The Exponential Function - Department of Mathematics | …deturck/m103/rogawski/chap7od… · · 2004-07-26CHAPTER7 The Exponential Function 7.1 Derivative of bx and the number

CHAPTER

7 The ExponentialFunction

7.1 Derivative of bx and the number e

Preliminary Questions

1. Which of the following equations is incorrect?(a) 32 · 35 = 37

(b) (√

5)4/3 = 52/3

(c) 32 · 23 = 1(d) (2−2)−2 = 16

2. To which of the following functions does the Power Rule apply?(a) x2

(b) 2e

(c) xe

(d) ex

3. For which values of b does bx have a negative derivative?

4. For which values of b is bx concave up?

5. Which point lies on the graph of y = bx for all b?

6. Which of the following statements is not true?(a) (ex)′ = ex

(b) limh→0

(eh − 1)/h = 1

(c) The tangent line to y = ex at x = 0 has slope e.(d) The tangent line to y = ex at x = 0 has slope 1.

1

2 Chapter 7 The Exponential Function

Exercises

1. Rewrite as a whole number (without using a calculator).(a) 70

(b) 102(2−2 + 5−2)

(c) (43)5/(45)3

(d) 274/3

(e) 8−1/3 · 85/3

(f) 3 · 41/4 − 12 · 2−3/2

(a) 70 = 1(b) 102(2−2 + 5−2) = 100(1/4 + 1/25) = 25 + 4 = 29(c) (43)5/(45)3 = 415/415 = 1(d) (27)4/3 = (271/3)4 = 34 = 81(e) 8−1/3 · 85/3 = (81/3)5/81/3 = 25/2 = 24 = 16(f) 3 · 41/4 − 12 · 2−3/2 = 3 · 21/2 − 3 · 22 · 2−3/2 = 0

In Exercises 2–10, solve for the unknown variable.2.

92x = 98

3. e2x = ex+1

If e2x = ex+1 then 2x = x + 1 if and only if x = 1.4.et2 = e4t−3

5. 3x = ( 13 )

x+1

3x = (1/3)x+1 is equivalent to 3x = 3−(x+1). Then x = −(x + 1) if and only ifx = −x − 1 if and only if 2x = −1 if and only if x = −1/2.6.(√

5)x = 1257. 4−x = 2x+1

Rewrite 4−x as (22)−x = 2−2x . Then 2−2x = 2x+1 or −2x = x + 1. Solving for xgives −3x = 1, or x = −1/3.8.b4 = 1012

9. k3/2 = 27

k3/2 = 27 is equivalent to (k1/2)3 = (91/2)3, so k = 9.10.(b2)x+1 = b−6

In Exercises 11–16, find the equation of the tangent line at the point indicated.

11. 4ex , x0 = 0

Let f (x) = 4ex . Then f ′(x) = 4ex and f ′(0) = 4. At x0 = 0, f (0) = 4. So thetangent line is y = 4(x − 0) + 4 = 4x + 4.12.e4x , x0 = 0

13. ex+2, x0 = −1

Let f (x) = ex+2 Then f ′(x) = ex+2 and f ′(−1) = e1. At x0 = −1, f (−1) = e,so the tangent line is y = e(x + 1) + e = ex + 2e.14.ex2

, x0 = 1

7.1 Derivative of bx and the number e 3

15. x2 ex , x0 = 2

Let f (x) = x2ex Then f ′(x) = 2xex + x2ex and f ′(2) = 8e2. At x0 = 2,f (2) = 4e2, so the tangent line is y = 8e2(x − 2) + 4e2 = 8e2x − 12e2.16.7e−2x + 5e7x , x0 = 1

In Exercises 17–40, find the derivative.

17. 7e2x + 3e4x

ddx 7e2x + 3e4x = 14e2x + 12e4x

18.e−5x

19. e−4x+9

ddx e−4x+9 = −4e−4x+9

20.4e−x + 7e−2x

21.ex2

x

ddx

ex2

x = −ex2

x2 + 2xex2

x = −ex2

x2 + 2ex2

22.x2e2x

23. x2ex

ddx x2ex = 2xex + x2ex

24.(1 + ex)4

25. (2e3x + 2e−2x)4

ddx (2e3x + 2e−2x)4 = 4(2e3x + 2e−2x)3(6e3x − 4e−2x

26.ex2+2x−3

27. e1/x

ddx e1/x = −e1/x

x228.e3

29. esin x

ddx esin x = cos xesin x

30.e(x2+2x+3)2

31. sin(ex)

ddx sin(ex) = ex cos ex

32.e

√t

33.1

1 − e−3t

ddt

11−e−3t = d

dx 1 − e−3t)−1 = −1(1 − e−3t)−2(3e3t) = −3e3t

(1−e−3t )234.

et+1/2t−1

35. cos(te−2t)

ddt cos(te−2t) = (e−2t − 2te−2t) sin(te−2t)e−2t(1 − 2t) sin(te−2t)

36.ex

3x + 137. tan(e5−6x)

ddx tan(e5−6x) = sec2(e5−6x)(−6e5−6x) = −6e5=6xsec2e5−6x

38.ex+1 + x

2ex − 139. eex

ddx eex = ex eex


40.exex

In Exercises 41–46, find the critical points and determine if they are local minima,maxima, or neither.

41. f (x) = ex − x

f ′(x) = ex − 1. Setting equal to zero and solving for x gives ex = 1, which istrue if and only if x = 0. f ′′(x) = ex , so f ′′(0) = e0 = 1 > 0. Therefore, x = 0is a local minimum.42.f (x) = x + e−x

43. f (x) = ex

xfor x > 0

f ′(x) = ex x−ex

x2 . Setting equal to zero and solving for x:ex (x−1)

x2 = 0 if and only if x = 1 which is our critical point.

f ′′(x) = (ex (x−1)+ex )x2+2x(ex (x−1))

x4 . f ′′(1) = (e(0)+e)+2(e(0))

1 = e > 0. Therefore,x = 1 is a local minimum.44.f (x) = x2ex

45. g(t) = et

t2 + 1

g′(x) = ex (x−1)2

(x2+1)2 and

g′′(x) = ex(x4 − 4x3 + 8x2 − 4x − 1)

(x2 + 1)3

so second derivative test fails. However, g′(x) doesn’t change sign, so x = 1 isnot a local extremum.46.g(t) = (t3 − 2t)et

In Exercises 47–52, find the critical points and points of inflection. Then sketch thegraph.

