Exercise 6, we revolved about the y-axis the region the arches upside down? between the curve y = 9x 2x3 + 9 and the x-axis from x = 0 = = + = N 87. a. Graph y sin x and the curves y ln sa sin xd for a 2, to x = 3 to generate a solid of volume 36p. What volume do you 4, 8, 20, and 50 together for 0 … x … 23. get if you revolve the region about the x-axis instead? (See b. Why do the curves flatten as a increases? (Hint: Find an Section 6.2, Exercise 6, for a graph.) a-dependent upper bound for ƒ y¿ ƒ . ) 77. Find the lengths of the following curves. 88. Does the graph of y = 1x - ln x, x 7 0, have an inflection a. y = sx2 8d - ln x, 4 … x … 8 > point? Try to answer the question (a) by graphing, (b) by using cal- b. x = sy 4d2 - 2 ln sy 4d, 4 … y … 12 culus. > > 78. Find a curve through the point (1, 0) whose length from x = 1 to 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 486

486 Chapter 7: Transcendental Functions 7.3 The

Having developed the theory of the function ln x, we introduce the exponential function exp x = ex as the inverse of ln x. We study its properties and compute its derivative and in- tegral. Knowing its derivative, we prove the power rule to differentiate xn when n is any , rational or irrational.

y The Inverse of ln x and the Number e q 8 –1 The function ln x, being an increasing function of x with domain s0, d and range y ln x -1 or s - q, q d, has an inverse ln x with domain s - q, q d and range s0, q d. The graph of - 7 x ln y ln 1 x is the graph of ln x reflected across the line y = x. As you can see in Figure 7.11, - - 6 lim ln 1 x = q and lim ln 1 x = 0. : q : - q x x - 5 The function ln 1 x is also denoted by exp x. = 4 In Section 7.2 we defined the number e by the equation ln sed 1, so - e = ln 1s1d = exp s1d. Although e is not a rational number, later in this section we see one e (1, e) way to express it as a . In Chapter 11, we will calculate its value with a computer to as many places of accuracy as we want with a different formula (Section 11.9, Example 6). 2 y ln x To 15 places, 1 e = 2.718281828459045. x –2 –1 0 1 2 e 4 The Function y ex We can raise the number e to a rational power r in the usual way:

2 # -2 1 1 2 e = e e, e = , e > = 2e, FIGURE 7.11 The graphs of y = ln x and 2 e - y = ln 1 x = exp x. The number e is - r r ln 1 1 = exp s1d. and so on. Since e is positive, e is positive too. Thus, e has a logarithm. When we take the logarithm, we find that

ln er = r ln e = r # 1 = r. - Since ln x is one-to-one and ln sln 1 rd = r, this equation tells us that - er = ln 1 r = exp r for r rational. (1) - We have not yet found a way to give an obvious meaning to ex for x irrational. But ln 1 x has meaning for any x, rational or irrational. So Equation (1) provides a way to extend the - definition of ex to irrational values of x. The function ln 1 x is defined for all x, so we use it x Typical values of e to assign a value to ex at every point where ex had no previous definition.

x e x (rounded) -1 0.37 DEFINITION The Natural Exponential Function x -1 01 For every real number x, e = ln x = exp x. 1 2.72 2 7.39 10 22026 For the first time we have a precise meaning for an irrational exponent. Usually the x x 100 2.6881 * 1043 exponential function is denoted by e rather than exp x. Since ln x and e are inverses of one another, we have 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 487

7.3 The Exponential Function 487

Inverse Equations for e x and ln x eln x = x sall x 7 0d (2) ln sexd = x sall xd (3)

