calculus one and several variables 10e salas solutions manual ch07

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-07 JWDD027-Salas-v1 November 25, 2006 15:41

SECTION 7.1 341

CHAPTER 7

SECTION 7.1

1. Suppose f(x1) = f(x2) x1 �= x2. Then

5x1 + 3 = 5x2 + 3 ⇒ x1 = x2;

f is one-to-one

f(y) = x

5y + 3 = x

5y = x− 3

y = 15 (x− 3)

f−1(x) = 15 (x− 3)

dom f−1 = (−∞,∞)

2. f−1(x) =13(x− 5)

dom f−1 = (−∞,∞)

3. f is not one-to-one; e.g. f(1) = f(−1) 4. f−1(x) = x1/5; dom f−1 = (−∞,∞)

5. f ′(x) = 5x4 ≥ 0 on (−∞,∞) and

f ′(x) = 0 only at x = 0; f is increasing.

Therefore, f is one-to-one.

f(y) = x

y5 + 1 = x

y5 = x− 1

y = (x− 1)1/5

f−1(x) = (x− 1)1/5

dom f−1 = (−∞,∞)

6. not one-to-one; e.g. f(0) = f(3)

7. f ′(x) = 9x2 ≥ 0 on (−∞,∞) and



f(y) = x

1 + 3y3 = x

y3 = 13 (x− 1)

y =[13 (x− 1)

]1/3f−1(x) =

[13 (x− 1)

]1/3dom f−1 = (−∞,∞)

8. f−1(x) = (x + 1)1/3

dom f−1 = (−∞,∞)



342 SECTION 7.1

9. f ′(x) = 3(1 − x)2 ≥ 0 on (−∞,∞) and



f(y) = x

(1 − y)3 = x

1 − y = x1/3

y = 1 − x1/3

f−1(x) = 1 − x1/3

dom f−1 = (−∞,∞)

10. not one-to-one; e.g. f(0) = f(2).

11. f ′(x) = 3(x + 1)2 ≥ 0 on (−∞,∞) and

f ′(x) = 0 only at x = −1; f is increasing.


f(y) = x

(y + 1)3 + 2 = x

(y + 1)3 = x− 2

y + 1 = (x− 2)1/3

y = (x− 2)1/3 − 1

f−1(x) = (x− 2)1/3 − 1

dom f−1 = (−∞,∞)

12. f−1(x) =14(x1/3 + 1)

dom f−1 = (−∞,∞)

13. f ′(x) =3

5x2/5> 0 for all x �= 0;

f is increasing on (−∞,∞)

f(y) = x

y3/5 = x

y = x5/3

f−1(x) = x5/3

dom f−1 = (−∞,∞)

14. f−1(x) = (1 − x)3 + 2

dom f−1 = (−∞,∞)

15. f ′(x) = −3(2 − 3x)2 ≤ 0 for all x and

f ′(x) = 0 only at x = 2/3; f is decreasing

f(y) = x

(2 − 3y)3 = x

2 − 3y = x1/3

3y = 2 − x1/3

y = 13 (2 − x1/3)

f−1(x) = 13 (2 − x1/3)

dom f−1 = (−∞,∞)

16. not one-to-one; e.g. f(1) = f(−1)



SECTION 7.1 343

17. f ′(x) = cosx ≥ 0 on [−π/2, π/2]; and f ′(x) = 0 only at x = −π/2, π/2. Therefore f is increasing

on [−π/2, π/2] and f has an inverse. The inverse is denoted by arcsin x; this function will be studied

in Section 7.7. The domain of arcsin x is [−1, 1] = range of sinx on [−π/2, π/2].

18. f is not one-to-one on (−π/2, π/2). For example, f(−π/4) = f(π/4) =√

22

.

19. f ′(x) = − 1x2

< 0 for all x �= 0;

f is decreasing on (−∞, 0) ∪ (0,∞)

f(y) = x =⇒ 1y

= x =⇒ y =1x

f−1(x) =1x

dom f−1: x �= 0

20. f−1(x) = 1 − 1x

dom f−1: x �= 0

21. f is not one-to-one; e.g. f(

12

)= f(2) 22. not one-to-one; e.g. f(1) = f(2)

23. f ′(x) = − 3x2

(x3 + 1)2≤ 0 for all x �= −1;

f is decreasing on (−∞,−1) ∪ (−1,∞)

f(y) = x

1y3 + 1

= x

y3 + 1 =1x

y3 =1x− 1

y =(

1x− 1)1/3

f−1(x) =(

1x− 1)1/3

dom f−1: x �= 0

24. f−1(x) =x

1 + xdom f−1: x �= −1

25. f ′(x) =−1

(x + 1)2< 0 for all x �= −1;

f is decreasing on (−∞,−1) ∪ (−1,∞)

f(y) = x

y + 2y + 1

= x

y + 2 = xy + x

y(1 − x) = x− 2

y =x− 21 − x

f−1(x) =x− 21 − x

dom f−1: x �= 1

26. not one-to-one; e.g. f(1) = f(−3)



344 SECTION 7.1

27. They are equal.

28. 29.

30. 31.

32. (a) Suppose f and g are one-to-one, and that f (g(x1)) = f (g(x2)). Then since f is one-to-one, g(x1) =

g(x2), and since g is one-to-one this implies x1 = x2.

(b) Since g−1(f−1 (f (g(x)))) = g−1 (g(x)) = x, we have (f ◦ g)−1 = g−1 ◦ f−1.

33. (a) f ′(x) = x2 + 2x + k: f will be increasing on (−∞,∞) if f ′ does not change sign. This will occur

if the discriminant of f ′, namely 4 − 4k is non-positive.

4 − 4k ≤ 0 =⇒ k ≥ 1

(b) g′(x) = 3x2 + 2kx + 1: g will be increasing on (−∞,∞) if g′ does not change sign. This will

occur if the discriminant of g′, namely 4k2 − 12 is non-positive.

4k2 − 12 ≤ 0 =⇒ k2 ≤ 3 =⇒ −√

3 ≤ k ≤√

3

34. (a)(f−1)′ (5) =

1f ′(2)

= − 43

(b) g′ =−1

(f−1)2· (f−1)′; g′(3) =

−122

· 12/3

= − 38

35. f ′(x) = 3x2 ≥ 0 on I = (−∞,∞) and f ′(x) = 0 only at x = 0; f is increasing on I and so

it has an inverse.

f(2) = 9 and f ′(2) = 12; (f−1)′(9) =1

f ′(2)=

112

36. f ′(x) = −2 − 3x2; (f−1)′ (4) =1

f ′ (f−1 (4))=

1f ′(−1)

= −15



SECTION 7.1 345

37. f ′(x) = 1 +1√x> 0 on I = (0,∞); f is increasing on I and so it has an inverse.

f(4) = 8 and f ′(4) = 1 +12

=32;(f−1)′ (8) =

1f ′(4)

=1

3/2=

23

38. f ′(x) = 1 + cosx; (f−1)′ (0) =1

f ′(f−1 (−1/2))=

1f ′(−π/6)

=12

39. f ′(x) = 2 − sinx > 0 on I = (−∞,∞); f is increasing on I and so it has an inverse.

f(π/2) = π and f ′(π/2) = 1; (f−1)′(π) =1

f ′(π/2)= 1

40. f ′(x) =−4

(x− 1)2; (f−1)′ (3) =

1f ′(f−1 (3))

=1

f ′(3)=

11/2

= 2

41. f ′(x) = sec2 x > 0 on I = (−π/2, π/2) ; f is increasing on I and so it has an inverse

f(π/3) =√

3 and f ′(π/3) = 4; (f−1)′(√

3) =1

f ′(π/3)=

14

42. f ′(x) = 5x4 + 6x2 + 2; (f−1)′ (−5) =1

f ′(f−1 (−5))=

1f ′(−1)

=113

43. f ′(x) = 3 +3x4

> 0 on I = (0,∞); f is increasing on I and so it has an inverse.

f(1) = 2 and f ′(1) = 6; f−1′(2) =

1f ′(1)

=16.

44. f ′(x) = 1 − sinx ≥ 0 on I = [0, π], with f(x) = 0 for only one value on I and so it has an inverse.

f(π) = −1 and f ′(π) = 1; f−1′(−1) =

1f ′(π)

= 1.

45. Let x ∈ dom (f−1) and let f(z) = x. Then

(f−1)′(x) =1

f ′(z)=

1f(z)

=1x

46. (f−1)′ (x) =1

f ′(f−1 (x))=

11 + [f (f−1(x))]2

=1

1 + x2

47. Let x ∈ dom (f−1) and let f(z) = x. Then

(f−1)′(x) =1

f ′(z)=

1√1 − [f(z)]2

=1√

1 − x2

48. (a) The figure indicates that f is one-to-one. (b)

f−1(x) =

{(x + 1)1/3, x < 0√x , x ≥ 0



346 SECTION 7.1

49. (a)f ′(x) =

(cx + d)a− (ax + b)c(cx + d)2

=ad− bc

(cx + d)2, x �= −d/c

Thus, f ′(x) �= 0 iff ad− bc �= 0.

(b)at + b

ct + d= x

at + b = ctx + dx

(a− cx)t = dx− b

t =dx− b

a− cx; f−1(x) =

dx− b

a− cx

50. f = f−1 =⇒ ax + b

cx + d=

dx− b

a− cx

=⇒ a2x + ab− acx2 = cdx2 + d2x− bd

=⇒ a = −d as long as either b or c �= 0.

If b = c = 0, then a = ±d.

51. (a) f ′(x) =√

1 + x2 > 0, so f is always increasing, hence one-to-one.

(b) (f−1)′ (0) =1

f ′(f−1(0))=

1f ′(2)

=1√5.

52. (a) f ′(x) =√

16 + (2x)4 (2) = 8√

1 + x4 > 0 for all x.

(b) Since f(1/2) = 0, we have

(f−1)′(0) =1

f ′(1/2)=

12√

17=

√17

34

53. (a) g′(x) =1

f ′[g(x)]; g′′(x) = − 1

(f ′[g(x)])2f ′′[g(x)]g′(x) = − f ′′[g(x)]

(f ′[g(x)])3

(b) If f ′ is increasing, then the graphs of f and g have opposite concavity;

if f ′ is decreasing then the graphs of f and g have the same concavity.

54. (a) No. If p is a polynomial of even degree, then limx→±∞

p(x) = ∞ or limx→±∞

p(x) = −∞.

(b) Yes, for instance P (x) = x3 has an inverse. P (x) = x3 − x does not have an inverse.

55. Let f(x) = sinx and let y = f−1(x). Then

sin y = x

cos ydy

dx= 1

dy

dx=

1cos y

(y �= ±π/2)

=1√

1 − sin2 y=

1√1 − x2

(x �= ±1)



SECTION 7.2 347

56. Let y = f−1(x). Then tan y = x, so sec2 ydy

dx= 1.

Thusdy

dx=

1sec2 y

= cos2 y =1

1 + x2

57. f−1(x) =x2 − 8x + 25

9, x ≥ 4 58. f−1(x) =

5x3 − 2x

59. f−1(x) = 16 − 12x + 6x2 − x3 60. f−1(x) =1 − x

1 + x= f(x)

61. f ′(x) = 3x2 + 3 > 0 for all x;


62. f ′(x) =35x−2/5 > 0 (x �= 0)


63. f ′(x) = 8 cos 2x > 0, x ∈ (−π/4, π/4);

f is increasing on [−π/4, π/4]

64. f ′(x) = 3 sin 3x > 0, x ∈ (0, π/3)

f is increasing on [0, π/3]

SECTION 7.2

1. ln 20 = ln 2 + ln 10 ∼= 2.99 2. ln 16 = ln 24 = 4 ln 2 ∼= 2.78

3. ln 1.6 = ln 1610 = 2 ln 4 − ln 10 ∼= 0.48 4. ln 34 = 4 ln 3 ∼= 4.39

5. ln 0.1 = ln 110 = ln 1 − ln 10 ∼= −2.30 6. ln 2.5 = ln 10

4 = ln 10 − ln 4 ∼= 0.92

7. ln 7.2 = ln 7210 = ln 8 + ln 9 − ln 10 ∼= 1.98 8. ln

√630 = 1

2 ln(9 · 7 · 10) = 12 (ln 9 + ln 7 +

ln 10) ∼= 3.22



348 SECTION 7.2

9. ln√

2 = 12 ln 2 ∼= 0.35 10. ln 0.4 = ln 4

10 = ln 4 − ln 10 ∼= −0.92

11. For any positive integer k:∫ 2k

k

1xdx = [ln x]2kk = ln 2k − ln k = ln

(2kk

)= ln 2.

12. Fix a positive integer m. For any positive integer k:

∫ km

k

1xdx = [ln x]kmk = ln km− ln k = ln

(km

k

)= ln m.

13. 12 [Lf (P ) + Uf (P )] = 1

2

[7631980 + 1691

3960

] ∼= 0.406 14. ln 2.5 ∼= 12

[Lf (P ) + Uf (P )] ∼= 0.921

15. (a) ln 5.2 ∼= ln 5 + 15 (0.2) ∼= 1.65 16. (a) ln 10.3 ∼= ln 10 + 1

10 (0.3) ∼= 2.33

(b) ln 4.8 ∼= ln 5 − 15 (0.2) ∼= 1.57 (b) (b) ln 9.6 ∼= ln 10 + 1

10 (−0.4) ∼= 2.26

(c) ln 5.5 ∼= ln 5 + 15 (0.5) ∼= 1.71 (c) ln 11 ∼= ln 10 + 1

10 (1) ∼= 2.40

17. x = e2 18. x =1e

19. 2 − lnx = 0 or lnx = 0. Thus x = e2 or x = 1.