47. xe−x

1 f ′(x) = e−x − xe−x = (1 − x)e−x = 0 at x = 1, acritical point. f ′′(x) = −e−x − e−x + xe−x = −2e−x + xe−x = 0 at x = 2 whichis a point of inflection.48.e−x + ex

49. e−x cos x on [−π

2 , π

2 ]

7.1 Derivative of bx and the number e 5

f ′(x) = −e−x(sin x + cos x), critical pointsare 3π/4 and 7π/4. And f ′′(x) = 2 sin xe−x and inflection points are atx = 0, π .50.e−x2

51. ex − x

f ′(x) = ex − 1 = 0 at x = 0, a critical point.f ′′(x) = ex does not equal zero anywhere, so there are no points of inflection.52.x2e−x on [0, 10]

53. Use Newton’s Method to find solutions of ex = 5x up to 3 decimal places.

1 2 3

10

20

Figure 1 Graphs of ex and 5x .

In other words, we solve for the roots of f (x) = ex − 5x = 0. We use theformula xn+1 = xn − f (xn)

f ′(xn)to find our solutions. f ′(x) = ex − 5, so

xn+1 = xn − exn −5x

exn −5 . From the graph, we pick the starting point as x0 = 1/3:

x1 = 1/3 − f (1/3)

f ′(1/3)x1 ≈ .25813

x2 = .25813 − f (.25813)

f ′(.25813)x2 ≈ .25917

x3 = .25917 − f (.25917)

f ′(.25917)x3 ≈ .25917

The root is approximately, x = .259.

In Exercises 54–66, evaluate the indefinite integral.54. ∫

(ex + 2) dx


55.∫

4ex dx∫4ex dx = e4x + c56. ∫e4x dx

57.∫

xex2dx∫

xex2dx = ex2

/2 + c58. ∫(e4x + 1) dx

59.∫

e−9x dx∫e−9x dx = −e−9x/9 + c60. ∫(ex + e−x) dx

61.∫(e−x − 4x) dx∫(e−x − 4x) dx = −e−x − 4x2/2 + c = −e−4 − 2x2 + c62. ∫(7 − e10x) dx

63.∫

xe−4x2dx∫

xe−4x2dx = e−4x2

/(−8) + c64. ∫ex cos(ex) dx

65.∫

ex

√ex + 1

dx∫ex

√ex + 1

dx = 2(ex + 1)1/2 + c66. ∫

ex(e2x + 1)3 dx67. Find an approximation to m3 using the limit definition and estimate the slope of

the tangent line to y = 3x at x = 0 and x = 2.

m3 ≈ limh→03x+h−3x

h = limh→03x (3h−1)

h = 3x . Therefore the slopes of the tangentlines are f ′(0) ≈ 1 and f ′(2) ≈ 968.Find the area between the curves y = ex and y = e2x over [0, 1].

69. Find the area between the curves y = ex and y = e−x over [0, 2].∫ 20 ex − e−x dx = ex + ex |20 = 2e2 − 270.

Find the area bounded by the curves y = e2 and y = ex and x = 0.

Further Insights and Challenges

71. Prove that ex is not a polynomial function. Hint: differentiation lowers the degreeof a polynomial by one.

Letting f (x) = ex , we see that f ′(x) = ex . The degree x has not changes, so ex

cannot be a polynomial.72.Calculate the first three derivatives of f (x) = xex . Then guess the formula forf (n)(x) (use induction to prove it if you are familiar with this method of proof).73. Consider the equation ex = λx where λ is a constant.(a) For which values of λ does the equation have at least one solution?(b) For which values of λ does the equation have a unique solution? For

intuition, draw a graph of ex and the line y = λx .

(a) For all λ ≥ e and λ ≤ 0(b) The solution is unique if y = λx is tangent to ex . If they intersect at x = a,

this gives: λa = ea and slope = λ = ea. This gives a = 1 and λ = e74.

Prove that the numbers ma and mb satisfy the relation

mab = ma + mb

in two different ways.

(a) First method: use the limit definition of mb and the relation

(ab)h − 1

h= bh

(ah − 1

h

)+ bh − 1

h

(b) Second method: use the relation (bx)′ = mbbx together with the ProductRule applied to ax bx = (ab)x .

7.2 Inverse Functions 7

7.2 Inverse Functions


1. Which of the following functions f (x) coincides with its own inverse? In otherwords, f −1(x) = f (x).(a) f (x) = x(b) f (x) = 1 − x(c) f (x) = 1(d) f (x) = √

x(e) f (x) = |x |(f) f (x) = x−1

2. The graph of a function looks like the track of a roller coaster. Is the functionone-to-one?

3. Consider the function f mapping teenagers in the U.S. to their last names. Explainwhy the inverse of f does not exist.

4. View the following fragment of a train schedule for the New Jersey Transit Systemas defining a function f from towns to times. Is f one-to-one? What is f −1(6:27)?

Trenton 6:21

Hamilton Township 6:27

Princeton Junction 6:34

New Brunswick 6:38

5. A homework problem asks for a sketch of the graph of the inverse of f (x) =x + cos x . Frank, after trying but failing to find a formula for f −1(x), says it’simpossible to graph the inverse. Sally hands in an accurate sketch without solvingfor f −1. How did Sally do it?

Exercises

In Exercises 1–8, find a domain on which f (x) is one-to-one and find a formula forthe inverse of f restricted to this domain. Sketch the graph of f (x) and f −1(x).

1. f (x) = 3x − 2


y = 3x − 2 if and only if x = (y + 2)/3 So, f −1(x) = (x + 2)/3 and its domainis all reals.