The domain of ln x is s0, q d and its range is s - q, q d. So the domain of ex is s - q, q d Transcendental Numbers and q Transcendental Functions and its range is s0, d. Numbers that are solutions of polynomial equations with rational coefficients are EXAMPLE 1 Using the Inverse Equations - called algebraic: 2 is algebraic because 2 = it satisfies the equation x + 2 = 0, and (a) ln e 2 -1 23 is algebraic because it satisfies the (b) ln e =-1 2 - = equation x 3 0. Numbers that are 1 (c) ln 2e = not algebraic are called transcendental, 2 like e and p. In 1873, Charles Hermite proved the transcendence of e in the (d) ln esin x = sin x sense that we describe. In 1882, C.L.F. (e) eln 2 = 2 Lindemann proved the transcendence 2 ln sx +1d = 2 + of p. (f ) e x 1 3 Today, we call a function y = ƒsxd (g) e3 ln 2 = eln 2 = eln 8 = 8 One way algebraic if it satisfies an equation of the (h) e3 ln 2 = seln 2d3 = 23 = 8 Another way form n Á P y + + P y + P = 0 n 1 0 EXAMPLE 2 Solving for an Exponent in which the P’s are polynomials in x 2k = with rational coefficients. The function Find k if e 10. y = 1 2x + 1 is algebraic because > it satisfies the equation Solution Take the natural logarithm of both sides: sx + 1dy 2 - 1 = 0. Here the e2k = 10 polynomials are P2 = x + 1, P1 = 0, 2k = and P0 =-1. Functions that are not ln e ln 10 algebraic are called transcendental. 2 k = ln 10 Eq. (3) 1 k = ln 10. 2

The General Exponential Function ax Since a = eln a for any positive number a, we can think of ax as seln adx = ex ln a . We there- fore make the following definition.

DEFINITION General Exponential Functions For any numbers a 7 0 and x, the exponential function with base a is ax = ex ln a .

# When a = e, the definition gives ax = ex ln a = ex ln e = ex 1 = ex . 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 488

488 Chapter 7: Transcendental Functions

HISTORICAL BIOGRAPHY EXAMPLE 3 Evaluating Exponential Functions Siméon Denis Poisson (a) 223 = e 23 ln 2 L e1.20 L 3.32 (1781–1840) (b) 2p = ep ln 2 L e2.18 L 8.8 We study the calculus of general exponential functions and their inverses in the next section. Here we need the definition in order to discuss the laws of exponents for ex.

Laws of Exponents - Even though ex is defined in a seemingly roundabout way as ln 1 x, it obeys the familiar laws of exponents from algebra. Theorem 3 shows us that these laws are consequences of the definitions of ln x and ex.

THEOREM 3 Laws of Exponents for e x x For all numbers x, x1, and x2, the natural exponential e obeys the following laws: + 1. ex1 # ex2 = ex1 x2

- 1 2. e x = ex x e 1 - 3. = ex1 x2 ex2 4. sex1dx2 = ex1 x2 = sex2dx1

Proof of Law 1 Let

= x1 = x2 y1 e and y2 e . (4) Then

x1 = ln y1 and x2 = ln y2 Take logs of both sides of Eqs. (4). x1 + x2 = ln y1 + ln y2

= ln y1 y2 Product Rule for logarithms + ex1 x2 = eln y1 y2 Exponentiate.

ln u = y1 y2 e = u = ex1 ex2 . The proof of Law 4 is similar. Laws 2 and 3 follow from Law 1 (Exercise 78).

EXAMPLE 4 Applying the Exponent Laws + (a) ex ln 2 = ex # eln 2 = 2ex Law 1

- 1 1 (b) e ln x = = Law 2 eln x x 2x e = 2x-1 (c) e e Law 3

(d) se3dx = e3x = sexd3 Law 4 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 489

7.3 The Exponential Function 489

Theorem 3 is also valid for ax, the exponential function with base a. For example,

ax1 # ax2 = ex1 ln a # ex2 ln a Definition of ax + = ex1 ln a x2 ln a Law 1 + = esx1 x2dln a Factor ln a + = ax1 x2 . Definition of ax

The Derivative and Integral of ex The exponential function is differentiable because it is the inverse of a differentiable func- tion whose derivative is never zero (Theorem 1). We calculate its derivative using Theorem 1 and our knowledge of the derivative of ln x. Let - - ƒsxd = ln x and y = ex = ln 1 x = ƒ 1sxd. Then,

dy d d - = sexd = ln 1 x dx dx dx

d - = ƒ 1sxd dx 1 = Theorem 1 ƒ¿sƒ -1sxdd 1 = ƒ -1sxd = ex ƒ¿sexd

= 1 ¿ = 1 = x ƒ szd z with z e 1 aex b = ex . That is, for y = ex, we find that dy dx = ex so the natural exponential function ex is its > own derivative. We will see in Section 7.5 that the only functions that behave this way are constant multiples of ex. In summary,

d ex = ex (5) dx

EXAMPLE 5 Differentiating an Exponential

d d s5exd = 5 ex dx dx = 5ex

The Chain Rule extends Equation (5) in the usual way to a more general form. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 490