20. lnx1/2 − ln(2x− 1) = 0 =⇒ ln√x

2x− 1= 0 =⇒

√x

2x− 1= 1 =⇒ x = 1

21. ln[(2x + 1)(x + 2)] = 2 ln(x + 2)

ln[(2x + 1)(x + 2)] = ln[(x + 2)2]

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

x2 + x− 2 = 0

(x + 2)(x− 1) = 0x = −2, 1

We disregard the solution x = −2 since it does not satisfy the initial equation.

Thus, the only solution is x = 1.

22. 2 ln(x + 2) − 12

lnx4 = ln(x + 2)2

x2= 1 = ln e =⇒ (x + 2)2

x2= e =⇒ x =

−2(1 ±√

e)

23. See Exercises 3.1, (3.1.6).

limx→1

lnx

x− 1=

d

dx(lnx)

∣∣∣∣x=1

=1x

∣∣∣∣x=1

= 1

24. Let n > 2 be a positive integer. Let P be the partition {1, 2,· · · , n}. Let f(x) = 1/x.

Then the lower sum12

+13

+ · · · + 1n< lnn.

Therefore, k = n.



SECTION 7.2 349

25. Continuing Exercise 24, the upper sum12

+13

+ · · · + 1n− 1

> lnn.

Therefore, k = n− 1.

26. (a) ln 1 − g(1) = 1 > 0, ln 2 − g(2) = ln 2 − 2 < 0, so by the intermediate-value theorem

ln r − g(r) = 0 for some r ∈ [1, 2].

(b) r ∼= 1.7915

1 2x

1

2

y

27. (a) Let G(x) = lnx− sinx. Then G(2) = ln 2 − sin 2 ∼= −0.22 < 0 and G(3) = ln 3 − sin 3 ∼= 0.96 > 0.

Thus, G has at least one zero on [2, 3] which implies that there is at least one number r ∈ [2, 3] such

that sin r = ln r.

(b) r ∼= 2.2191

28. (a) ln 1 − 112

= −1 < 0, ln 2 − 122

∼= 0.69 − 14> 0, so by the intermediate-value theorem

ln r − 1r2

= 0 for some r ∈ [1, 2].

(b) r ∼= 1.5316

29. L = 1 30. L = 0



350 SECTION 7.3

SECTION 7.3

1. dom (f) = (0,∞) , f ′(x) =14x

(4) =1x

2. dom (f) =(−1

2,∞), f ′(x) =

22x + 1

3. dom (f) = (−1,∞) , f ′(x) =1

x3 + 1d

dx

(x3 + 1

)=

3x2

x3 + 1

4. dom (f) = (−1,∞) , f ′(x) =3

x + 1

5. dom (f) = (−∞,∞), f(x) =12

ln (1 + x2) so f ′(x) =12

[1

1 + x2(2x)

]=

x

1 + x2

6. dom (f) = (0,∞) , f ′(x) =3 (lnx)2

x

7. dom (f) = {x |x �= ±1} , f ′(x) =1

x4 − 1d

dx(x4 − 1) =

4x3

x4 − 1

8. dom (f) = (1,∞), f ′(x) =1

x lnx

9. dom (f) =(− 1

2 ,∞),

f ′(x) = (2x + 1)2d

dx[ln(2x + 1)] + 2(2x + 1)(2) ln (2x + 1)

= (2x + 1)22

2x + 1+ 4(2x + 1) ln (2x + 1)

= 2(2x + 1) + 4(2x + 1) ln (2x + 1) = 2(2x + 1) [1 + 2 ln (2x + 1)]

10. dom(f) = (−∞,−2)∪ (−2, 1)∪ (1,∞), f ′(x) =1

x + 2− 3x2

x3 − 1(rewrite f(x) as ln|x + 2|− ln|x3−1|)

11. dom (f) = (0, 1) ∪ (1,∞) , f(x) = (lnx)−1 so f ′(x) = − (lnx)−2 d

dx(lnx) = − 1

x (lnx)2

12. dom (f) = (−∞,∞) , f ′(x) =x

2(x2 + 1)(rewrite f(x) as

14

ln(x2 + 1).)

13. dom (f) = (0,∞) , f ′(x) = cos (lnx)(

1x

)=

cos (lnx)x

14. dom (f) = (0,∞) , f ′(x) = − sin(lnx)x

15.∫

dx

x + 1= ln |x + 1| + C



SECTION 7.3 351

16.∫

dx

3 − x= −

∫dx

x− 3= − ln |x− 3| + C

17.

{u = 3 − x2

du = −2x dx

};

∫x

3 − x2dx = −1

2

∫du

u= −1

2ln |u| + C = −1

2ln |3 − x2| + C

18.∫

x + 1x2

dx =∫ (

1x

+1x2

)dx = ln |x| − 1

x+ C

19.

{u = 3x

du = 3dx

};

∫tan 3x dx =

13

∫tanu du =

13

ln |secu| + C =13

ln | sec 3x | + C

20.∫

secπ

2x dx =

2π

ln∣∣∣sec π

2x + tan

π

2x∣∣∣+ C

21.

{u = x2

du = 2x dx

};∫

x secx2 dx =12

∫secu du =

12

ln | secu + tanu| + C

= 12 ln | secx2 + tanx2| + C

22.

{u = 2 + cotx

du = − csc2 x dx

};

∫csc2 x

2 + cotxdx = −

∫du

u= − ln |u| + C = − ln |2 + cotx| + C.

23.

{u = 3 − x2

du = −2x dx

};

∫x

(3 − x2)2dx = −1

2

∫du

u2=

12u

+ C =1

2 (3 − x2)+ C

24.

⎧⎪⎨⎪⎩

u = ln(x + a)

du =1

x + adx

⎫⎪⎬⎪⎭ ;

∫ln(x + a)x + a

dx =∫

u du =u2

2+ C =

12

[ln(x + a)]2 + C

25.

{u = 2 + cosx

du = − sinx dx

};

∫sinx

2 + cosxdx = −

∫1udu = − ln |u| + C = − ln |2 + cosx| + C

26.

{u = 4 − tan 2x

du = −2 sec2 2x dx

};∫

sec2 2x4 − tan 2x

dx = −12

∫du

u= −1

2ln |u| + C = −1

2ln |4 − tan 2x| + C

27.{u = lnx, du =

dx

x

};∫

dx

x lnx=∫

du

u= lnu + C = ln | lnx| + C

28.

{u = 2x3 − 1

du = 6x2 dx

};

∫x2

2x3 − 1dx =

16

∫du

u=

16

ln |u| + C =16

ln |2x3 − 1| + C

29.{u = lnx, du =

dx

x

};∫

dx

x (lnx)2=∫

du

u2= − 1

u+ C = − 1

lnx+ C



352 SECTION 7.3

30.

{u = 1 + sec 2x

du = 2 sec 2x tan 2x dx

};

∫sec 2x tan 2x1 + sec 2x

dx =12

∫1

1 + udu =

12

ln |1 + u| + C =12

ln |1 + sec 2x| + C

31.

{u = sinx + cosx

du = (cosx− sinx) dx

};

∫sinx− cosxsinx + cosx

dx = −∫

1udu = − ln |u| + C = − ln | sinx + cosx| + C

32.

⎧⎨⎩

u =√x

du =1

2√xdx

⎫⎬⎭ ;

∫1√

x(1 +√x)

dx = 2∫

du

1 + u= 2 ln |1 + u| + C = 2 ln |1 +

√x| + C

33.{u = 1 + x

√x , du =

32x1/2 dx

};∫ √

x

1 + x√x

dx =23

∫du

u=

23

ln |u| + C

=23

ln∣∣1 + x

√x∣∣+ C

34.{u = lnx, du =

1xdx

};∫

tan(lnx)x

dx =∫

tanu du = ln | secu| + C = ln | sec(lnx)| + C

35.∫

(1 + secx)2 dx =∫ (

1 + 2 secx + sec2 x)dx = x + 2 ln | secx + tanx| + tanx + C

36.∫

(3 − cscx)2 dx =∫

(9 − 6 cscx + csc2 x) dx = 9x− 6 ln | cscx− cotx| − cotx + C

37.∫ e

1

dx

x= [lnx]e1 = ln e− ln 1 = 1 − 0 = 1

38.∫ e2

1

dx

x= [ ln |x| ]e

2

1 = ln e2 − ln 1 = 2

39.∫ e2

e

dx

x= [ lnx ]e

2

e = ln e2 − ln e = 2 − 1 = 1

40.∫ 1

0

(1

x + 1− 1

x + 2

)dx =

[ln∣∣∣∣x + 1x + 2

∣∣∣∣]10

= ln23− ln

12

= ln43

41.∫ 5

4

x

x2 − 1dx =

[12

ln |x2 − 1|]54

=12

(ln 24 − ln 15) =12

ln85

42.∫ 1/3

1/4

tanπx dx =1π

[ln | secπx|]1/31/4 =1π

(ln 2 − ln

√2)

=ln 22π

.

43.

{u = 1 + sinx

du = cosx dx

x = π/6 =⇒ u = 3/2

x = π/2 =⇒ u = 2

};

∣∣∣∣∫ π/2

π/6

cosx1 + sinx

dx =∫ 2

3/2

du

u= [lnu]23/2 = ln

43



SECTION 7.3 353

44.∫ π/2

π/4

(1 + cscx)2 dx =∫ π/2

π/4

(1 + 2 cscx + csc2 x) dx = [x + 2 ln | cscx− cotx| − cotx]π/2π/4

=π

4+ 1 − 2 ln(

√2 − 1)

45.∫ π/2

π/4

cotx dx = [ ln |sinx| ]π/2π/4 = ln 1 − ln√

22

= ln√

2 =12

ln 2

46.{u = lnx, du =

dx

x

};∫ e

1

lnx

xdx =

[(lnx)2

2

]e1

=12

47. The integrand1

x− 2is not defined at x = 2.

48. Let f(x) = lnx. Then f ′(x) =1x

and f ′(1) = 1.

By the definition of the derivative at x = 1, we have

f ′(1) = limh→0

ln(1 + h) − ln(1)h

= limh→0

ln(1 + h)h

= limx→0

ln(1 + x)x

= 1

49. ln |g(x)| = 2 ln(x2 + 1) + 5 ln |x− 1| + 3 lnx

g′(x)g(x)

= 2(

2xx2 + 1

)+

5x− 1

+3x

g′(x) = (x2 + 1)2 (x− 1)5 x3

(4x

x2 + 1+

5x− 1

+3x

); g′(1) = 0

50. ln |g(x)| = ln |x| + ln |x + a| + ln |x + b| + ln |x + c|

g′(x)g(x)

=1x

+1

x + a+

1x + b

+1

x + c

g′(x) = x(x + a)(x + b)(x + c)(

1x

+1

x + a+

1x + b

+1

x + c

); g′(−b) = −b(a− b)(c− b)

51. ln |g(x)| = 4 ln |x| + ln |x− 1| − ln |x + 2| − ln(x2 + 1)

g′(x)g(x)

=4x

+1

x− 1− 1

x + 2− 2x

x2 + 1

g′(x) =x4(x− 1)

(x + 2) (x2 + 1)

(4x

+1

x− 1− 1

x + 2− 2x

x2 + 1

); g′(0) = 0

52. ln |g(x)| = 12 (ln |x− 1| + ln |x− 2| − ln |x− 3| − ln |x− 4|)

g′(x)g(x)

=12

(1

x− 1+

1x− 2

− 1x− 3

− 1x− 4

)

g′(x) =12

(√(x− 1) (x− 2)(x− 3) (x− 4)

)2 (1

x− 1+

1x− 2

− 1x− 3

− 1x− 4

); g′(2) = 0



354 SECTION 7.3

53.A =

∫ π/6

0

(2 − secx) dx

= [2x− ln | secx + tanx| ]π/60

=π

3− ln

∣∣∣∣ 2√3

+1√3

∣∣∣∣ = π

3− 1

2ln 3

54.A =

∫ 1

1/2

(csc

π

2x− x

)dx

=[

2π

ln | csc π

2x− cot

π

2x| − x2

2

]11/2

= − 2π

ln(√

2 − 1) − 38

=2π

ln(1 +√

2) − 38

55.A =

∫ π/4

0

(1 − tanx) dx

= [x− ln | secx|]π/40

=π

4− ln

√2 =

π

4− 1

2ln 2

56.A =

∫ π/4

0

(secx− cosx) dx

= [ln | secx + tanx| − sinx]π/40

= ln(1 +√

2) −√

22

57. A =∫ 4

1

[5 − x

4− 1

x

]dx =

[54x− 1

8x2 − lnx

]41

=158

− ln 4

58. A =∫ 2

1

(3 − x− 2

x

)dx =

[3x− x2

2− 2 lnx

]21

=32− 2 ln 2

59. V =∫ 8

0

π

(1√

1 + x

)2

dx = π

∫ 8

0

11 + x

dx = π [ ln |1 + x| ]80 = π ln 9

60. By shells: V =∫ 3

0

2πx · 31 + x2

dx =[3π ln(1 + x2)

]30

= 3π ln 10



SECTION 7.3 355

61.V =

∫ π/3

−π/3

π(√

secx)2

dx

= 2π∫ π/3

0

secx dx

= 2π [ln | secx + tanx|]π/30 = 2π ln (2 +√

3)

62.V =

∫ π/4

0

π tan2 x dx

= π

∫ π/4

0

(sec2 x− 1) dx

= π [tanx− x]π/40

= π(1 − π

4

)

63. v (t) =∫

a (t) dt =∫

− (t + 1)−2 dt =1

t + 1+ C.

Since v (0) = 1, we get 1 = 1 + C so that C = 0. Then

s =∫ 4

0

|v(t)| dt =∫ 4

0

dt

t + 1= [ ln (t + 1) ]40 = ln 5.

The particle traveled ln 5 ft.