Function Inverse of Function

2.f (x) = 4 − x

3. f (x) = 1

x + 1

y = 1

x + 1if and only if y(x + 1) = 1 if and only if x + 1 = 1

yif and only if

x = 1

y− 1. So, f −1(x) = 1

x− 1 and its domain is (−∞, 0) ∪ (0, ∞)

–4 –2 2 4

–4

–2

2

4

–4 –2 2 4

–4

–2

2

4

Function Inverse

4.f (x) = 1

7x − 35. f (x) = 1

x2

y = 1

x2if and only if x2 y = 1 if and only if x2 = 1

yif and only if x = 1√

y.


Hence, f −1(x) = 1√x

and its domain is (0, ∞).

–4 –2 2 4

–4

–2

2

4

1 2 3 4 5

–4

–2

2

4

Function Inverse

6.f (x) = 1√

x2+17. f (x) = x3

y = x3 if and only if y1/3 = x . So, f −1(x) = (x + 1)1/3 and its domain is allreals.

–4 –2 2 4

–4

–2

2

4

–4 –2 2 4

–4

–2

2

4

Function Inverse

8.f (x) = √

x3 + 99. For each function shown in Figure 1, sketch the graph of the inverse (restrict the

function’s domain if necessary).


(A)

(B)

(C)

(D)

(E)

(F)

Figure 1

(a)

(b)

(c)

(d)


(e)

(f)10.

Find the inverse of f (x) = (x − 2)/(x + 3).11. Let f (x) = x7 + x + 1.

(a) Show that f −1 exists (but do not attempt to find the inverse). Hint: show thatf is increasing.

(b) What is the domain of f −1?(c) Find f −1(3). Hint: solve f (x) = 3.

(a)Looking at the graph, we see that f (x) is one-to-one.

(b) The domain is (−∞, ∞)

(c) x = 1, since f (1) = 17 + 1 + 1 = 312.

Show that the inverse of f (x) = e−x exists (without finding the inverseexplicitly). What is the domain of f −1?13. Show that f (x) = (x2 + 1)−1 is one-to-one on (−∞, 0] and find a formula forf −1 for this domain.


This is what the graph from (−∞, 0) looks like. It is clearly one to one.y = (x2 + 1)−1 if and only if y = 1/(x2 + 1) if and only if x2 + 1 =1/y if and only if x2 = 1

y − 1 if and only if x = ±√

1y − 1 So,

f −1(x) = −√

1x − 1.

14.Let f (x) = x2 − 2x . Determine a domain on which f −1 exists (Hint: draw agraph) and find a formula for f −1 on this domain (Hint: use the quadraticformula).

15. Show that f (x) = x + x−1 is one-to-one on [1, ∞) and find a formula for f −1 onthis domain. What is the domain of f −1?

Looking at the graph, we see that f (x) is one-to-one on the given domain. Thedomain of f −1(x) is [2, ∞) since that is the range of f (x). We solve for f −1(x):y = x + (1/x) if and only if y = x2+1

x2 if and only if y − 1 = 1/x2 if and only ifx2 = 1/(y − 1) if and only if x = √

1/(y − 1). Therefore,f −1(x) = √

1/(x − 1).16.

Let f (x) = √x2 − 9 (x ≥ 9) and let g(x) be its inverse.

(a) Determine the domain and range of f and g.(b) Calculate g(x).(c) Calculate g′(x) in two ways: first using Theorem ?? and then by direct

calculation.

17. Let g(x) be the inverse of f (x) = x3 + 1.(a) Find a formula for g(x).(b) Calculate g(x) in two ways: first using Theorem ?? and then by direct

calculation.

(a) To find g(x), we solve y = x3 + 1 for x :

y = x3 + 1

y − 1 = x3

Therefore x = (y − 1)13 and the inverse is g(x) = (y − 1)

13 .

(b) We have f ′(x) = 3x2. According to Theorem ??,

g′(x) = 1

f ′(g(x))= 1

3g(x)2= 1

3(y − 1)23

= 13 (y − 1)− 2

3

This agrees with the answer we get by differentiating directly:

d

dx(y − 1)

13 = 1

3 (y − 1)− 23

In Exercises 18–23, use Theorem ?? to calculate g′(x) where g = f −1.


18.7x + 6

19.√

3 − x

Let f (x) = (3 − x)1/2 then f ′(x) = (1/2)(3 − x)−1/2(−1) = −12(3−x)1/2 Then

solving y = √3 − x for x and switching variables, we obtain the inverse

g(x) = 3 − x2. Thus, g′(x) = 1/ −12(3−3x+x2)1/2 = −2x

20.x−5

21. 4x3 − 1

Let f (x) = 4x3 − 2 then f ′(x) = 12x2. Then solving y = 4x3 − 1 for x and thenswitching variables, we obtain the inverse g(x) = ( x+1

4 )1/3. Thus,g′(x) = (1/12)( x+1

4 )−2/3

22. x

x + 123. 2 + x−1

let f (x) = 2 + (1/x) then f ′(x) = −1/x2. Then solving y = 2 + x−1 for x andswitching variables, we obtain the inverse g(x) = 1/(x − 2). Thusg′(x) = 1/( −1

(1/((x−2)2)= −1/((x − 2)2)

In Exercises 24–29, calculate g(b) and g′(b) where g = f −1 (without calculating g(x)

explicitly).24.

f (x) = x + cos x , b = 125. f (x) = 4x3 − 2x , b = −2

f (−1) = −2, so g(−2) = −1. . f ′(x) = 12x2 − 2 sof ′(g(−2)) = f ′(−1) = 12 − 2 = 10. Thus, g′(−2) = 1/1026.f (x) = √

x2 + 6x , b = 427. f (x) = 1/(x + 1), b = 1

4

f (3) = 1/4, so g(1/4) = 3. f ′(x) = −1(x+1)2 so

f ′(g(1/4)) = f ′(3) = −1(3+1)2 = −1/16. Thus, g′(x) = 1/( −1

(3+1)2 ) = −1628.

f (x) = ex , b = e29. f (x) = cos(x2), b = −1

f ′(x) = −2x sin(x2) and f (√

π) = −1, so g′(−1) = 1/ f ′(√

π).30.