490 Chapter 7: Transcendental Functions

If u is any differentiable function of x, then d du eu = eu . (6) dx dx

EXAMPLE 6 Applying the Chain Rule with Exponentials

d - - d - - (a) e x = e x s -xd = e xs -1d =-e x Eq. (6) with u =-x dx dx

d sin x sin x d sin x # (b) e = e ssin xd = e cos x Eq. (6) with u = sin x dx dx

The integral equivalent of Equation (6) is

eu du = eu + C. L

EXAMPLE 7 Integrating Exponentials 1 u = 3x, du = dx, us0d = 0, ln 2 ln 8 3 3x u # 1 (a) e dx = e du usln 2d = 3 ln 2 = ln 23 = ln 8 L0 L0 3 1 ln 8 = eu du 3L0 1 ln 8 = eu 3 d 0 1 7 = s8 - 1d = 3 3

p 2 p 2 > > (b) esin x cos x dx = esin x from Example 6 L0 d 0 = e1 - e0 = e - 1

EXAMPLE 8 Solving an Initial Value Problem Solve the initial value problem dy e y = 2x, x 7 23; ys2d = 0. dx

Solution We integrate both sides of the with respect to x to obtain e y = x2 + C. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 491

7.3 The Exponential Function 491

We use the initial condition ys2d = 0 to determine C: C = e0 - s2d2 = 1 - 4 =-3. This completes the formula for e y : e y = x2 - 3. To find y, we take logarithms of both sides:

ln e y = ln sx2 - 3d

y = ln sx2 - 3d. Notice that the solution is valid for x 7 23. Let’s check the solution in the original equation. dy d e y = e y ln sx2 - 3d dx dx

y 2x = e Derivative of ln sx2 - 3d x2 - 3

ln sx2 -3d 2x = e y = ln sx2 - 3d x2 - 3 2x = sx2 - 3d eln y = y x2 - 3 = 2x. The solution checks.

The Number e Expressed as a Limit

We have defined the number e as the number for which ln e = 1, or the value exp (1). We see that e is an important constant for the logarithmic and exponential functions, but what is its numerical value? The next theorem shows one way to calculate e as a limit.

THEOREM 4 The Number e as a Limit The number e can be calculated as the limit 1 x e = lim s1 + xd > . x:0

Proof If ƒsxd = ln x, then ƒ¿sxd = 1 x, so ƒ¿s1d = 1. But, by the definition of derivative, > ƒs1 + hd - ƒs1d ƒs1 + xd - ƒs1d ƒ¿s1d = lim = lim h:0 h x:0 x ln s1 + xd - ln 1 1 = lim = lim ln s1 + xd ln 1 = 0 x:0 x x:0 x

= lim ln s1 + xd1/x = ln lim s1 + xd1/x ln is continuous. x:0 cx:0 d 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 492

492 Chapter 7: Transcendental Functions

Because ƒ¿s1d = 1, we have

1 x ln lim s1 + xd > = 1 cx:0 d Therefore,

1 x lim s1 + xd > = e ln e = 1 and ln is one-to-one x:0 By substituting y = 1 x, we can also express the limit in Theorem 4 as > 1 y e = lim 1 + . (7) y: q a y b

At the beginning of the section we noted that e = 2.718281828459045 to 15 decimal places.

The Power Rule (General Form) We can now define xn for any x 7 0 and any real number n as xn = en ln x. Therefore, the n in the equation ln xn = n ln x no longer needs to be rational—it can be any number as long as x 7 0:

n n ln x ln x = ln se d = n ln x ln eu = u, any u Together, the law ax ay = ax-y and the definition xn = en ln x enable us to establish > the Power Rule for differentiation in its final form. Differentiating xn with respect to x gives d d xn = en ln x Definition of xn, x 7 0 dx dx d = en ln x # sn ln xd Chain Rule for eu dx = n # n x x The definition again = nxn-1 . In short, as long as x 7 0,

d - xn = nxn 1 . dx The Chain Rule extends this equation to the Power Rule’s general form.