64. v(t) =∫

a(t) dt =∫

−(t + 1)−2 dt =1

t + 1+ C, v(0) = 2 =⇒ v(t) =

1t + 1

+ 1

Then s =∫ 4

0

v(t) dt =∫ 4

0

(1

1 + t+ 1)

dt = [ln(t + 1) + t ]40 = 4 + ln 5 ft

65. d

dx(lnx) =

1x

d2

dx2(lnx) = − 1

x2

d3

dx3(lnx) =

2x3

d4

dx4(lnx) = −2 · 3

x4

...

dn

dxn(lnx) = (−1)n−1 (n− 1)!

xn

66.d

dx(ln(1 − x)) =

−11 − x

d2

dx2(ln(1 − x)) =

−1(1 − x)2

d3

dx3(ln(1 − x)) =

−2(1 − x)3

d4

dx4(ln(1 − x)) =

−2 · 3(1 − x)4

...

dn

dxn[ln(1 − x)] = − (n− 1)!

(1 − x)n



356 SECTION 7.3

67.∫

cscx dx =∫

cscx(cscx− cotx)cscx− cotx

dx =∫

csc2 − cscx cotxcscx− cotx

dx

{u = cscx− cotx

du = (− cscx cotx + csc2 x) dx

};∫

cscx dx =∫

du

u= ln |u| + C

= ln | cscx− cotx| + C

68. (a) If g(x) = g1(x)g2(x), (7.3.7) gives g′(x) = g(x)[g

′1(x)g1(x)

+g

′2(x)g2(x)

]

=⇒ g′(x) = g1(x)g2(x)[g

′1(x)g1(x)

+g

′2(x)g2(x)

]= g

′1(x) g2(x) + g1(x) g

′2(x).

(b) If g(x) =g1(x)g2(x)

= g1(x)(

1g2(x)

), then

=⇒ g′(x) = g1(x)(

1g2(x)

)⎛⎜⎝g′1(x)g1(x)

+

d

dx(1/g2(x))

1/g2(x)

⎞⎟⎠ =

g′1(x)g2(x) − g1(x)g

′2(x)

[g2(x)]2

69. f(x) = ln (4 − x), x < 4

f ′(x) =1

x− 4

f ′′(x) =−1

(x− 4)2

(i) domain (−∞, 4)

(ii) decreases throughout

(iii) no extreme values

(iv) concave down throughout: no pts of inflection

70. f ′(x) = 1 − 1x

f ′′(x) =1x2

(i) domain (0,∞)

(ii) decreases on (0, 1], increases on [1,∞)

(iii) f(1) = 1 local and absolute min

(iv) concave up on (0,∞); no pts of inflection



SECTION 7.3 357

71. f(x) = x2 lnx, x > 0

f ′(x) = 2x lnx + x

f ′′(x) = 2 lnx + 3

(i) domain (0, ∞)

(ii) decreases on (0, 1/√e] , increases on [1/

√e, ∞)

(iii) f(1/√e) = −1/2e local and absolute min

(iv) concave down on (0, 1/e3/2),

concave up on (1/e3/2, ∞);

pt of inflection at (1/e3/2, −3/2e3)

72. f ′(x) = − 2x4 − x2

f ′′(x) = −2(4 + x2)(4 − x2)2

(i) domain (−2, 2)

(ii) increases on (−2, 0], decreases on [0, 2)

(iii) f(0) = ln 4 local and absolute max

(iv) concave down on (−2, 2);

no pts of inflection

73. f(x) = ln[

x

1 + x2

], x > 0

f ′(x) =1 − x2

x + x3

f ′′(x) =x4 − 4x2 − 1

(x + x3)2

(i) domain (0, ∞)

(ii) increases on (0, 1], decreases on [1,∞)

(iii) f(1) = local and absolute max

(iv) concave down on (0, 2.0582) ,

concave up on (2.0582,∞) ;

pt of inflection at (2.0582,−0.9338) (approx.)

1 5 10x

-2

-1

y



358 SECTION 7.3

74. f(x) = ln[

x3

x− 1

]

f ′(x) =2x− 3x2 − x

f ′′(x) =−(2x2 − 6x + 3

)x2(x− 1)2

(i) domain (−∞, 0) ∪ (1, ∞)

(ii) increases on[32 ,∞

), decreases on (−∞, 0) ∪

(1, 3

2

](iii) f(3/2) ∼= 1.91 local and absolute min

(iv) concave down on (−∞, 0) ∪ (2.366,∞)) , concave up on (1, 2.366) ;

pt of inflection at (2.366, 2.272) (approx.)

-1-2 1 2 3 4x

5y

75. Average slope =1

b− a

∫ b

a

1xdx =

1(b− a)

lnb

a

76. (a) f ′(x) =12x

(2) =1x

; g′(x) =13x

(3) =1x

(b) F ′(x) =1kx

(k) =1x

(c) F (x) = ln kx = ln k + lnx, so F ′(x) = 0 +d

dx(lnx) =

1x

.

77. 78.

x-intercept: 1; abs min at x = 1/e2; x-intercept at x=1; abs min at x ∼= 0.7165;

abs max at x = 10 abs max at x = 2



SECTION 7.3 359

79.

x-intercepts: 1, 23.1407; abs min at x = 100;

abs max at x = 4.8105

80.

x-intercept at x = π/2; abs max at x = π/2;

local min at x ∼= 0.7269; abs min at x = 2;

81. (a) v(t) − v(0) =∫ t

0

a(u)du, 0 ≤ t ≤ 3

v(t) =∫ t

0

[4 − 2(u + 1) +

3u + 1

]du + 2

=[2u− u2 + 3 ln |u + 1|

]t0

= 2 + 2t− t2 + 3 ln(t + 1)

(b) (c) max velocity at t = 1.5811; min velocity at t = 0

82. (a) v(t) =∫

a(t) dt =∫ [

2 cos 2(t + 1) +2

t + 1

]dt = sin 2(t + 1) + 2 ln(t + 1) + C

v(0) = 2 =⇒ v(t) = sin 2(t + 1) + 2 ln(t + 1) + 2 − sin 2

(b) (c) max at t ∼= 6.1389; min at t ∼= 1.1092

1 2 3 4 5 6 7x

2

4

6

y

83. (b) x-coordinates of the points of intersection: x = 1, 3.3028

(c) A ∼= 2.34042



360 SECTION 7.4

84. (b) x-coordinates of the points of intersection: x1∼= −0.6180, x2

∼= 1.6180, x3∼= 2.6180

(c) A =∫ x3

x2

[f(x) − g(x)] dx ∼= 0.2549

85. (a) f(x) =lnx

x2, f ′(x) =

1 − 2 lnx

x3, f ′′(x) =

−5 + 6 lnx

x4

(b) f(1) = 0, f ′(e1/2) = 0, f ′′(e5/6) = 0

(c) f(x) > 0 on (1,∞) f ′(x) > 0 on (0, e1/2) f ′′(x) > 0 on (e5/6,∞)

f(x) < 0 on (0, 1) f ′(x) < 0 on (e1/2,∞) f ′′(x) > 0 on(0, e5/6

)(d) f(e1/2) local and absolute maximum

86. (a) f(x) =1 + 2 lnx

2√

lnx, f ′(x) =

2 lnx− 14x(lnx)3/2

, f ′′(x) =3 − 4(lnx)2

8x2(lnx)5/2

(b) f(x) �= 0, f ′(e1/2) = 0, f ′′(e√

3/2) = 0

(c) f(x) > 0 on (1,∞) f ′(x) > 0 on (e1/2,∞) f ′′(x) > 0 on(1, e

√3/2)

f ′(x) < 0 on (1, e1/2) f ′′(x) < 0 on(e√

3/2,∞)

(d) f(e1/2) local and absolute maximum

SECTION 7.4

1.dy

dx= e−2x d

dx(−2x) = −2e−2x 2.

dy

dx= 3e2x+1 · 2 = 6e2x+1

3.dy

dx= ex

2−1 d

dx(x2 − 1) = 2xex

2−1 4.dy

dx= 2e−4x(−4) = −8e−4x

5.dy

dx= ex

d

dx(lnx) + lnx

d

dx(ex) = ex

(1x

+ lnx

)

6.dy

dx= 2xex + x2ex

7.dy

dx= x−1 d

dx(e−x) + e−x d

dx(x−1) = −x−1e−x − e−xx−2 = −(x−1 + x−2)e−x

8.dy

dx= e

√x+1

(1

2√x

)

9.dy

dx=

12(ex − e−x) 10.

dy

dx=

12(ex − e−x(−1)) =

12(ex + e−x)

11.dy

dx= e

√x d

dx

(ln√x)

+ ln√x

d

dx(e

√x) = e

√x

(1√x· 12√x

)+ ln

√xe√x

2√x

=12e√x

(1x

+ln√x√x

)

12.dy

dx= 3(3 − 2e−x)2(−2e−x(−1)) = 6e−x(3 − 2e−x)2



SECTION 7.4 361

13.dy

dx= 2(ex

2+ 1)

d

dx(ex

2+ 1) = 2(ex

2+ 1)ex

2 d

dx(x2) = 4xex

2(ex

2+ 1)

14.dy

dx= 2(e2x − e−2x) · (2e2x + 2e−2x) = 4(e4x − e−4x)

15.dy

dx= (x2 − 2x + 2)

d

dx(ex) + ex(2x− 2) = x2ex

16.dy

dx= 2xex + x2ex − ex

2 − xex22x = (x2 + 2x)ex − (2x2 + 1)ex

2

17.dy

dx=

(ex + 1) ex − (ex − 1) ex

(ex + 1)2=

2ex

(ex + 1)2

18.dy

dx=

2e2x(e2x + 1) − (e2x − 1)2e2x

(e2x + 1)2=

4e2x

(e2x + 1)2

19. y = e4 lnx = (elnx)4 = x4 sody

dx= 4x3. 20. y = ln e3x = 3x =⇒ dy

dx= 3

21. f ′(x) = cos(e2x) e2x · 2 = 2e2x cos(e2x) 22. f ′(x) = esin 2x · 2 cos 2x

23. f ′(x) = e−2x(− sinx) + e−2x(−2) cosx = − e−2x(2 cosx + sinx)

24. f ′(x) = − 1cos e2x

· sin e2x(2e2x) = −2e2x tan(e2x)

25.∫

e2x dx =12e2x + C 26.

∫e−2x dx = −1

2e−2x + C

27.∫

ekx dx =1kekx + C 28.

∫eax+b dx =

1aeax+b + C

29.{u = x2, du = 2x dx

};∫

xex2dx =

12

∫eu du =

12eu + C =

12ex

2+ C

30.∫

xe−x2dx = −1

2

∫eu du = −1

2eu + C = −1

2e−x2

+ C

31.{u =

1x, du = − 1

x2dx

};∫

e1/x

x2dx = −

∫eu du = −eu + C = −e1/x + C

32.∫

e2√x

√x

dx = e2√x + C 33.

∫ln ex dx =

∫x dx =

12x2 + C

34.∫

elnx dx =∫

x dx =x2

2+ C 35.

∫4√ex

dx =∫

4e−x/2 dx = −8e−x/2 + C



362 SECTION 7.4

36.∫

ex

ex + 1dx =

∫du

u= ln |u| + C = ln(ex + 1) + C

37.

{u = ex + 1

du = ex dx

};

∫ex√ex + 1

dx =∫

du√u

=∫

u−1/2 du = 2u1/2 + C = 2√ex + 1 + C

38.

{u = eax

2+ 1

du = 2axeax2dx

};

∫xeax

2

eax2 + 1dx =

12a

∫du

u=

12a

ln |u| + C =12a

ln(eax2+ 1) + C

39.

{u = 2e2x + 3

du = 4e2x dx

};

∫e2x

2e2x + 3dx =

14

∫du

u=

14

lnu + C =14

ln(2e2x + 3) + C

40.

{u = e−2x

du = −2e−2x dx

};

∫sin(e−2x)

e2xdx = −1

2

∫sinu du =

12

cosu + C =12

cos (e−2x) + C

41. {u = sinx, du = cosx dx};∫

cosx esinx dx =∫

eu du = eu + C = esinx + C

42. {u = e−x, du = − e−x dx };∫e−x [ 1 + cos (e−x)] dx = −

∫(1 + cosu) du = −u− sinu + C = −e−x − sin (e−x) + C

43.∫ 1

0

ex dx = [ ex ]10 = e− 1

44.∫ 1

0

e−kx dx = −1k

[e−kx]10 =1k

(1 − e−k)

45.∫ lnπ

0

e−6x dx =[− 1

6e−6x

]lnπ

0= −1

6e−6 lnπ +

16e0 =

16(1 − π−6

)

46.∫ 1

0

xe−x2dx = −1

2[e−x2

]10 =12

(1 − 1

e

)

47.∫ 1

0

ex + 1ex

dx =∫ 1

0

(1 + e−x) dx = [x− e−x]10 =(1 − e−1

)− (0 − 1) = 2 − 1

e

48.∫ 1

0

4 − ex

exdx =

∫ 1

0

(4e−x − 1) dx = [−4e−x − x]10 = 3 − 4e−1

49.∫ ln 2

0

ex

ex + 1dx = [ ln (ex + 1) ]ln 2

0 = ln (eln 2 + 1) − ln (e0 + 1) = ln 3 − ln 2 = ln32

50.∫ 1

0

ex

4 − exdx = [− ln |ex − 4| ]10 = ln

(3

4 − e

)



SECTION 7.4 363

51.∫ 1

0

x(ex2+ 2) dx =

∫ 1

0

(xex2+ 2x) dx =

[12ex

2+ x2

]10

=(

12e + 1

)−(

12 + 0

)= 1

2 (e + 1)

52.∫ ln π

4

0

ex sec ex dx = [ln | sec ex + tan ex|]lnπ4

0 = ln

(√2 + 12

).