R & W Use graphical reasoning to determine if the following statements aretrue or false. If false, modify to make a correct statement.

(a) If f (x) is increasing, then f −1 is decreasing.(b) If f (x) is decreasing, then f −1 is decreasing.(c) If f (x) is concave up, then f −1 is concave up.(d) If f (x) is concave down, then f −1 is concave up.(e) Linear functions f (x) = ax + b (a �= 0) are always one-to-one.(f) Quadratic polynomials f (x) = ax2 + bx + c (a �= 0) are always one-to-one.(g) sin x is not one-to-one.


31. Show that the inverse of an odd function is again an odd function. On the otherhand, explain why an even function does not have an inverse.

Graphically speaking, f −1(x) is simply f (x) reflected across the line x = y.Therefore, if f (x) is odd, then f −1(x) is odd as well. On the other hand, an evenfunction is never one-to-one as it always fails the horizontal line test.32.Use the Intermediate Value Theorem to show that if f (x) is continuous andone-to-one on an interval [a, b], then f (x) is either an increasing or decreasingfunction. Give an example of a one-to-one function (necessarily discontinuous)which is neither increasing nor decreasing on [a, b].


7.3 Logarithms and their derivatives


1. When is ln x negative?

2. What is the logarithm of −3?

3. How do you explain the fact that the functions ln x and ln(2x) have the samederivative? In fact, ln(ax) and ln x have the same derivative for any constant a.

4. Why is the area under the hyperbola y = 1/x between 1 and 3 equal to the areaunder the same hyperbola between 5 and 15?

5. The slope of the tangent line to the curve y = bx at x = 0 is 4. What is b?

Exercises

In Exercises 1–11, calculate directly (without using a calculator).

1. log3(27)

We set equal to x . xx = 37 if and only if x = 3.2.log2(8

3/5)3. log5(

125)

5x = 1/25 if and only if 5x = 5−2 if and only if x = −24.log64(4)

5. log7(492)

7x = 292 if and only if 7x = (72)2 if and only if 7x = 74 if and only if x = 4.6.log8(2) + log4(2)

7. log25(30) + log25(56 )

This is equivalent to log25(30) · (5/6) = log25 25 = 18.log4(48) − log4(12)

9. ln(√

e · e7/5)

ln(√

e · e7/5) = ln(e1/2 · e7/5) = ln(e1/2+7/5) = ln(e19/10) = 19/1010.ln(e3) + ln(e4)

11. log2(4) + log2(21) − log2(24) − log2(7)

log2(4 · 21) − log2(24) − log2(7) = log2(4 · 21) − log2(24 · 27) = log2(4·2124·7) =

log2(3/6) = log2(1/2) = −112.Find the equation of the tangent line to y = ln x at x = 3.

13. Find the equation of the tangent line to y = 2x at x = 3.

y′ = ln 2 · 2x . At x = 3, the slope is 8 ln 2 and y = 8. The tangent line isy = 8 ln 2(x − 3) + 8

In Exercises 14–31, find the derivative.14.

x ln x

7.3 Logarithms and their derivatives 15

15. (ln x)2

ddx (ln x)2 = 2 ln x(1/x) = (2/x) ln x

16.ln(x2)

17. ln

(x + 1

x3 + 1

)

We rewrite: ln(x + 1) as ln(x3 + 1). Thus, ddx ln(x + 1)− ln(x3 + 1) = 1

x+1 − 3x2

x3+118.ln(x3 + 3x + 1)

19. ln(2x)

We rewrite ln(2x) as x ln 2 Thus, ddx x ln 2 = ln 2

20.ln(ex + 1)

21. ln(sin x + 1)

ddx ln(sin x + 1) = cos x

sin x+122.ln(ln x)

23. x2 ln xd

dx x2 ln x = 2x ln x + x2

x = 2x ln x + x24.

ln x

x25. (ln(ln x))3

ddx (ln(ln x))3 = 3(ln(ln x))2(1/ ln x)(1/x) = 3(ln(ln x))2

x ln x26.ln((ln x)3)

27. x2x

We rewrite x2x as eln x2x = e2x ln x . Thusd

dx e2x ln x = (2 ln x + 2x/x)e2x ln x = (2 ln x + 2)e2x ln x

28.xex

29. xx2

We rewrite xx2as eln x x2 = ex2

ln x Thus,d

dx ex2 ln x = (2x ln x + x2/x)ex2 ln x = (2x ln x + x)ex2 ln x

30.x2x

31. ex x

We rewrite ex xas eeln xx = eex ln x

. Thus, ddx eex ln x = (ln x + 1)ex ln x eeex ln x

In Exercises 32–41, find the equation of the tangent line at the point indicated.32.

4x , x = 333. (

√2)x , x = √

2

Let f (x) = (√

2)x . Then f (√

2) = (√

2)√

2. f ′(x) = ln√

2 · (√2)x . So,f ′(

√2) = ln

√2 · (

√2)

√2 is the slope of the tangent line

y = (ln√

2)(√

2√

2)(x − √

2) + √234.

37t , t = 235. π(3x+9), x = 1

Let f (x) = π3x+9. Then f (1) = π12. f ′(x) = 3 ln π · π3x+9 sof ′(1) = 3π12 ln π is the slope of the tangent line y = 3π12 ln π(x − 1) + π12

36.5y2−2y+9, y = 1

37. ln t , t = 5

Let f (t) = ln t . Then f (5) = ln 5. f ′(t) = 1/t so f ′(5) = 1/5 is the slope of thetangent line y = (1/5)(x − 5) + ln 5.