Power Rule (General Form) If u is a positive differentiable function of x and n is any real number, then un is a differentiable function of x and

d - du un = nun 1 . dx dx 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 493

7.3 The Exponential Function 493

EXAMPLE 9 Using the Power Rule with Irrational Powers d 2 2 - (a) x 2 = 22x 2 1 sx 7 0d dx

d - (b) s2 + sin 3xdp = ps2 + sin 3xdp 1scos 3xd # 3 dx - = 3ps2 + sin 3xdp 1scos 3xd. 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 493

7.3 The Exponential Function 493

EXERCISES 7.3

- 2 - Algebraic Calculations with the Exponential 27. y = cos se u d 28. y = u3e 2u cos 5u - - and Logarithm 29. y = ln s3te td 30. y = ln s2e t sin td Find simpler expressions for the quantities in Exercises 1–4. u 2 = e = u ln 7.2 -ln x2 ln x-ln y 31. y ln u 32. y ln 1. a. e b. e c. e a1 + e b a1 + 2u b 2 + 2 - - + 2. a. eln sx y d b. e ln 0.3 c. eln px ln 2 33. y = escos t ln td 34. y = esin tsln t2 + 1d - 2 - 2 2x 3. a. 2 ln 2e b. ln sln eed c. ln se x y d ln x e 35. y = sin et dt 36. y = ln t dt sec u sexd 2 ln x 4. a. ln se d b. ln se d c. ln se d L0 Le41x In Exercises 37–40, find dy dx. > Solving Equations with Logarithmic y x+y 37. ln y = e sin x 38. ln xy = e or Exponential Terms 39. e2x = sin sx + 3yd 40. tan y = ex + ln x In Exercises 5–10, solve for y in terms of t or x, as appropriate. 5. ln y = 2t + 4 6. ln y =-t + 5 Integrals 7. ln sy - 40d = 5t 8. ln s1 - 2yd = t Evaluate the integrals in Exercises 41–62. 9. ln sy - 1d - ln 2 = x + ln x 41. se3x + 5e-xd dx 42. s2ex - 3e-2xd dx 10. ln sy2 - 1d - ln sy + 1d = ln ssin xd L L ln 3 0 - In Exercises 11 and 12, solve for k. 43. ex dx 44. e x dx Lln 2 L-ln 2 2k 10k k 1000 11. a. e = 4 b. 100e = 200 c. e > = a 1 sx+1d s2x-1d 12. a. e5k = b. 80ek = 1 c. esln 0.8dk = 0.8 45. 8e dx 46. 2e dx 4 L L ln 9 ln 16 In Exercises 13–16, solve for t. x 2 x 4 47. e > dx 48. e > dx Lln 4 L0 -0.3t kt 1 sln 0.2dt 13. a. e = 27 b. e = c. e = 0.4 2 -2 2 e r e r 49. dr 50. dr L 2 L 2 - 1 1 r r 14. a. e 0.01t = 1000 b. ekt = c. esln 2dt = 10 2 - 2 4 51. 2t e t dt 52. t 3est d dt 2 15. e2t = x2 16. esx des2x+1d = et L L 1 x -1 x2 e > e > 53. dx 54. dx Derivatives L x2 L x3 p 4 p 2 In Exercises 17–36, find the derivative of y with respect to x, t, or u, as > > 55. s1 + etan ud sec2 u du 56. s1 + ecot ud csc2 u du appropriate. L0 Lp 4 -5x 2x 3 > 17. y = e 18. y = e > sec pt - s 1 + 2d 57. e sec pt tan pt dt 19. y = e5 7x 20. y = e 4 x x L = x - x = + -2x 21. y xe e 22. y s1 2xde + 58. ecsc sp td csc sp + td cot sp + td dt 23. y = sx2 - 2x + 2dex 24. y = s9x2 - 6x + 2de3x L - 25. y = eussin u + cos ud 26. y = ln s3ue ud 4100 AWL/Thomas_ch07p466-552 9/2/04 10:33 AM Page 494