53. (a) f(x) = eax, f ′(x) = aeax, f ′′(x) = a2eax, . . . , f (n)(x) = aneax

(b) f(x) = e−ax, f ′(x) = −ae−ax, f ′′(x) = a2e−ax, . . . , f (n)(x) = (−1)n ane−ax

54. (a) x′(t) = kAekt − kBe−kt

x′(t) = 0 =⇒ kAekt − kBe−kt = 0 =⇒ e2kt =B

A=⇒ t =

lnB − lnA

2k

The particle is closest to the origin at time t =lnB − lnA

2k.

(b) x′′(t) = k2Aekt + k2Be−kt = k2 x(t); k2 is the constant of proportionality.

55. A = 2xe−x2

A′ = 2x(−2xe−x2) + 2e−x2

= 2e−x2(1 − 2x2) = 0 =⇒ x = ±

√12

and y =1√e.

Put the vertices at(± 1√

2,

1√e

).

56. y = ex =⇒ x = ln y. The area of the rectangle is given by: A = y ln y.

dA

dt= (1 + ln y)

dy

dt.

At y = 3,dy

dt=

12. Thus

dA

dt=

12(1 + ln 3) square units per minute.

57. f(x) = e−x2

f ′(x) = −2xe−x2

f ′′(x) = (4x2 − 2)e−x2

(a) symmetric with respect to the y-axis

f(−x) = f(x)

(b) increases on (−∞, 0 ] , decreases on [ 0, ∞)

(c) f(0) = 1 local and absolute max

(d) concave up on(−∞,−1/

√2)∪(1/√

2, ∞), concave down on

(−1/

√2, 1/

√2)

58. (a) V =∫ 1

0

π (ex)2 dx = π

∫ 1

0

e2x dx = π[

12 e

2x]10

= 12π[e2 − 1

]

(b) V =∫ 1

0

2πx ex dx = 2π∫ 1

0

x ex dx.



364 SECTION 7.4

59. (a) V =∫ 1

0

2π xe−x2dx = π

[− e−x2]1

0= π[1 − 1

e

]

(b) V =∫ 1

0

π [e−x2]2 dx = π

∫ 1

0

e−2x2dx

60.A =

∫ 0

− ln 4

(4 − e−y) dy +∫ 1

2 ln 4

0

(4 − e2y) dy

=[4y + e−y

]0− ln 4

+[4y − 1

2e2y

] 12 ln 4

0

= 12 ln 2 − 92

61.A =

∫ 2

0

(e2x − ex) dx +∫ 4

2

(e4 − ex) dx

=[12e

2x − ex]20

+[e4x− ex

]42

=(

12e

4 − e2 − 12 + 1

)+ (4e4 − e4 − 2e4 + e2)

= 12 (3e4 + 1)

62. A = triangle − upper left corner

=12e2 −

∫ 1

0

(e− ex) dx

=12e2 − [ex− ex]10

=12e2 − 1

63.A =

∫ 2

1

(ey − 2) dy

= [ey − 2y]21 = e2 − e− 2



SECTION 7.4 365

64. f ′(x) = −xex

f ′′(x) = −ex − xex = −ex(1 + x)

(i) domain (−∞, ∞)

(ii) increases on (−∞, 0 ] , decreases on [ 0, ∞)

(iii) f(0) = 1 local and absolute max

(iv) concave up on (−∞,−1), concave down on (−1,∞)

pt of inflection (−1, 2/e)

65. f(x) = e(1/x)2

f ′(x) =−2x3

e(1/x)2

f ′′(x) =6x2 + 4

x6e(1/x)2

(i) domain (−∞, 0) ∪ (0, ∞)

(ii) increases on (−∞, 0) , decreases on (0, ∞)

(iii) no extreme values

(iv) concave up on (−∞, 0) and on (0, ∞)

66. f ′(x) = (2x− x2)e−x

f ′′(x) = e−x(2 − 4x + x2)

(i) domain (−∞, ∞)

(ii) decreases on (−∞, 0 ], and on [2,∞), increases on [0, 2]

(iii) f(0) = 0 local and absolute min,

f(2) = 4e−2 local max

(iv) concave up on (−∞, 2 −√

2) and (2 +√

2,∞),

concave down on (2 −√

2, 2 +√

2);

points of inflection at x = 2 ±√

2.



366 SECTION 7.4

67. f(x) = x2 lnx

f ′(x) = 2x lnx + x

f ′′(x) = 3 + 2 lnx

(i) domain (0, ∞)

(ii) decreases on (0, e−1/2), increases on (e−1/2, ∞)

(iii) f(e−1/2) = −1/2e is a local and absolute min

(iv) concave down on (0, e−3/2) and concave up

on (e−3/2, ∞); (e−3/2, −3/2e3) is a point of inflection

68. f(x) = (x− x2)e−x

f ′(x) = (x2 − 3x + 1)e−x

f ′′(x) = −(x2 − 5x + 4)e−x

(i) domain (−∞, ∞)

(ii) decreases on (r1, r2), where r1 =3 −

√5

2, r2 =

3 +√

52

increases on (−∞, r1) ∪ (r2,∞)

(iii) f (r1)) is a local and absolute max; f (r2) is a local min

(iv) concave down on (−∞, 1) ∪ (4,∞); concave up on (1, 4)

pts of inflection at (1, 0) and (4,−12e−4)

-2 -1 1 2 3x

-8

-4

y

69.∫ xn

0

ex dx = [ex]xn0 = exn − 1; exn − 1 = n =⇒ xn = ln (n + 1).

70. (a) f(x) = xk ln x, f ′(x) = xk−1 + kxk−1 ln x = xk−1(1 + k ln x).

f ′(x) = 0 : 1 + k ln x = 0 =⇒ ln x = −1/k =⇒ x = e−1/k.

f ′(x) < 0 on (0, e−1/k) and f ′(x) > 0 on (e−1/k,∞); f has a local and absolute

minimum at x = e−1/k.

(b) f(x) = xke−x, f ′(x) = −xke−x + kxk−1e−x = xk−1e−x(k − x).

f ′(x) = 0 : k − x = 0 =⇒ x = k.

f ′(x) > 0 on (0, k) and f ′(x) < 0 on (k,∞); f has a local and absolute maximum

at x = e−1/k.



SECTION 7.4 367

71. (a) For y = eax we have dy/dx = aeax. Therefore the line tangent to the curve y = eax

at an arbitrary point (x0, eax0) has equation

y − eax0 = aeax0 (x− x0) .

The line passes through the origin iff eax0 = (aeax0)x0 iff x0 = 1/a. The point of tangency

is (1/a, e). This is point B. By symmetry, point A is (−1/a, e).

(b) The tangent line at B has equation y = aex. By symmetry

AI = 2∫ 1/a

0

(eax − aex) dx = 2[1aeax − 1

2aex2

]1/a0

=1a

(e− 2) .

(c) The normal at B has equation

y − e = − 1ae

(x− 1

a

).

This can be written

y = − 1ae

x +a2e2 + 1

a2e.

Therefore

AII = 2∫ 1/a

0

(− 1ae

x +a2e2 + 1

a2e− eax

)dx =

1 + 2a2e

a3e.

72. By induction. True for n = 0 : ex > 1 for x > 0.

Assume true for n. Then

ex = 1 +∫ x

0

et dt > 1 +∫ x

0

(1 + t +

t2

2!+ · · · + tn

n!

)dt

= 1 +[t +

t2

2+

t3

3!+ · · · + tn+1

(n + 1)!

]x0

= 1 + x +x2

2!+

x3

3!+ · · · + xn+1

(n + 1)!

So the result is true for n + 1

73. For x > (n + 1)!

ex > 1 + x + · · · + xn+1

(n + 1)!>

xn+1

(n + 1)!= xn

[x

(n + 1)!

]> xn.

74. (a) (b) Intersect at x ∼= ±0.7531

(c) Area ∼= 0.98



368 SECTION 7.4

75. (a) (b) x1∼= − 1.9646, x2

∼= 1.0580

(c) A ∼=∫ 1.0580

−1.9646

[4 − x2 − ex

]dx =

[4x− 1

3x3 − ex

]1.0580−1.9646

∼= 6.4240

76. 77.

f (g(x)) = e2 ln√x = e2(1/2) lnx = elnx = x f (g(x)) = e

(√lnx)2

= elnx = x

78.

f (g(x)) = e2+lnx−2 = elnx = x

79. (a) f(x) = sin (ex); f(x) = 0 =⇒ ex = nπ =⇒ x = ln nπ, n = 1, 2, · · · .

(b)

π--π

2 2

x

-1

1

y

π π



SECTION 7.4 369

80. (a) f(x) = esin x − 1; f(x) = 0 =⇒ sin x = 0 =⇒ x = nπ, n any integer.

(b)

-2π -π π 2πx

-1

1

2

y

81. (a) (b) x ∼= 1.3098

(c) f ′(1.3098) ∼= −0.26987; g′(1.3098) ∼= 0.76348

(d) the tangent lines are not perpendicular1 2

x

-0.5

0.5

y

82. (a)

1 2x

2

4

6

8

10y

(b) The x-coordinates of the points of intersection are: x = ln 2 and x = ln 5.

(c) A =∫ ln 5

ln 2

[7 − ex − 10e−x

]dx ∼= 0.4140

83. (a)∫

11 − ex

dx = x− ln |ex − 1| + C (c)∫

etanx

cos2 xdx = etanx + C

(b)∫

e−x

(1 − ex

ex

)4

dx = −15e−5x + e−4x − 2 e−3x + 2 e−2x − e−x + C

PROJECT 7.4

Step 1. ln(

1 +1n

)=∫ 1+ 1

n

1

dt

t≤∫ 1+ 1

n

1

1 dt =1n

(since1t≤ 1 throughout the interval of integration) = 1.

ln(

1 +1n

)=∫ 1+ 1

n

1

dt

t≥∫ 1+ 1

n

1

dt

1 + 1n

=1

1 + 1n

1n

=1

n + 1.

(since1t≥ 1

1 + 1n

throughout the interval of integration)



370 SECTION 7.5

Step 2. From Step 1, we get

1 +1n≤ e1/n =⇒

(1 +

1n

)n

≤ e and e1/n+1 ≤ 1 +1n

=⇒ e ≤(

1 +1n

)n+1

Combining these two inequalities, we have(1 +

1n

)n

≤ e ≤(

1 +1n

)n+1

SECTION 7.5

1. log2 64 = log2 (26) = 6 2. log2

164

= log2 2−6 = −6

3. log64 (1/2) =ln (1/2)ln 64

=− ln 26 ln 2

= −16

4. log10 0.01 = log10 10−2 = −2

5. log5 1 = log5

(50)

= 0 6. log5 0.2 = log5 5−1 = −1

7. log5 (125) = log5

(53)

= 3 8. log2 43 = log2 26 = 6

9. logp xy =lnxy

ln p=

lnx + ln y

ln p=

lnx

ln p+

ln y

ln p= logp x + logp y

10. logp1x

=ln 1

x

ln p= − lnx

ln p= − logp x.

11. logp xy =

lnxy

ln p= y

lnx

ln p= y logp x

12. logpx

y=

ln xy

ln p=

lnx− ln y

ln p= logp x− logp y

13. 10x = ex =⇒(eln 10x)

= ex =⇒ ex ln 10 = ex

=⇒ x ln 10 = x =⇒ x(ln 10 − 1) = 0 =⇒ Thus, x = 0.

14. log5 x = 0.04 =⇒ x = 50.04

15. logx 10 = log4 100 =⇒ ln 10lnx

=ln 100ln 4

=⇒ ln 10lnx

=2 ln 102 ln 2

=⇒ lnx = ln 2 Thus, x = 2.

16. logx 2 = log3 x =⇒ ln 2lnx

=lnx

ln 3=⇒ lnx = ±

√(ln 2)(ln 3) =⇒ x = e±

√(ln 2)(ln 3)

17. The logarithm function is increasing. Thus,

et1 < a < et2 =⇒ t1 = ln et1 < ln a < ln et2 = t2.



SECTION 7.5 371

18. Since the exponential function is increasing, elnx1 < eb < elnx2 , so x1 < eb < x2

19. f ′(x) = 32x(ln 3)(2) = 2(ln 3)32x 20. g′(x) = 43x2(ln 4) 6x

21. f ′(x) = 25x(ln 2)(5)3lnx + 25x3lnx(ln 3)1x

= 25x3lnx

(5 ln 2 +

ln 3x

)

22. F ′(x) = 5−2x2+x(ln 5)(−4x + 1)

23. g′(x) = 12 (log3x)−1/2

(1

ln 3

)1x

=1

2(ln 3)x√

log3x

24. h′(x) = 7 sinx2(ln 7)(cosx2) 2x

25. f ′(x) = sec2 (log5x) (ln 5)1x

=sec2 (log5x)

x ln 5

26. g′(x) =1

ln 10

(1x

)x2 − lnx(2x)

x4=

x− 2x lnx

x4 ln 10=

1 − 2 lnx

x3 ln 10.

27. F ′(x) = − sin (2x + 2−x) [2x ln 2 − 2−x ln 2] = ln 2 (2−x − 2x) sin (2x + 2−x)

28. h′(x) = a−x ln a(−1) cos bx + a−x(− sin bx)b = −(ln a)a−x cos bx− ba−x sin bx

29.∫

3x dx =3x

ln 3+ C 30.

∫2−x dx = −2−x

ln 2+ C

31.∫

(x3 + 3−x) dx =14x4 − 3−x

ln 3+ C

32.∫

x10−x2dx = −1

2

∫10u du = − 10u

2 ln 10+ C = − 10−x2

2 ln 10+ C

33.∫

dx

x ln 5=

1ln 5

∫dx

x=

ln |x|ln 5

+ C = log5 |x| + C

34.∫

log5 x

xdx =

1ln 5

∫lnx

xdx =

1ln 5

12(lnx)2 + C =

(lnx)2

2 ln 5+ C

35.∫

log2 x3

xdx =

1ln 2

∫lnx3

xdx =

3ln 2

∫lnx

xdx

=3

ln 2

[12

(lnx)2]

+ C =3

ln 4(lnx)2 + C

36. Write c = blogb c Then loga c = loga(blogb c

)= (logb c)(loga b).