38.ln(8 − 4t), t = 1

39. ln(x2), x = 4

Let f (x) = ln x2 = 2 ln x . Then f (4) = 2 ln 4. f ′(x) = 2/x so f ′(4) = 1/2which is the slope of the tangent line y = (1/2)(x − 4) + 2 ln 440.4 ln(9x + 2), x = 2

41. ln(sin x), x = π

4

Let f (x) = ln sin x then f (π/4) = ln(√

2/2). f ′(x) = cos x/ sin x = cotx sof ′(π/4) = 1 is the slope of the tangent line y = (x − π/4) + ln(

√2/2)

In Exercises 42–45, find the local extreme values in the domain {x : x > 0} anduse the Second Derivative Test to determine whether these values are local minima ormaxima.42.

ln x

x43.ln x

x2

Let f (x) = ln xx2 . Then f ′(x) = x2/x−2x ln x

x4 = x−2x ln xx4 = 1−2 ln x

x3 . To find criticalpoints, f ′(x) = 1−2 ln x

x3 = 0 at ln x = 1/2 or x = e1/2.

f ′′(x) = (−2/x)x3−(1−2 ln x)3x2)

x6 = −5x2−6x2 ln xx6 = −5−6 ln x

x4 sof ′′(e1/2) = −5−6(1/2)

(e1/2)4 = −8/e2 < 0. Thus, x = e1/2 is a local maximum44.

x − ln x45. x ln x

Let f (x) = x − ln x . Then f ′(x) = ln x + 1. To find critical pointsf ′(x) = ln x + 1 = 0 at x = 1/e. f ′′(x) = 1/x sof”(e)=1/(1/e)=e¿0.T hus,x=1/eisalocalminimum.

In Exercises 46–51, evaluate the definite integral.46. ∫ 2

1

1

xdx

47.∫ e

1

1

xdx

∫ e

1

1

xdx = ln x |e1 = ln e − ln 1 = 1

48. ∫ 12

4

1

xdx

49.∫ e3

e

1

tdt

∫ e3

e

1

tdt = ln t |e3

e = ln e3 − ln e = 3 − 1 = 250. ∫ −e

−e2

1

tdt

51.∫ en

em

1

tdt (any n, m)

∫ en

em

1

tdt = ln t |en

em = ln en − ln em = n − m

In Exercises 52–69, evaluate the indefinite integral, using substitution if necessary.


52. ∫dx

2x + 453.∫

t dt

t2 + 4

Let u = t2 + 4. Then du = 2t dt or 12 du = t dt , and∫

t

t2 + 4dt = 1

2

∫1

udu = 1

2ln

(t2 + 4

) + C

54. ∫x2dx

x3 + 255.∫

(3x − 1)dx

9 − 2x + 3x2

Let u = 9 − 2x + 3x2 then du = −2 + 6x = 6x − 2 = 2(3x − 1)dx . So∫(3x − 1)dx

9 − 2x + 3x2= (1/2)

∫duu = (1/2) ln(9 − 2x + 3x2) + C .

56. ∫tan(4x + 1) dx

57.∫

cot x dx

We rewrite∫

cot x dx as∫

cos xsin x dx . Let u = sin x then du = cos x dx so∫

cos xsin x dx = ∫

duu = ln sin x + c .58. ∫

cos x

2 sin x + 3dx

59.∫

ln x

xdx

Let u = ln x then du = (1/x)dx . So,∫ln x

xdx =

∫udu = u2/2 = (ln x)2

2+ C

60. ∫4 ln x + 5

xdx

61.∫

(ln x)2

xdx

Let u = ln x then du = (1/x)dx . So∫(ln x)2

xdx =

∫u2du = (u3/3) = (ln x)3

3+ c

62. ∫dx

x ln x63.∫

dx

(4x − 1) ln(8x − 2)

Let u = 4x − 1. Then du = 4 dx and the integral becomes∫

du

4(u ln(2u)).

Further substitute w = ln(2u) = ln 2 + ln u. This makes dw = 1

u, so the integral

becomes∫

dw

4w= 1

4ln w + C . Back substituting twice, we get

1

4ln w + C = 1

4ln(ln 2u) + C = 1

4ln(ln(8x − 2)) + C .

64. ∫ln(ln x)

x ln xdx65.

∫cot x dx∫

cot x dx = ∫ cos x

sin xdx . Let u = sin x , so that du = cos x dx and the integral

becomes∫ du

u= ln |u| + C = ln | sin x | + C .


66. ∫3x dx

67.∫

x3x2dx

Let u = x2 then du = 2xdx . So∫

x3x2dx

= (1/2)∫

3udu = (1/2) 3u

ln 3 + C = 3x2

2 ln 3 + C68. ∫

cos x 3sin x dx69.

∫( 1

2 )3x+2 dx

Let u = 3x + 2 then du = 3dx . So,∫( 1

2 )3x+2 dx = (1/3)

∫(1/2)udu = (1/3)

(1/2)u

ln 1/2 + C = (1/2)3x+2

2 ln(1/2)+ C .

In Exercises 70–75, evaluate the derivative using logarithmic differentiation as in Ex-ample ??.70.

x(x + 1)3

(3x − 1)271.x(x2 + 1)√

x + 1

Let f (x) = x(x2+1)√x+1

. Then

ln f (x) = ln x(x2 + 1) − ln(x + 1)1/2 = ln x + ln(x2 + 1) − (1/2) ln(x + 1).d

dx ln f (x) = 1/x+

2xx2+1 − 1

2(x+1)= f ′(x)

f (x)We multiply through by f (x) to get

f ′(x) = x(x2+1)√(x+1)

( 1x + 2x

x2+1 − 12(x+1)

)72.

(2x + 1)(4x2)√

x − 9

73.

√x(x + 2)

(2x + 1)(2x + 2)

let f (x) =√

x(x+2)

(2x+1)(2x+2)Then

ln f (x) = (1/2)[ln(x) + ln(x + 2) − ln(2x + 1) − ln(2x + 2)].d

dx ln f (x) = (1/2)( 1x + 1

x+2 + −22x+1 + −2

2x+2) = f ′(x)

f (x)We multiply through by

f (x) to get f ′(x) = (1/2)(

√x(x+2)

(2x+1)(2x+2)) · ( 1

x + 1x+2 + −2

2x+1 + −22x+2)74.

(x2 + 1)(x2 + 2)(x2 + 3)2

75.x cos x

(x + 1) sin x

Let f (x) = x cos x(x+1) sin x . Then

ln f (x) = ln x cos x − ln(x + 1) sin x = ln x + ln cos x − ln(x + 1) − ln sin x .d

dx ln f (x) = 1x − sin x

cos x − 1x+1 − cos x

sin x = 1x − tan x − 1

x+1 − cotx . We multiplythrough by f (x) to get f ′(x) = x cos x

(x+1) sin x ( 1x − tan x − 1

x+1 − cotx)76.