494 Chapter 7: Transcendental Functions

ln sp 2d 2ln p y y –y > 2 2 e e y y x x x 59. 2e cos e dy 60. 2x e cos se d dx 2 Lln sp 6d L0 > er dx ln 2 61. dr 62. L 1 + er L 1 + ex Initial Value Problems 0 Solve the initial value problems in Exercises 63–66. 1 x dy 63. = et sin set - 2d, ysln 2d = 0 dt 75. a. Show that 1 ln x dx = x ln x - x + C. dy - - 64. = e t sec2 spe td, ysln 4d = 2 p b. Find the average value of ln x over [1, e]. dt > 76. Find the average value of ƒsxd = 1 x on [1, 2]. 2 > d y x 65. = 2e -x, ys0d = 1 and y¿s0d = 0 77. The linearization of e at x = 0 dx2 a. Derive the linear approximation ex L 1 + x at x = 0. d 2y = - 2t s d =- ¿s d = b. Estimate to five decimal places the magnitude of the error 66. 2 1 e , y 1 1 and y 1 0 T dt involved in replacing x by 1 + xe on the interval [0, 0.2]. x + - … … T c. Graph e and 1 x together for 2 x 2. Use different Theory and Applications colors, if available. On what intervals does the approximation 67. Find the absolute maximum and minimum values of ƒsxd = appear to overestimate ex ? Underestimate ex ? x - e 2x on [0, 1]. 78. Laws of Exponents sin sx 2d 68. Where does the periodic function ƒsxd = 2e > take on its ex- + a. Starting with the equation ex1ex2 = ex1 x2 , derived in the text, treme values and what are these values? show that e -x = 1 ex for any real number x. Then show that - > ex1 ex2 = ex1 x2 for any numbers x and x . y > 1 2 x1 x2 x1 x2 x2 x1 b. Show that se d = e = se d for any numbers x1 and x2 . sin (x/2) y 2e T 79. A decimal representation of e Find e to as many decimal places as your calculator allows by solving the equation ln x = 1. x x T 80. The inverse relation betweene and ln x Find out how good 0 your calculator is at evaluating the composites

eln x and ln sexd. 69. Find the absolute maximum value of ƒsxd = x2 ln s1 xd and say > where it is assumed. 81. Show that for any number a 7 1 2 x T 70. Graph ƒsxd = sx - 3d e and its first derivative together. Com- a ln a ment on the behavior of ƒ in relation to the signs and values of ƒ¿ . ln x dx + e y dy = a ln a. Identify significant points on the graphs with calculus, as neces- L1 L0 sary. (See accompanying figure.) 71. Find the area of the “triangular” region in the first quadrant that is = 2x = x y bounded above by the curve y e , below by the curve y e , y ln x and on the right by the line x = ln 3. 72. Find the area of the “triangular” region in the first quadrant that is ln a x 2 bounded above by the curve y = e > , below by the curve -x 2 y = e > , and on the right by the line x = 2 ln 2. x 0 1 a 73. Find a curve through the origin in the xy-plane whose length from x = 0 to x = 1 is

1 1 L = 1 + ex dx. A 4 L0 82. The geometric, logarithmic, and arithmetic mean 74. Find the area of the surface generated by revolving the curve a. Show that the graph of ex is concave up over every interval of - x = sey + e yd 2, 0 … y … ln 2, about the y-axis. x-values. > 4100 AWL/Thomas_ch07p466-552 8/20/04 10:02 AM Page 495

7.3 The Exponential Function 495

b. Show, by reference to the accompanying figure, that if American Mathematical Monthly, Vol. 94, No. 6, June–July 0 6 a 6 b then 1987, pp. 527–528.)

ln b ln a + ln b x sln a+ln bd 2 # x e e # y e e > sln b - ln ad 6 e dx 6 sln b - ln ad. Lln a 2 c. Use the inequality in part (b) to conclude that F C b - a a + b 2ab 6 6 . ln b - ln a 2 E M This inequality says that the geometric mean of two positive B numbers is less than their , which in turn is A D less than their arithmetic mean. x ln a ln a ln b ln b (For more about this inequality, see “The Geometric, 2 Logarithmic, and Arithmetic Mean Inequality” by Frank Burk, NOT TO SCALE