37. f ′(x) =1

x ln 3so f ′(e) =

1e ln 3



372 SECTION 7.5

38. f(x) = x log3 x; f ′(x) = log3 x + x · 1x ln 3

=lnx + 1

ln 3; f ′(e) =

2ln 3

39. f ′(x) =1

x lnxso f ′(e) =

1e ln e

=1e

40. f(x) = log3 (log2 x) =ln(

lnxln 2

)ln 3

=ln(lnx) − ln(ln 2)

ln 3; f ′(x) =

1ln 3

· 1lnx

· 1x

=⇒ f ′(e) =1

e ln 3

41. f(x) = px

ln f(x) = x ln p

f ′(x)f(x)

= ln p

f ′(x) = f(x) ln p

f ′(x) = px ln p

42. f(x) = pg(x)

ln f(x) = g(x) ln p

f ′(x)f(x)

= g′(x) ln p

f ′(x) = f(x)g′(x) ln p = pg(x)g′(x) ln p

43. y = (x + 1)x

ln y = x ln (x + 1)1y

dy

dx=

x

x + 1+ ln (x + 1)

dy

dx= (x + 1)x

[x

x + 1+ ln (x + 1)

]

44. y = (lnx)x

ln y = x ln(lnx)

1y

dy

dx= ln(lnx) + x

1lnx

· 1x

dy

dx= (lnx)x

[ln(lnx) +

1lnx

]

45. y = (lnx)lnx

ln y = lnx [ln (lnx)]

1y

dy

dx= lnx

[1

x lnx

]+

1x

[ln (lnx)]

dy

dx= (lnx)lnx

[1 + ln (lnx)

x

]

46. y =(

1x

)xln y = x ln 1

x = −x lnx

1y

dy

dx= − lnx− x · 1

x= − lnx− 1

dy

dx= −

(1x

)x

[1 + lnx]

47. y = xsinx

ln y = (sinx)(lnx)

1y

dy

dx= (cosx)(lnx) + sinx

(1x

)

dy

dx= xsinx

[(cosx)(lnx) +

sinx

x

]

48. y = (cosx)x2+1

ln y = (x2 + 1) ln(cosx)

1y

dy

dx= 2x ln(cosx) + (x2 + 1)

(− sinx

cosx

)

dy

dx= (cosx)x

2+1[2x ln(cosx)− (x2 + 1) tanx

]



SECTION 7.5 373

49. y = (sinx)cosx

ln y = (cosx)(ln[sinx])

1y

dy

dx= (− sinx)(ln[sinx]) + (cosx)

(1

sinx

)(cosx)

dy

dx= (sinx)cosx

[cos2 xsinx

− (sinx)(ln[sinx])]

50. y = xx2

ln y = x2 lnx

1y

dy

dx= 2x lnx + x2 · 1

x

dy

dx= xx2+1 (2 lnx + 1)

51. y = x2x

ln y = 2x lnx

1y

dy

dx= 2x ln 2 lnx + 2x

(1x

)

dy

dx= x2x

[2x ln 2 lnx +

2x

x

]

52. y = (tanx)secx

ln y = secx ln(tanx)

1y

dy

dx= secx tanx ln(tanx) + secx · sec2 x

tanx

dy

dx= (tanx)secx [secx tanx ln(tanx)

+ sec3 x cotx]

53. From the definition of the derivative, the derivative of f(x) = ln x at x = 1 is

f ′(1) = limh→0

ln (1 + h) − ln 1h

.

Since f ′(1) = 1, we have

ln (1 + h) − ln 1h

=1h

ln (1 + h) = ln (1 + h)1/h → 1 as h → 0.

Set x = 1/h. Then h → 0 =⇒ x → ∞ and

ln(

1 +1x

)x

→ 1 as x → ∞Therefore (

1 +1x

)x

= eln (1+1/x)x → e1 = e as x → ∞.

54. 55. 56.

57. 58.



374 SECTION 7.6

59.∫ 2

1

2−x dx =[−2−x

ln 2

]21

=1

4 ln 260.

∫ 1

0

4x dx =[

4x

ln 4

]10

=3

ln 4

61.∫ 4

1

dx

x ln 2= [log2 x]41 = log2 4 − 0 = 2 62.

∫ 2

0

px/2 dx =[2px/2

ln p

]20

=2(p− 1)

ln p

63.∫ 1

0

x101+x2dx =

[1

2 ln 10101+x2

]10

=1

2 ln 10(100 − 10) =

45ln 10

64.∫ 1

0

5p√x+1

√x + 1

dx =

[10p

√x+1

ln p

]1

0

=10ln p

(p√

2 − p)

65.∫ 1

0

(2x + x2

)dx =

[2x

ln 2+

x3

3

]10

=13

+1

ln 2

66. 71/ ln 7 ∼= 2.71828. 71/ ln 7 =(eln 7)1/ ln 7

= e1 ∼= 2.71828.

67. approx 16.99999; 5ln 17/ ln 5 =(eln 5)ln 17/ ln 5 = eln 17 = 17

68. approx 54.59815; 161/ ln 2 =(eln 16

)1/ ln 2 = eln 16/ ln 2 = e4 ln 2/ ln 2 = e4 ∼= 54.59815

69. (b) the x-coordinates of the points of intersection are: x1∼= −1.198, x2 = 3 and x3

∼= 3.408

(c) for the interval [−1.198, 3], A ∼= 5.5376; for the interval [3, 3.408], A ∼= 0.01373

70. (b) the x-coordinates of the points of intersection are: x1∼= −0.7667, x2 = 2 and x3 = 4

(c) The area of the region bounded by the two graphs is: A =∫ 4

2

(2−x − x−2

)dx ∼= 0.0205

SECTION 7.6

1. We begin with A(t) = A0ert

and take A0 = $500 and t = 10. The interest earned is given by

A(10) −A0 = 500(e10r − 1

).

Thus, (a) 500(e0.6 − 1

) ∼= $411.06 (b) 500(e0.8 − 1

) ∼= $612.77

(c) 500 (e− 1) ∼= $859.14.

2. We want A(t) = A0ert = 2A0, so ert = 2 =⇒ rt = ln 2 =⇒ t =

ln 2r

(a) t =ln 20.06

∼= 11.55 years. (b) t =ln 20.08

∼= 8.66 years. (c) t =ln 20.1

∼= 6.93 years.



SECTION 7.6 375

3. In general

A(t) = A0ert.

We set

3A0 = A0e20r

and solve for r:

3 = e20r, ln 3 = 20r, r =ln 320

∼= 5 12%.

4. We want A(10) = A0e10r = 2A0, so r =

ln 210

∼= 6.9%

5. P (t) = 92 e

t20 ln (4/3) = 9

2 eln (4/3)t/20

= 92

(43

)t/20.

6. P (t) = P0ekt. P (4) = P0e

k4 = 3P0 =⇒ k =ln 34

(a) P (12) = P0eln34 12 = P0e

3 ln 3 = 27P0 = 1 =⇒ P0∼= 0.037 square inches

(b) P (t) = P0eln34 t = 2P0 =⇒ ln 3

4t = ln 2 =⇒ t =

4 ln 2ln 3

∼= 2.52 hours.

7. (a) P (t) = 10, 000et ln 2 = 10, 000(2)t

(b) P (26) = 10, 000(2)26, P (52) = 10, 000(2)52

8. qC = Cekp =⇒ q = ekp =⇒ p =1k

ln q

9. (a) P (10) = P (0)e0.035(10)t = P (0)e0.35t. Thus it increases by e0.35.

(b) 2P (0) = P (0)e15k =⇒ k =ln 215

.

10. Let P (t) be the world population at time t years, and let 1990 correspond to t = 0.

Then P (t) = 249 ekt; P (10) = 249 e10k = 281 =⇒ e10k =281249

=⇒ k ∼= 0.0121.

Thus P (t) = 249 e0.0121 t.

According to this model, the population in 1980 was P (−10) = 249 e−0.121 ∼= 221 million.

11. Using the data from Ex. 10, the growth constant k ∼= 0.0121. Therefore

P (20) ∼= 249 e20k ∼= 249 e0.242 ∼= 317.1 million.

P (11) ∼= 249 e11k ∼= 249 e0.1331 ∼= 284.4 million.



376 SECTION 7.6

12. Pekt = 2P =⇒ kt = ln 2.

Since k =110

ln(

281249

), we get t =

10 ln 2

ln(

281249

) ∼= 57.3 years.

13. 4.5e0.0143t = 30 =⇒ 0.0143t = ln304.5

=⇒ t � 115.7 years.

Thus maximum population will be reached in 2095.

14.ds

dx= − s

V=⇒ s(x) = Ce−x/V = s0e

−x/V . We want s(x) = 12s0, so e−x/V = 1

2 ,

hence x = −V ln 12 = V ln 2; V ln 2 = 10, 000 ln 2 ∼= 6931 gallons

15. V ′(t) = kV (t)

V ′(t) − ktV (t) = 0

e−ktV ′(t) − ke−kt V (t) = 0

d

dt

[e−kt V (t)

]= 0

e−kt V (t) = C

V (t) = Cekt.

Since V (0) = C = 200, V (t) = 200 ekt.

Since V (5) = 160, 200 e5k = 160, e5k = 45 , ek =

(45

)1/5and therefore V (t) = 200

(45

)t/5 liters.

16. A(t) = A0ekt. A(5) = A0e

5k =23A0 =⇒ k =

ln(2/3)5

=⇒ A(t) = A0eln(2/3)

5 t

A(t) =12A0 =⇒ ln(2/3)

5t = ln

12

=⇒ t =5 ln(1/2)ln(2/3)

∼= 8.55 years.

17. Take two years ago as time t = 0. In general

(∗) A(t) = A0ekt.

We are given that

A0 = 5 and A(2) = 4.Thus,

4 = 5e2k so that 45 = e2k or ek =

(45

)1/2.

We can writeA(t) = 5

(45

)t/2and compute A(5) as follows:

A(5) = 5(

45

)5/2 = 5e52 ln(4/5) ∼= 5e−0.56 ∼= 2.86.

About 2.86 gm will remain 3 years from now.



SECTION 7.6 377

18. Let t = 0 correspond to a year ago. Then A(t) = 4ekt, and A(1) = 3 =⇒ 4ek = 3 =⇒ k = ln(3/4)

Therefore, A(t) = 4eln(3/4)t = 4(

34

)t. Ten years ago, t = −9; A(−9) = 4

(34

)−9 ∼= 53.27 grams.

19. A fundamental property of radioactive decay is that the percentage of substance that decays during

any year is constant:

100[A (t) −A (t + 1)

A (t)

]= 100

[A0e

kt −A0ek(t+1)

A0ekt

]= 100(1 − ek)

If the half-life is n years, then

12A0 = A0e

kn so that ek =(

12

)1/n.

Thus, 100[1 −(

12

)1/n]% of the material decays during any one year.

20. A(t) = nekt. A(5) = ne5k = m =⇒ 5k = ln(m/n) =⇒ k =15

ln(m/n) and A(t) = neln(m/n)

5 t.

A(10) = ne2 ln(m/n) = n(mn

)2

=m2

ngrams.

21. (a) A(1620) = A0e1620k =

12A0 =⇒ k =

ln12

1620� −0.00043.

Thus A(500) = A0e500k = 0.807A0. Hence 80.7% will remain.

(b) 0.25A0 = A0ekt =⇒ t = 3240 years.

22. Ae5.3k =12A =⇒ k � −0.1308.

(a) A(8) = A0e8k � 0.351A0. Thus 35.1% will remain.

(b) 100 = A0e3k =⇒ A0 � 148 grams.

23. (a) x1 (t) = 106t, x2 (t) = et − 1

(b)d

dt[x1(t) − x2(t)] =

d

dt[106t− (et − 1)] = 106 − et

This derivative is zero at t = 6 ln 10 ∼= 13.8. After that the derivative is negative.

(c) x2(15) < e15 = (e3)5 ∼= 205 = 25(105) = 3.2(106) < 15(106) = x1(15)

x2(18) = e18 − 1 = (e3)6 − 1 ∼= 206 − 1 = 64(106) − 1 > 18(106) = x1(18)

x2(18) − x1(18) ∼= 64(106) − 1 − 18(106) ∼= 46(106)

(d) If by time t1 EXP has passed LIN, then t1 > 6 ln 10. For all t ≥ t1 the speed of EXP is greaterthan the speed of LIN:

for t ≥ t1 > 6 ln 10, v2 (t) = et > 106 = v1 (t) .



378 SECTION 7.6

24. (a) x1(t) = t, x3(t) = 106 ln(t + 1)

(b)d

dt[x3(t) − x1(t)] =

d

dt

[106 ln(t + 1) − t

]=

106

t + 1− 1

This derivative is 0 at t = 106 − 1. After that the derivative is negative.

(c) x1

(107 − 1

)= 107 − 1 < 7(ln 10)106 = 106 ln 107 = x3(107 − 1)

x3(108 − 1) = 106 ln 108 = (106)8 ln 10 < (106)24 < 108 − 1 = x1(108 − 1)

(d) If by time t1 LIN had passed LOG, then t1 > 106 − 1. For all t ≥ t1 the speed of LIN is greater

than the speed of LOG:

for t ≥ t1 > 106 − 1, v1(t) = t >106

t + 1= v3(t).

25. Let p (h) denote the pressure at altitude h. The equationdp

dh= kp gives

(∗) p (h) = p0ekh

where p0 is the pressure at altitude zero (sea level).Since p0 = 15 and p (10000) = 10,

10 = 15e10000k, 23 = e10000k, 1

10000 ln 23 = k.