What is the relationship between logb x and log1/b x?


77. (a) Show that

d

dxln( f (x)g(x)) = f ′(x)

f (x)+ g′(x)

g(x)(1)

for any differentiable functions f and g.


(b) Use Eq. (1) to compute the derivative of ln(x cos x) and ln((x − 3)√

x + 6).(c) Give a new proof of the Product Rule by observing that the right-hand side of

Eq. (1) is equal to ( f (x)g(x))′/( f (x)g(x)).

(a) ddx ln f (x)g(x) = d

dx (ln f (x) + ln g(x)) = f ′(x)

f (x)+ g′(x)

g(x)

(b) ddx ln x cos x = 1

x + − sin xcos x and

ddx ln(x − 3)

√x + 6 = 1

x−3 + 12(x+6)1/2(x+6)1/2 = 1

x−3 + 12(x+6)1/2

(c) f ′(x)

f (x)+ g′(x)

g(x)= f ′(x)g(x)+ f (x)g′(x)

f (x)g(x)= ( f (x)g(x))′

f (x)g(x)78.

Show that loga b · logb a = 1.79. Verify the formula

logb x = loga x

loga b

for any positive numbers a, b.

Equivalently, we cross multiply and prove that loga b · logb x = loga x . Letloga b = m and logb x = n then am = b and bn = x . Substituting, we have(bm)n = x if and only if bmn = x . So, loga x = mn. Since m = loga b andn = logb x , that implies that loga b · logb x = loga x80.Use Exercise 79 to verify the formula

d

dxlogb x = 1

(ln b)x

81. Compute the derivative of log10 x at x = 2.d

dx log10 x = ddx

ln xln 10 = 1

x ln 10 at x = 2 ⇒ 12 ln 1082.

Find the equation of the tangent line to y = logb x at x = b (any b > 0).83. Defining ln x as an Integral. Define a function ϕ(x) in the domain x > 0:

ϕ(x) =∫ x

1

1

tdt

This exercise proceeds as if we didn’t know that ϕ(x) = ln x and deduces thebasic properties of ln x from the integral expression. Prove the followingstatements:

(a)∫ b

1

1

tdt =

∫ ab

a

1

tdt for all a, b > 0.

Hint: use the substitution u = t/a.(b) ϕ(ab) = ϕ(a) + ϕ(b). Hint: break up the integral from 1 to ab into two

integrals and use (a).(c) ϕ(1) = 0 and ϕ(a−1) = −ϕ(a) for a > 0.(d) ϕ(an) = nϕ(a) for all a > 0 and integers n.(e) ϕ(a1/n) = 1

n ϕ(a) for all a > 0 and integers n.(f) ϕ(ar ) = rϕ(a) for all a > 0 and rational number r .(g) There exists x such that ϕ(x) > 1. Hint: show that ϕ(a) > 0 for any a > 1.

Then take x = am for m > 1/ϕ(a).(h) Show that ϕ(t) is increasing and use the Intermediate Value Theorem to

show that there exists a unique number e such that ϕ(e) = 1.(i) Show that ϕ(x) is a continuous function satisfying ϕ(er ) = r for any rational

number r .


(a) Firstly,∫ b

11t dt = ln t |b1 = ln b − ln 1 = ln b = ϕ(b). Then∫ ab

a 1/tdt = ln ab − ln a = ln aba = ln b = ϕ(b)

(b) Using part a:ϕ(ab) = ∫ ab

1 1/tdt = ∫ a1 1/tdt + ∫ ab

a 1/tdt = ∫ a1 1/tdt + ∫ b

1 1/tdt .

(c) ϕ(1) = ∫ 11 1/tdt = 0 and using part a,

ϕ(a−1) = ϕ(1/a) = ∫ 1/a1 1/tdt = ∫ 1

a 1/tdt = −ϕ(a)

(d) Using part b: ϕ(an) = ϕ(a · a · a · . . . ) a multiplied n times, which is equal toϕ(a) + ϕ(a) + ϕ(a) + . . . n times, which is equal to nϕ(a)

(e) ϕ(a1/n) = ∫ a1/n

1 1/tdt = ln(a1/n = ln(a)/n = (1/n)ϕ(a)

(f) Let r = m/n where m and n are integers. Thusϕ(ar ) = ϕ(am/n = ϕ(am · a1/n) = ϕ(am) + ϕ(a1/n) =mϕ(a) + (1/n)ϕ(a) = (m/n)ϕ(a) = rϕ(a)

(g) Firstly, ϕ(x) = ln x is greater than 0 for all values of x > 1. We take x = am

for m > 1/ϕ(a). Then ϕ(x) = ϕ(am) = mϕ(a) > 1 since m > 1/ϕ(a).Therefore, such x exists.

(h) ϕ(x) is increasing since ϕ′(x) = x−1 > 0. Since ϕ(1) = 0 and ϕ(x) > 1 forsome x > 0, the Intermediate Value Theorem implies that there exists somenumber which we call e between 1 and x such that ϕ(e) = 1. It is uniquesince ϕ is increasing.

(i) This follows from part f and part h.84.

Let g(x) be the inverse of a function f (x) satisfying f (xy) = f (x) + f (y).Show that g(x)g(y) = g(x + y).

7.4 Exponential growth and decay


1. Two quantities increase exponentially with growth constants k = 1.2 and k = 3.4,respectively. Which quantity doubles more rapidly?

2. If you are given the doubling time and the growth constant of a quantity thatincreases exponentially, can you determine the initial amount?

3. Does it make sense to speak of the doubling time of a quantity P that satisfies alinear growth law P(t) = P0t? Why or why not?

4. Referring to his popular book, A Brief History of Time, the renowned physicistStephen Hawking said “Someone told me that each equation I included in thebook would halve its sales.” If n denotes the number of equations (and we treatn as a continuous variable), which differential equation is satisfied by S(n), thenumber of books sold?