Thus, (∗) can be writtenp (h) = 15

(23

)h/10000.

(a) p (5000) = 15(

23

)1/2 ∼= 12.25 lb/in.2.

(b) p (15000) = 15(

23

)3/2 ∼= 8.16 lb/in.2.

26. P = 20,000 e−(0.06)(4) ∼= $15, 732.56.

27. From Exercise 26, we have 6000 = 10,000e−8r. Thus

e−8r =6000

10,000=

35

⇒ −8r = ln(3/5) and r ∼= 0.064 or r = 6.4%

28. (a) P = 50,000 e−(0.04)(20) ∼= $22, 466.45

(b) P = 50,000 e−(0.06)(20) ∼= $15, 059.71

(c) P = 50,000 e−(0.08)(20) ∼= $10, 094.83

29. The future value of $25, 000 at an interest rate r, t years from now is given by Q(t) = 25, 000 ert.

Thus

(a) For r = 0.05 : P (3) = 25, 000 e(0.05)3 ∼= $29, 045.86.

(b) For r = 0.08 : P (3) = 25, 000 e(0.08)3 ∼= $31, 781.23.

(c) For r = 0.12 : P (3) = 25, 000 e(0.12)3 ∼= $35, 833.24.



SECTION 7.6 379

30.dv

dt= −kv =⇒ v = ce−kt, v(0) = ce0 = c, so c is velocity when power is shut off.

31. By Exercise 30

(∗) v (t) = Ce−kt, t in seconds.

We use the initial conditions

v (0) = C = 4 mph = 1900 mi/sec and v (60) = 2 = 1

1800 mi/sec

to determine e−k:

11800 = 1

900e−60k , e60k = 2, ek = 21/60.

Thus, (∗) can be written

v (t) = 1900 2−t/60.

The distance traveled by the boat is

s =∫ 60

0

1900

2−t/60 dt =1

900

[−60ln 2

2−t/60

]600

=1

30 ln 2mi =

176ln 2

ft (about 254 ft).

32. Since the amount A(t) of raw sugar present after t hours decreases at a rate proportional

to A, we have

A(t) = A0ekt.

We are given A0 = 1000 and A(10) = 800. Thus,

800 = 1000e10k, 45 = e10k, ek =

(45

)1/10so that

A(t) = 1000(

45

)t/10.

Now,

A(20) = 1000(

45

)20/10

= 640;

after 10 more hours of inversion there will remain 640 pounds.

33. Let A(t) denote the amount of 14C remaining t years after the organism dies. Then A(t) = A(0)ekt

for some constant k. Since the half-life of 14C is 5700 years, we have

12

= e5700k ⇒ k = − ln 25700

∼= 0.000122 and A(t) = A(0)e−0.000122t

If 25% of the original amount of 14C remains after t years, then

0.25A(0) = A(0)e−0.000122t ⇒ t =ln 0.25

−0.000122∼= 11, 400 (years)

34. A(t) = A0e− ln2

5700 t; A(2000) = A0 e− ln2

5700 ·2000 ∼= 0.78A0; 78% remains



380 SECTION 7.7

35. f ′(t) = tf(t)

f ′(t) − tf(t) = 0

e−t2/2 f ′(t) − te−t2/2 f(t) = 0

d

dt[e−t2/2 f(t)] = 0

e−t2/2 f(t) = C

f(t) = Cet2/2

36. f ′(t) = sin t f(t)

f ′(t) − sin t f(t) = 0

ecos t f ′(t) − sin t ecos t f(t) = 0

d

dt

[ecos t f(t)

]= 0

ecos t f(t) = C

f(t) = Ce− cos t

37. f ′(t) = cos t f(t)

f ′(t) − cos t f(t) = 0

e− sin t f ′(t) − cos t e− sin t f(t) = 0

d

dt

[e− sin t f(t)

]= 0

e− sin t f(t) = C

f(t) = Cesin t

38. Write the equation as f ′(t) − g(t)f(t) = 0; and set h(t) = −∫

g(t) dt. Then

f ′(t) − g(t)f(t) = 0

eh(t)f ′(t) − g(t)eh(t)f(t) = 0

[eh(t)f(t)]′ = 0

eh(t)f(t) = C

f(t) = Ce−h(t) = Ce∫

g(t)dt

SECTION 7.7

1. (a) 0 (b) −π/3 2. (a) π/3 (b) π/3 3. (a) 2π/3 (b) 3π/4

4. (a) −2/√

3 (b) −2 5. (a) 1/2 (b) π/4 6. (a) −π/6 (b) −π/4



SECTION 7.7 381

7. (a) does not exist

(b) does not exist

8. (a) 4/5 (b) 5/3 9. (a)√

3/2 (b) −7/25

10. (a) arc cosine; domain: [−1, 1], range: [0, π]

(b) arc cotangent: domain: (−∞,∞), range: [0, π]

11.dy

dx=

11 + (x + 1)2

=1

x2 + 2x + 212.

dy

dx=

11 + (

√x)2

· 12√x

=1

2√x(1 + x)

13. f ′(x) =1

|2x2|√

(2x2)2 − 1d

dx

(2x2)

=2

x√

4x4 − 1

14. f ′(x) = ex arcsinx + ex1√

1 − x2= ex

[arcsinx +

1√1 − x2

]

15. f ′(x) = arcsin 2x + x1√

1 − (2x)2d

dx(2x) = arcsin 2x +

2x√1 − 4x2

16. f ′(x) = earctanx · 11 + x2

17.du

dx= 2 (arcsinx)

d

dx(arcsinx) =

2 arcsinx√1 − x2

18.dy

dx=

11 + (ex)2

· ex =ex

1 + e2x

19.dy

dx=

x

(1

1 + x2

)− (1) arctanx

x2=

x−(1 + x2

)arctanx

x2 (1 + x2)

20.dy

dx=

1

|√x2 + 2|

√(√x2 + 2

)2 − 1· x√

x2 + 2=

x

(x2 + 2)√x2 + 1

21. f ′(x) =12

(arctan 2x)−1/2 d

dx(arctan 2x) =

12

(arctan 2x)−1/2 21 + (2x)2

=1

(1 + 4x2)√

arctan 2x

22. f ′(x) =1

arctanx· 11 + x2

=1

(1 + x2) arctanx

23.dy

dx=

11 + (lnx)2

d

dx(lnx) =

1x[1 + (lnx)2]

24. g′(x) =− sinx

| cosx + 2|√

(cosx + 2)2 − 1



382 SECTION 7.7

25.dθ

dr=

1√1 −(√

1 − r2)2 d

dr(√

1 − r2) =1√r2

· −r√1 − r2

= − r

|r|√

1 − r2

26.dθ

dr=

1√1 − [r/(r + 1)]2

· 1(r + 1)2

=1

(r + 1)√

2r + 1

27. g′(x) = 2x arcsec(

1x

)+ x 2 · 1∣∣∣∣ 1x

∣∣∣∣√

1x 2

− 1·(− 1

x 2

)= 2x sec−1

(1x

)− x 2

√1 − x 2

28.dθ

dr=

11 + [1/(1 + r2)]2

· −2r(1 + r2)2

=−2r

r4 + 2r2 + 2

29.dy

dx= cos [arcsec (ln x )] · 1

| lnx |√

(lnx)2 − 1· 1x

=cos [arcsec (ln x )]

x | lnx |√

(lnx)2 − 1

30. f ′(x) = earcsec x · 1|x|

√x2 − 1

=earcsec x

|x|√x2 − 1

31. f ′(x) =−x√c2 − x2

+c√

1 − (x/c)2·(

1c

)=

c− x√c2 − x2

=√

c− x

c + x

32.dy

dx=

√c2 − x2 (1) − x

( −x√c2 − x2

)(√

c2 − x2)2 − 1√

1 − (x/c)2

(1c

)

=c2

(c2 − x2)3/2− 1

(c2 − x2)1/2=

x2

(c2 − x2)3/2

33. (a) sin (arcsin x) = x (b) cos (arcsin x) =√

1 − x2 (c) tan (arcsin x) =x√

1 − x2

(d) cot (arcsin x) =√

1 − x2

x(e) sec (arcsin x) =

1√1 − x2

(f) csc (arcsin x) =1x

34. (a) tan (arctan x) = x (b) cot (arctan x) = 1x (c) sin (arctan x) =

x√1 + x2

(d) cos (arctan x) =1√

1 + x2(e) sec (arctan x) =

√1 + x2 (f) csc (arctan x) =

√1 + x2

x



SECTION 7.7 383

35.

{au = x + b

a du = dx

};∫

dx√a2 − (x + b)2

=∫

a du√a2 − a2u2

=∫

du√1 − u2

= arcsinu + C = sin−1

(x + b

a

)+ C

36.

{au = x + b

a du = dx

};∫

dx

a2 + (x + b)2=

1a2

∫a du

1 + u2=

1a

arctanu + C =1a

arctan(x + b

a

)+ C

37.

{au = x + b

a du = dx

};∫

dx

(x + b)√

(x + b)2 − a2

=∫

a du

au√a2u2 − a2

=1a

∫du

u√u2 − 1

=1a

arcsec |u| + C =1a

arcsec( |x + b|

a

)+ C

38. (a) sec (arcsecx) = sec (arccos 1/x) =1

1/x= x; csc (arccscx) = csc (arcsin 1/x) =

11/x

= x

(b) arc secant: range: [0, π/2) ∪ (π/2, π) (c) arc cosecant: range: [−π/2, 0) ∪ (0, π/2]

39.∫ 1

0

dx

1 + x2= [arctanx]10 =

π

440.

∫ 1

−1

dx

1 + x2= [arctanx]1−1 =

π

4−(−π

4

)=

π

2

41.∫ 1/

√2

0

dx√1 − x2

= [arcsinx]1/√

20 =

π

442.

∫ 1

0

dx√4 − x2

=[arcsin

x

2

]10

=π

6

43.∫ 5

0

dx

25 + x2=[15

arctanx

5

]50

=π

20

44.∫ 8

5

dx

x√x2 − 16

=14

[arcsec

x4

]85

=14

(arcsec 2 − arcsec

54

)=

π

12− 1

4arcsec

54

45.

{3u = 2x x = 0 =⇒ u = 0

3 du = 2 dx x = 3/2 =⇒ u = 1

};

∣∣∣∣∫ 3/2

0

dx

9 + 4x2=

16

∫ 1

0

du

1 + u2=

16

[arctanu]10 =π

24

46.∫ 5

2

dx

9 + (x− 2)2=

13

[arctan

(x− 2

3

)]52

=13

(π4− 0)

=π

12

47.

{u = 4x x = 3/2 =⇒ u = 6

du = 4 dx x = 3 =⇒ u = 12

};

∣∣∣∣∫ 3

3/2

dx

x√

16x2 − 9=∫ 12

6

du/4(u/4)

√u2 − 9

=13

[arcsec

( |u|3

)]126

=13arcsec 4 − π

9

48.∫ 6

4

dx

(x− 3)√x2 − 6x + 8

=∫ 6

4

dx

(x− 3)√

(x− 3)2 − 1= [arcsec (x − 3)]64 = arcsec 3



384 SECTION 7.7

49.∫ −2

−3

dx√4 − (x + 3)2

=[arcsin

(x + 3

2

)]−2

−3

=π

6

50.∫ ln 3

ln 2

e−x

√1 − e−2x

dx =[− arcsin e−x

]ln 3

ln 2= arcsin

(12

)− arcsin

(13

)=

π

6− arcsin

(13

)

51.

{u = ex x = 0 =⇒ u = 1

du = ex dx x = ln 2 =⇒ u = 2

};

∣∣∣∣∫ ln 2

0

ex

1 + e2xdx =

∫ 2

1

du

1 + u2= [arctanu]21 = arctan 2 − π

4∼= 0.322

52.∫ 1/2

0

dx√3 − 4x2

=1√3

∫ 1/2

0

dx√1 − 4

3x2

=12

[arcsin

2x√3

]1/20

=12

arcsin

(√3

3

)

53.

{u = x2

du = 2x dx

};∫

x√1 − x4

dx =12

∫du√

1 − u2=

12

arcsinu + C =12

arcsinx2 + C

54.∫

sec2 x√9 − tan2 x

dx =∫

du√9 − u2

= arcsin(u

3

)+ C = arcsin

(tanx

3

)+ C

55.

{u = x2

du = 2x dx

};

∫x

1 + x4dx =

12

∫du

1 + u2=

12

arctanu + C =12

arctanx2 + C

56.∫

dx√4x− x2

=∫

dx√4 − (x− 2)2

= arcsin(x− 2

2

)+ C

57.

{u = tanx

du = sec2 x dx

};

∫sec2 x

9 + tan2 xdx =

∫du

9 + u2=

13

arctan(u

3

)+ C =

13

arctan(

tanx

3

)+ C

58.∫

cosx3 + sin2 x

dx =∫

du

3 + u2=

1√3

arctan(

u√3

)+ C =

1√3

arctan(

sinx√3

)+ C

59.

⎧⎨⎩

u = arcsinx

du =1√

1 − x2dx

⎫⎬⎭ ;

∫arcsinx√

1 − x2dx =

∫u du =

12u2 + C =

12

(arcsinx)2 + C

60.∫

arctanx

1 + x2dx =

∫u du =

u2

2+ C =

12

(arctanx)2 + C

61.

⎧⎨⎩

u = lnx

du =1xdx

⎫⎬⎭ ;

∫dx

x√

1 − (lnx)2=∫

du√1 − u2

= arcsinu + C = arcsin(lnx) + C



SECTION 7.7 385

62.∫

1x· 11 + (lnx)2

dx =∫

du

1 + u2= arctanu + C = arctan(lnx) + C

63. A =∫ 1

−1

1√4 − x2

dx = 2∫ 1

0

1√4 − x2

dx

= 2[arcsin

(x2

)]10

=π

3 -1 1x

0.5

y

64. A =∫ 3

−3

39 + x2

dx = 3[13

arctan(x

3

)]3−3

=π

4+

π

4=

π

2

65.8

x2 + 4=

14x2 ⇒ x = ±2

A =∫ 2

−2

(8

x2 + 4− 1

4x2

)dx = 2

∫ 2

0

(8

x2 + 4− 1

4x2

)dx

= 2[8 · 1

2arctan

(x2

)− 1

12x3

]20

= 2π − 43 -2 2

x

1

2

y

66. V =∫ 2

0

π1

4 + x2dx =

π

2

[arctan(x/2)

]20

=π2

8

67. V =∫ 2

0

2πx1√

4 + x2dx = π

∫ 2

0

2x√4 + x2

dx = 2π[√

4 + x2]20

= 4π(√

2 − 1)

68. V =∫ 6

2√

3

2πx · 1x2

√x2 − 9

dx = 2π∫ 6

2√

3

dx

x√x2 − 9

= 2π[13arcsec

(x3

)]62√

3

=π2

9

69. Let x be the distance between the motorist and the point on the road where the line determined

by the sign intersects the road. Then, from the given figure,

θ = arctan(s + k

x

)− arctan

s

x, 0 < x < ∞

and dθ

dx=

1

1 +(s + k)2

x2

(− s + k

x2

)− 1

1 +s2

x2

(− s

x2

)

=−(s + k)

x2 + (s + k)2+

s

x2 + s2=

s2k + sk2 − kx2

[x2 + (s + k)2] [x2 + s2]

Setting dθ/dx = 0 we get x =√s2 + sk. Since θ is essentially 0 when x is close to 0 and when

x is “large,” we can conclude that θ is a maximum when x =√s2 + sk .

70. y = arctanx

30;

dy

dt=

11 + (x/30)2

· 130

· dxdt

Ifdy

dt= 6 and x = 50 then

dy

dt=

30900 + (50)2

· 6 =9

170rad/sec



386 SECTION 7.7

71. (b)∫ a

−a

√a2 − x2 dx =

[x

2

√a2 − x2 +

a2

2arcsin

(xa

)]a−a

=π a2

2

The graph of f(x) =√a2 − x2 on the interval [−a, a] is the upper half of the circle of radius a centered

at the origin. Thus, the integral gives the area of the semi-circle: A = 12 πa

2.

72. (a) f ′(x) =1

1 +(

a+x1−ax

)2 · 1 + a2

(1 − ax)2=

1 + a2

(1 + a2)(1 + x2)=

11 + x2

(b) limx→(1/a)−

f(x) = π/2; limx→(1/a)+

f(x) = −π/2

(c) Let g(x) = arctanx. For x = 0 <1a, g(0) = 0, f(0) = arctan(a), so

f(x) = g(x) + arctan a for x <1a, i.e., C1 = arctan a

For x >1a, note that lim

x→∞arctanx =

π

2, lim

x→∞arctan

(a + x

1 − ax

)= arctan(−1/a) so

f(x) = g(x) + arctan(−1/a) − π

2for x >

1a, i.e., C2 = arctan(−1/a) − π

2

73. Set y = arccotx. Then cot y = x and, by the hint, tan(

12π − y

)= x. Therefore

12π − y = arctan x, arctan x + y =

12π, arctan x + arccotx =

12π.

74. Set y = arccscx. Then csc y = x and sec(

12π − y

)= x

[sec(

12π − θ)

]= csc θ

]. Therefore

12π − y = arcsecx, arcsecx + y =

12π, arcsecx + arccscx =

12π.

75. The integrand is undefined for x ≥ 1.

76. Numerical work suggests limit ∼= 1. One way to see this is to note that the limit is the derivative

of f(x) = arcsinx at x = 0 and this derivative is 1:

f ′(x) =1√

1 − x2, f ′(0) = 1

77. I =∫ 0.5

0

1√1 − x2

dx ∼= 110

[f(0.05) + f(0.15) + f(0.25) + f(0.35) + f(0.45)] ∼= 0.523;

and sin(0.523) ∼= 0.499. Explanation: I = arcsin(0.5) and sin [arcsin(0.5)] = 0.5.

78. (a) 1.5698, 1.5704, 1.5706, 1.5707 (b) 1.5708 ∼= π

2(c) 1.570796 ∼= π

2



SECTION 7.8 387

PROJECT 7.7

1. (a) n1 sin θ1 = n sin θ = n2 sin θ2

2. (a) Think of n and θ as functions of altitude y. Then

n sin θ = C

Differentiation with respect to y gives

n cos θdθ

dy+

dn

dysin θ = 0, cot θ

dθ

dy+

1n

dn

dy= 0

and so when α =π

2− θ,

1n

dn

dy= − cot θ

dθ

dy= −dy

dx

(−dα

dy

)=

dα

dx

Now, α = arctan(dy

dx

)and

dα

dx=

d2y

d2x

1 +(dy

dx

)2 .

(b) 1 +(dy

dx

)2

= 1 + tan2 α = 1 + cot2 θ = csc2 θ =n2

C2= (a constant) · [n(y)]2.

(c) n(y) =k

|y + b| , with b, k constants, k > 0.

SECTION 7.8

1.dy

dx= coshx2 d

dx

(x2)

= 2x coshx2 2.dy

dx= sinh(x + a)

3.dy

dx=

12

(cosh ax)−1/2 (a sinh ax) =a sinh ax

2√

cosh ax

4.dy

dx= a cosh2 ax + a sinh2 ax = a(cosh2 ax + sinh2 ax)

5.dy

dx=

(coshx− 1) (coshx) − sinhx (sinhx)(coshx− 1)2

=1

1 − coshx

6.dy

dx=

x coshx− sinhx

x2



388 SECTION 7.8

7.dy

dx= ab cosh bx− ab sinh ax = ab (cosh bx− sinh ax)

8.dy

dx= ex(coshx + sinhx) + ex(sinhx + coshx) = 2ex(coshx + sinhx)

9.dy

dx=

1sinh ax

(a cosh ax) = a coth ax 10.dy

dx=

11 − cosh ax

(− sinh ax)a =−a sinh ax

1 − cosh ax

11.dy

dx= cosh (e2x)e2x(2) = 2e2x cosh (e2x) 12.

dy

dx= sinh (lnx3) · 1

x3· 3x2 =

3 sinh (lnx3)x

13.dy

dx= −e−x cosh 2x + 2e−x sinh 2x 14.

dy

dx=

11 + sinh2 x

· coshx =1

coshx

15.dy

dx=

1coshx

(sinhx) = tanhx 16.dy

dx=

1sinhx

· coshx = cothx

17. ln y = x ln sinhx;1y

dy

dx= ln sinhx + x

coshx

sinhxand

dy

dx= (sinhx)x [ ln sinhx + x cothx ]

18. y = x coshx =⇒ ln y = coshx lnx =⇒ 1y

dy

dx= sinhx lnx + coshx

1x

=⇒ dy

dx= x coshx

(sinhx lnx +

coshx

x

)

19. cosh2 t− sinh2 t =(et + e−t

2

)2

−(et − e−t

2

)2

=14{(

e2t + 2 + e−2t)−(e2t − 2 + e−2t

)}=

44

= 1

20. sinh t cosh s + cosh t sinh s =12(et − e−t)

12(es + e−s) +

12(et + e−t)

12(es − e−s)

=12(es+t − e−(s+t)

)= sinh(s + t)

21. cosh t cosh s + sinh t sinh s =(et + e−t

2

)(es + e−s

2

)+(et − e−t

2

)(es − e−s

2

)

=14{2et+s + 2e−(t+s)

}=

et+s + e−(t+s)

2= cosh (t + s)

22. Follows from Exercise 20, with s = t.

23. Set s = t in cosh (t + s) = cosh t cosh s + sinh t sinh s to get cosh(2t) = cosh2 t + sinh2 t.

Then use Exercise 19 to obtain the other two identities.

24. cosh(−t) =12(e−t + e−(−t)

)=

12(e−t + et

)= cosh t

25. sinh(− t) =e(−t) − e−(−t)

2= − et − e−t

2= − sinh t



SECTION 7.8 389

26. y = 5 coshx + 4 sinhx =52(ex + e−x) +

42(ex − e−x) =

92ex +

12e−x

dy

dx=

92ex − 1

2e−x =

e−x

2(9e2x − 1);

dy

dx= 0 =⇒ e2x =

19

=⇒ x = − ln 3.

d2y

dx2=

92ex +

12e−x > 0 for all x, so abs min occurs at x = − ln 3.

At x = − ln 3, y = 92 ( 1

3 ) + 12 (3) = 3.

27. y = −5 coshx + 4 sinhx = − 52 (ex + e−x) + 4

2 (ex − e−x) = − 12e

x − 92e

−x

dy

dx= −1

2ex +

92e−x =

e−x

2(9 − e2x);

dy

dx= 0 =⇒ ex = 3 or x = ln 3

d2y

dx2= −1

2ex − 9

2e−x < 0 for all x so abs max occurs at x = ln 3.

The abs max is y = − 12e

ln 3 − 92e

− ln 3 = − 12 (3) − 9

2

(13

)= −3.

28. y = 4 coshx + 5 sinhx =42(ex + e−x) +

52(ex − e−x) =

92ex − 1

2e−x

dy

dx=

92ex +

12e−x > 0 always increasing, so no extreme values.

29. [coshx + sinhx]n =[ex + e−x

2+

ex − e−x

2

]n

= [ex]n = enx =enx + e−nx

2+

enx − e−nx

2= coshnx + sinhnx

30. y = A cosh cx + B sinh cx, y′ = Ac sinh cx + Bc cosh cx, y′′ = Ac2 cosh cx + Bc2 sinh cx

=⇒ y′′ = c2y

31. y = A cosh cx + B sinh cx; y (0) = 2 =⇒ 2 = A.

y′ = Ac sinh cx + Bc cosh cx; y′ (0) = 1 =⇒ 1 = Bc.

y′′ = Ac2 cosh cx + Bc2 sinh cx = c2y; y′′ − 9y = 0 =⇒(c2 − 9

)y = 0.

Thus, c = 3, B = 13 , and A = 2.

32. From Exercise 30, y′′ = c2y, so c = 12 .

1 = y(0) = A cosh 0 + B sinh 0 = A =⇒ A = 1

2 = y′(0) = Ac sinh 0 + Bc cosh 0 = Bc =⇒ B =2c

= 4

33.1a

sinh ax + C 34.1a

cosh ax + C 35.13a

sinh3 ax + C

36.13a

cosh3 ax + C 37.1a

ln (cosh ax) + C 38.1a

ln | sinh ax| + C



390 SECTION 7.8

39. − 1a cosh ax

+ C

40.∫

sinh2 x dx =∫

14(e2x − 2 + e−2x) dx =

14

(12e2x − 2x− 1

2e−2x

)+ C

=14

sinh 2x− 12x + C =

12

sinhx coshx− 12x + C

41. From the identity cosh 2t = 2 cosh2 t− 1 (Exercise 23), we get

cosh2 t = 12 (1 + cosh 2t) . Thus,∫

cosh2 x dx =12

∫(1 + cosh 2x) dx

=12

(x +

12

sinh 2x)

+ C

=12

(x + sinhx coshx) + C

42.∫

sinh 2xecosh2x dx =12

∫eu du =

12eu + C =

12e cosh2x + C

43.

{u =

√x

du = dx/2√x

};∫

sinh√x√

xdx = 2

∫sinhu du = 2 coshu + C = 2 cosh

√x + C

44.∫

sinhx

1 + coshxdx =

∫du

1 + u= ln |1 + u| + C = ln(1 + coshx) + C

45. A.V. =1

1 − (−1)

∫ 1

−1

coshx dx =12

[sinhx]1−1 =e2 − 1

2e∼= 1.175

46. A.V. =14

∫ 4

0

sinh 2x dx =18

[cosh 2x]40 =18

[e8 + e−8

2− 1]∼= 186.185

47. A =∫ ln 10

0

sinhx dx =[coshx

]ln 10

0=

eln 10 + e− ln 10

2− 1 =

8120

48. A =∫ ln 5

− ln 5

cosh 2x dx = 12 [sinh 2x]ln 5

− ln 5 =31225

49. V =∫ 1

0

π(cosh2 x− sinh2 x

)dx =

∫ 1

0

π dx = π



SECTION 7.9 391

50. V =∫ ln 5

0

π[sinhx]2 dx = π

∫ ln 5

0

sinh2 x dx

= 14π

∫ ln 5

0

(e2x − 2 + e−2x

)dx

= 14π[12e

2x − 2x− 12e

−2x]ln 5

0= 1

4π [sinh(2 ln 5) − 2 ln 5]

51. V =∫ ln 5

− ln 5

π[cosh 2x]2 dx = 2π∫ ln 5

0

cosh2 2x dx

= 12π

∫ ln 5

0

(e4x + 2 + e−4x

)dx

= 12π[14e

4x + 2x + 14e

−4x]ln 5

0= π

[14 sinh(4 ln 5) + ln 5

]52. (a) lim

x→∞sinhx

ex= lim

x→∞ex − e−x

2ex= lim

x→∞

(12− e−2x

2

)=

12

(b) limx→∞

coshx

eax= lim

x→∞ex + e−x

2eax= lim

x→∞12(ex−ax + e−x−ax

)

For 0 < a < 1, limit = ∞. For a > 1, limit = 0.