5. Carbon dating is based on the assumption that the ratio R of C12–C14 in the atmo-sphere has been constant over the past 50,000 years. If R were actually smaller inthe past than it is today, would the age estimates produced by carbon dating be tooancient or too recent?

7.4 Exponential growth and decay 21

Exercises

1. A certain bacteria population P obeys the exponential growth law P(t) =2000e1.3t (t in hours).(a) How many bacteria are present initially?(b) At what time will there be 10, 000 bacteria?

(a) P(0) = 2000e0 = 2000 bacteria initially.(b) We solve for t. 2000e1.3t = 10, 000 if and only if e1.3t =

5 if and only if ln e1.3t = ln 5 if and only if 1.3t = ln 5. Sot = ln 5/1.3 = 1.24 hours.

2.

A quantity P obeys the exponential growth law P(t) = e5t (t in years).(a) At what time t is P = 10?(b) At what time t is P = 20?(c) What is the doubling time for P?

3. A certain bacteria colony doubles in size every hour. Find the formula for thenumber of bacteria present at time t , assuming that initially the colony contains1500 bacteria.

The doubling time is given as one hour. Therefore, 1 = ln 2/k. The initial colonyis 1500, so P(0) = 1500. Thus we obtain the formulaP(t) = 1500eln 2t = 1500(2t)4.

A quantity P obeys the exponential growth law P(t) = Cekt (t in years).(a) What is the growth constant for P if the doubling time is 7 years?(b) Find the formula for P(t), assuming that the doubling time is 7 years as in

(a) and that P(0) = 100.

5. The decay constant of cobalt is .13. What is its half-life?

half life=ln 2/.13 = .0536.Find the decay of constant of radium, given that its half-life is 1,622 years.

7. (a) Find all solutions to the differential equation y′ = −5y.(b) Find the solution satisfying the initial condition y(0) = 3.4.

(a) y′ = −5y, so the solutions are all equations of the form y(t) = ce−5t forsome constant c.

(b) Initial condition is given as y(0) = 3.4 which determines the constant c.Therefore, y(t) = 3.4e−5t .

8.

The population of a city is P(t) = 2 · e.06t (in millions), where t is measured inyears.

(a) Calculate the doubling time of the population.(b) How long does it take for the population to triple in size?(c) How long does it take for the population to quadruple in size?

9. The population of Washington state increased from 4.86 million in 1990 to 5.89million in 2000. Assuming exponential growth,(a) What will the population be in 2010?(b) What is the doubling time?

We let 1990 be our starting point.(a) P(0) = 4.86. Therefore, P(t) = 4.86ekt . In 2000, 10 years have gone by, so

P(10 = 5.89 = 4.86ek10. We use this to solve for k.5.89 = 4.86ek10 if and only if 1.212 = ek10 if and only if ln 1.212 =k10 if and only if k = ln 1.212

10 = .019. Therefore in 2010, t = 20 andp(20 = 4.86e.019(20) = 7.11 million people.

(b) Doubling time = ln 2/.019 ≈ 36 years.10.

Light intensity The intensity of light passing through an absorbing mediumdecreases exponentially with the distance traveled. Suppose the decay constantfor a certain plastic block is 2 when the distance is measured in feet. How thickmust the block be to reduce the intensity by a factor of one-third?


11. An insect population triples in size after 5 months. Assuming exponentialgrowth, when will it quadruple in size?

Suppose we have P(t) = P0ekt . We know at t = 5 months, p(t + 5) = 3P(t).We solve for k using this. P(t + 5) = 3P(t) if and only if P0ek(t+5) =3P0ekt if and only if P0ekt+5k = 3P0ekt if and only if e5k =3 if and only if ln e5k = ln 3 if and only if k = ln 3/5 = .22. Therefore,P(t) = P0e.22t . We see how long it takes to quadruple in size, but solving for T.P(t + T ) = 4P(t) if and only if P0e.22(t+T ) = 4P0e.22t if and only if e.22t e.22T =4e.22t if and only if e.22T = 4 if and only if ln e.22T = ln 4 if and only if T =ln 4/.22 if and only if T = 6.301 months.12.A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find thedecay constant of the isotope.13. Measurements showed that a sample of sheepskin parchment discovered byarchaeologists had a C14 to C12 ratio equal to 40% of that found in theatmosphere. Approximately how old is the parchment?

The C14 − C12 ratio at time t is Re−.000121t .40 =e−.00012t if and only if − .00012t = ln .40 if and only if t = ln .40

−.00012 = 7635.76years old.14.Chauvet Caves In 1994, rock climbers in southern France stumbled upon acave containing prehistoric cave paintings. C14 analysis of charcoal specimenscarried by French archeologist Helene Valladas showed that the paintings arebetween 29,700 and 32,400 years old, much older than any previously knownhuman art. Given that the C14–C12 ratio of the atmosphere is R = 10−12, whatC14–C12 ratio did Valladas find in the charcoal specimens?

15. Holocene Ice Age On the basis of diverse sources of data, a scientist believesthat the Holocene ice age, which stretches to the present time, should have begunbetween 10,000 and 12,000 years ago. She then learns that remains of animalsbelieved to have died at the onset of this ice age have been discovered. Whatrange of C14–C12 ratio would she expect to find in the remains if her theory iscorrect?

We have 10−12e−.000121(10000) = .2981 and 10−12e−.000121(12000) = .2341, so it isbetween 23% and 30%.16.Find the formula for the function f (t) that satisfies the differential equationy′ = −.7y and the initial condition y(0) = 10.17. Atmospheric Pressure Let P(h) be the atmospheric pressure (in pounds persquare inch) at a height h (in miles) above sea level on earth. It can be shown thatP satisfies a differential equation P ′ = −k P for some positive constant k.(a) Suppose measurements with a barometer show that P(0) = 14.7 and

P(10) = 2.13. What is the decay constant k?(b) Determine the atmospheric pressure 15 miles above sea level.

(a) Since P ′ = −k P for some positive constant k, the P(k) = Ce−kh whereC = P(0) = 14.7. Therefore, P(h) = 14.7e−kh . We know thatP(10 = 14.7e−k10 = 2.13. Solving for k,2.13 = 14.7e−k10 if and only if 2.10

14.7 = e−k10 if and only if ln( 2.1314.7 =

−k10 if and only if k = .193(b) P(15) = 14.7e−.193(15) = .813 pounds per square inch.

18.

A certain quantity P increases according to the square law P(t) = t2.(a) Starting at time t0 = 1, how long will it take for P to double in size? How

long will it take starting at t0 = 2 or 3?(b) In general, starting at time t0, how long will it take for P to double in size? In

other words, find � such that P(t0 + �) = 2P(t0).(c) How is the answer to (b) different from the answer you would obtain if P

grew exponentially?

19. Moore’s Law In 1965, Gordon Moore predicted that the number N oftransistors on a microchip would increase exponentially.(a) Does the following table of data confirm Moore’s prediction for the period

1972–2000? If so, estimate the growth constant k.(b) Let N (t) be the number of transistors t years after 1972. Find an

approximate formula N (t) ≈ Cekt where t is the number of years after 1972.

7.4 Exponential growth and decay 23

(c) Estimate the doubling time in Moore’s Law for the period 1972–2000.(d) If Moore’s Law continues to hold until the end of the decade, how many

transistors will there be in 2010?(e) Can Moore have expected his prediction to hold indefinitely?

Year No. Transistors

Transistors 4004 1971 2,2508008 1972 2,5008080 1974 5,0008086 1978 29,000286 1982 120,000386 processor 1985 275,000486 DX processor 1989 1,180,000Pentium processor 1993 3,100,000Pentium II processor 1997 7,500,000Pentium III processor 1999 24,000,000Pentium 4 processor 2000 42,000,000

0

5,000,000

10,000,000

15,000,000

20,000,000

25,000,000

30,000,000

35,000,000

40,000,000

45,000,000

1965 1970 1975 1980 1985 1990 1995 2000 2005

Year

No.

of

tran

sist

ors

Figure 1

(a) Yes, the graph looks like an exponential graph especially towards the latteryears. We estimate the growth constant by setting 1972 as our starting point,so P0 = 2250. Therefore, P(t) = 2250ekt . In 2000, t = 28. Therefore,P(28) = 2250e28k = 42000000 if and only if e28k =1866667 if and only if k = ln 18666.67

28 = .351. Remark: one can find a betterestimate by calculating k for each time t and then averaging the k values.

(b) N (t) = 2250e.351t

(c) doubling time is ln 2/.351 = 1.97(d) in 2010, t = 38 years. Therefore, N (38) = 2250e.351(38) = 1, 395, 732, 906(e) No, you can’t make a microchip smaller than an atom.

20.

Economic Model An economist wishes to test the hypothesis that in a certaincountry, the monthly rate at which jobs increase is proportional to the number ofpeople who already have jobs. Let J (t) be the number of jobs (in millions) attime t (in months).

(a) How would he formulate his hypothesis as a differential equation?(b) Suppose there are 15 million jobs at t = 0 and 17 million jobs six months

later. If his hypothesis is correct, how many jobs will there be at the end ofthe year?



21. Verify that the half-life of a quantity that decays exponentially with decay constantk is equal to ln 2/k.

Let y = Ce−kt be an exponential decay function. Let t be the half-life of the

quantity y, that is, the time t where y = C

2. Solve for t :

C

2= Ce−kt , so

1

2= e−kt . Taking logarithms of both sides, we get − ln 2 = −kt , so t = ln 2/k.

22.Isotopes for Dating Which of the following isotopes would be most suitablefor dating extremely old rocks: Carbon-14 (half-life 5570 years), Lead-210(half-life 22.26 years), Potassium-49 (half-life 1.3 billion years)? Explain why.

23. Two bacteria colonies are cultivated in a laboratory. The first colony has adoubling time of 2 hours and the second colony has a doubling time of 3 hours.Suppose that initially the first colony contains 1000 bacteria and the secondcolony contains 3000 bacteria. At what time will the two colonies be of equalsize?

Colony 1 has a half life of 2 = ln 2k1. Therefore, k1 = .347. ThusP1(t) = 1000e.347t . Colony 2 has a half life of 3 = ln 2/k2 = .231. Therefore,k2 = .231. Thus, P2(t) = 3000e.231t . We set P1(t) = P2(t) and solve for t :1000e.347t = 3000e.231t if and only if e.347t = 3e.231t if and only ifln e.347t = ln 3e.231t if and only if .347t = ln 3 + .231t if and only if .116t = ln 3if and only if t = 9.47 hours.24.Show that if a quantity P grows quadratically (that is, P(t) = P0t2), then P doesnot have a doubling time. In other words, there is no fixed time interval T suchthat P(t + T ) = 2P(t).

25. Let P = P(t) be a quantity that obeys an exponential growth law with growthconstant k. Show that P increases m-fold after an interval of ln m/k-years.

P(t + ln m/k) = Cek(t+ln m/k) = Cekt+ln m = Cekt eln m = eln mCekt = mcekt =m P(t)26.Inversion of Sugar When cane sugar is dissolved in water, a reaction takesplace that transforms the cane sugar into so-called invert sugar. The reactiontakes place over a period of several hours. The percentage f (t) of unconvertedcane sugar at time t is known to decrease exponentially. Suppose thatf ′ = −.2 f . What percentage of cane sugar remains unconverted after 5 hours?After 10 hours?

27. Suppose that a quantity P increases exponentially with doubling time 6 hours.After how many hours has P increased by 50%?

Doubling time is 6 = ln 2/k so k = .116. P(t) = Ce.116t Solve for t such thatP(t + T ) = P(t) + P(t)/2. e.116t e.116T = e.116t + (e.116t/2 if and only if e.116T =1 + 1/2 if and only if.116T = ln 1.5 if and only if T = 3.49 ≈ 3.5hours

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The Exponential Function - Department of Mathematics | …deturck/m103/rogawski/chap7od… · · 2004-07-26CHAPTER7 The Exponential Function 7.1 Derivative of bx and the number