53. (a) (0.69315, 1.25) (b) A ∼= 0.38629 54. (a) (±1.06128, 0.6180) (b) A ∼= 1.388

SECTION 7.9

1.dy

dx= 2 tanhx sech2x 2.

dy

dx= 2 tanh 3x sech 23x · 3 = 6 tanh 3x sech 3x

3.dy

dx=

1tanhx

sech2x = sechx cschx 4.dy

dx= sech 2(lnx) · 1

x

5.dy

dx= cosh

(arctan e2x

) d

dx

(arctan e2x

)=

2e2x cosh(arctan e2x

)1 + e4x

6.dy

dx= − sech (3x2 + 1) tanh(3x2 + 1)(6x) = −6x sech (3x2 + 1) tanh(3x2 + 1)

7.dy

dx= − csch2

(√x2 + 1

) d

dx

(√x2 + 1

)= − x√

x2 + 1csch2

(√x2 + 1

)

8.dy

dx=

1sechx

· (− sechx)(tanhx) = − tanhx



392 SECTION 7.9

9.dy

dx=

(1 + coshx) (− sechx tanhx) − sechx (sinhx)(1 + coshx)2

=− sechx (tanhx + coshx tanhx + sinhx)

(1 + coshx)2=

− sechx (tanhx + 2 sinhx)(1 + coshx)2

10.dy

dx=

sinhx(1 + sechx) − coshx(− sechx) tanhx

(1 + sechx)2=

sinhx + 2 tanhx

(1 + sechx)2

11. d

dx(cothx) =

d

dx

[coshx

sinhx

]=

sinhx (sinhx) − coshx (coshx)sinh2 x

= −cosh2 x− sinh2 x

sinh2 x=

−1sinh2 x

= − csch2x

12.d

dx( sechx) =

d

dx

(1

coshx

)=

−1(coshx)2

· sinhx = − 1coshx

· sinhx

coshx= − sechx tanhx

13.d

dx( cschx) =

d

dx

[1

sinhx

]= − coshx

sinh2 x= − cschx cothx

14. tanh(t + s) =sinh(t + s)cosh(t + s)

=sinh t cosh s + cosh t sinh s

cosh t cosh s + sinh t sinh s=

tanh t + tanh s

1 + tanh t tanh s

15. (a) By the hint sech2x0 =925

. Take sechx0 =35

since sechx =1

coshx> 0 for all x.

(b) coshx0 =1

sechx0=

53

(c) sinhx0 = coshx0 tanhx0 =(

53

)(45

)=

43

(d) cothx0 =coshx0

sinhx0=

5/34/3

=54

(e) cschx0 =1

sinhx0=

34

16. sech 2t0 = 1 − tanh2 t0 = 1 − 25144

=119144

=⇒ sech t0 =√

11912

; cosh t0 =1

sech t0=

12√119

;

sinh t0 = cosh t0 tanh t0 =12√119

· −512

=−5√119

; coth t0 =1

tanh t0=

−125

;

csch t0 =1

sinh t0= −

√1195

.

17. If x ≤ 0, the result is obvious. Suppose then that x > 0. Since x2 ≥ 1, we have x ≥ 1.Consequently

x− 1 =√x− 1

√x− 1 ≤

√x− 1

√x + 1 =

√x2 − 1

and therefore x−√x2 − 1 ≤ 1.

18. We will show that

tanh[12

ln(

1 + x

1 − x

)]= x for all x ∈ [−1, 1].



SECTION 7.9 393

First we observe that

tanh s =es − e−s

es + e−sand therefore tanh(ln t) =

t− 1/tt + 1/t

=t2 − 1t2 + 1

It follows that

tanh[12

ln(

1 + x

1 − x

)]= tanh

(ln

√1 + x

1 − x

)=

1+x1−x − 11+x1−x + 1

=2x2

= x.

19. By Theorem 7.9.2,

d

dx

(sinh−1 x

)=

d

dx

[ln(x +

√x2 + 1

)]=

1x +

√x2 + 1

(1 +

x√x2 + 1

)=

1√x2 + 1

.

20.d

dx(cosh−1 x) =

d

dx

[ln(x +

√x2 − 1)

]=

1x +

√x2 − 1

(1 +

x√x2 − 1

)=

1√x2 − 1

.

21. By Theorem 7.9.2

d

dx(arctanx) =

d

dx

[12

ln(

1 + x

1 − x

)]=

12

1(1 + x

1 − x

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

)

=1(

1 + x

1 − x

)(1 − x)2

=1

1 − x2.

22. y = sech −1x =⇒ sech y = x =⇒ cosh y =1x

=⇒ y = cosh−1

(1x

),

sody

dx=

1√(1/x)2 − 1

·(−1x2

)=

−1x√

1 − x2.

23. Let y = csch −1x. Then csch y = x and sinh y =1x.

sinh y =1x

cosh ydy

dx= − 1

x2

dy

dx= − 1

x2 cosh y= − 1

x2√

1 + (1/x)2= − 1

|x|√

1 + x2

24. y = coth−1 x =⇒ coth y = x =⇒ tanh y =1x

=⇒ y = tanh−1

(1x

), so

dy

dx=

11 − (1/x)2

·(− 1x2

)=

−1x2 − 1

=1

1 − x2



394 SECTION 7.9

25.

(a)dy

dx= − sechx tanhx = − sinhx

cosh2 xdy

dx= 0 at x = 0;

dy

dx> 0 if x < 0;

dy

dx< 0 if x > 0

f is increasing on (−∞, 0] and decreasing on [0,∞); f(0) = 1 is the absolute maximum of f .

(b)d2y

dx2= − cosh2 x− 2 sinh2 x

cosh3 x=

sinh2 x− 1cosh3 x

d2y

dx2= 0 ⇒ sinhx = ±1

sinhx = 1 ⇒ ex − e−x

2= 1 ⇒ e2x − 2ex − 1 = 0 ⇒ x = ln (1 +

√2) ∼= 0.881;

sinhx = −1 ⇒ ex − e−x

2= −1 ⇒ e2x + 2ex − 1 = 0 ⇒ x = − ln (1 +

√2) = −0.881

(c) The graph of f is concave up on (−∞,−0.881) ∪ (0.881,∞) and concave down on (−0.881, 0.881);

points of inflection at x = ±0.881

26. (a)

-1-2-3 1 2 3x

-10

-5

5

10y (b)

-1-2 1 2x

-10

-5

5

10y

27. y = sinhx;dy

dx= coshx;

d2y

dx2= sinhx.

d2y

dx2= 0 ⇒ sinhx = 0 ⇒ x = 0.

y = sinh−1 x = ln(x +

√x2 + 1

);

dy

dx=

1√x2 + 1

;d2y

dx2= − x

(x2 + 1)3/2.

d2y

dx2= 0 ⇒ − x

(x2 + 1)3/2= 0 ⇒ x = 0.



SECTION 7.9 395

It is easy to verify that (0, 0) is a point of inflection for both graphs.

28. (a) (b)

29. (a) tanφ = sinhx (b) sinhx = tanφ

φ = arctan(sinhx) x = sinh−1(tanφ)

dφ

dx=

coshx

1 + sinh2 x= ln

(tanφ +

√tan2 φ + 1

)=

coshx

cosh2 x=

1coshx

= sechx = ln(tanφ + secφ)

= ln(secφ + tanφ)

(c) x = ln (secφ + tanφ)

dx

dφ=

secφ tanφ + sec2 φ

tanφ + secφ= secφ

30. V =∫ 1

−1

π sech 2x dx = [π tanhx]1−1 = π

(e− e−1

e + e−1− e−1 − e

e−1 + e

)= 2π

(e2 − 1e2 + 1

)

31.∫

tanhx dx =∫

sinhx

coshxdx{

u = coshx

du = sinhx dx

};∫

sinhx

coshxdx =

∫1udu = ln |u| + C = ln coshx + C



396 SECTION 7.9

32.∫

cothx dx =∫

coshx

sinhxdx =

∫du

u= ln |u| + C = ln | sinhx| + C

33.∫

sechx dx =∫

1coshx

dx =∫

2ex + e−x

dx =∫

2ex

e2x + 1dx

{u = ex

du = ex dx

};∫

2ex

e2x + 1dx = 2

∫1

u2 + 1du = 2 arctan u + C = 2 arctan (ex) + C

34.∫

cschx dx =∫

2ex − e−x

dx = 2∫

ex

e2x − 1dx = −2

∫du

1 − u2(u = ex)

=

{−2 tanh−1 ex + C, ex < 1

−2 coth−1 ex + C, ex > 1.

35.

{u = sechx

du = − sech s tanhx dx

};∫

sech 3x tanhx dx = −∫

u2 du = − 13u3 + C

= − 13 sech 3x + C

36.∫

x sech 2x2 dx =12

∫sech 2u du =

12

tanhu + C =12

tanhx2 + C

37.

{u = ln (coshx)

du = tanhx dx

};∫

tanhx ln (coshx) dx =∫

u du =12u2 + C =

12

[ln (coshx)]2 + C

38.∫

1 + tanhx

cosh2 xdx =

∫ (sech 2x +

sinhx

cosh3 x

)dx = tanhx− 1

2 cosh2 x+ C = tanhx− 1

2sech 2x + C

39.

{u = 1 + tanhx

du = sech 2x dx

};∫

sech 2x

1 + tanhxdx =

∫1udu = ln |u| + C = ln |1 + tanhx| + C

40.∫

tanh5 x sech 2x dx =16

tanh6 x + C

41.

{x = a sinhu

dx = a coshu du

};

∫dx√

a2 + x2dx =

∫a coshu√

a2 + a2 sinh2 udu

=∫

du = u + C = sinh−1(xa

)+ C

42.∫

1√x2 − a2

dx =1a

∫1√

(x/a)2 − 1dx =

∫du√u2 − 1

= cosh−1(u) + C = cosh−1(xa

)+ C



REVIEW EXERCISES 397

43. Suppose |x| < a.{x = a tanhu

dx = a sech 2u du

};∫

dx

a2 − x2dx =

∫a sech 2u

a2 − a2 tanh2 udu

=1a

∫du =

u

a+ C =

1a

tanh−1(xa

)+ C

The other case is done in the same way.

44. (a) v(0) =√

mg

ktanh 0 = 0

v′(t) =√

mg

ksech 2

(√gk

mt

)(√gk

m

)= g sech 2

(√gk

mt

)

mg − kv2 = mg − kmg

ktanh2

(√gk

mt

)= mg

[1 − tanh2

(√gk

mt

)]

= mg sech 2

(√gk

mt

)= m

dv

dt

(b) limt→∞

v(t) = limt→∞

√mg

ktanh

(√gk

mt

)=√

mg

k.

REVIEW EXERCISES

1. f ′(x) = 13x

−2/3 > 0 except at x = 0; f is increasing, it is one-to-one; f−1(x) = (x− 2)3.

2. f is not one-to-one; f(3) = f(−2) = 0

3. Suppose f(x1) = f(x2). Thenx1 + 1x1 − 1

=x2 + 1x2 − 1

x1x2 + x2 − x1 − 1 = x1x2 + x1 − x2 − 1

2x2 = 2x1

x1 = x2

Thus f is one-to-one. f−1(x) =x + 1x− 1

.

4. f ′(x) = 6(2x + 1)2 > 0 except at x = − 12 ; f is increasing, it is one-to-one; f−1(x) =

x1/3 − 12

.

5. f ′(x) = − 1x2

e1x < 0 except at x = 0; f is decreasing, it is one-to-one; f−1(x) = 1

lnx .

6. f(x) is not one-to-one; f(π/2) = f(3π/2) = 0, .

7. f is not one-to-one. Reason: f ′(x) = 1 + lnx =⇒ f decreases on (0, 1/e] and increases on [1/e,∞).

There exist horizontal lines that intersect the graph in two points.



398 REVIEW EXERCISES

8. Suppose f(x1) = f(x2). Then

2x1 + 13 − 2x1

=2x2 + 13 − 2x2

−4x1x2 + 6x1 − 2x2 + 3 = −4x1x2 + 6x2 − 2x1 + 38x1 = 8x2

x1 = x2

Thus f is one-to-one. f−1(x) =3x− 12x + 2

.

9. f ′(x) =−ex

(1 + ex)2< 0 for all x; f is one to one and has an inverse function.

Since f(0) = 12 , (f−1)′

(12

)=

1f ′(0)

= − 4

10. f ′(x) = 3 +3x4

> 0 for all x > 0; f is one-to-one and has an inverse function.

f(1) = 2; therefore (f−1)′(2) =1

f ′(1)=

16

11. f ′(x) =√

4 + x2 > 0 =⇒ f has an inverse.

Since f(0) = 0,(f−1)′ (0) =

1f ′(0)

=12

12. f ′ = 1 − sinx ≥ 0 for all x; f has an inverse function.

Since f(π) = −1; (f−1)′(−1) =1

f ′(π)= 1

13. f ′(x) = 3(lnx2

)2 1x2

(2x) =6(lnx2

)2x

=24 (lnx)2

x

14. y′ = 2(cos e3x)(e3x)(3) = 6e3x cos e3x

15. g′(x) =

(1 + e2x

)ex − ex

(2e2x

)(1 + e2x)2

=ex(1 − e2x)(1 + e2x)2

16. ln f(x) = sinhx ln(x2 + 1);f ′(x)f(x)

= sinhx2x

x2 + 1+ ln(x2 + 1) coshx;

f ′(x) = (x2 + 1)sinhx

[coshx ln(x2 + 1) +

2x sinhx

x2 + 1

]

17.dy

dx=

1x3 + 3x

(3x2 + 3x ln 3

)=

3x2 + 3x ln 3x3 + 3x

18. g′(x) =sinhx

1 + cosh2 x




19. ln f(x) =1x

ln coshx =ln coshx

x

f ′(x)f(x)

=x

sinhx

coshx− ln coshx

x2=

sinhx

x coshx− ln coshx

x2.

Therefore f ′(x) = (coshx)1/x[

sinhx

x coshx− ln coshx

x2

]

20. f ′(x) = 2x3 2x√1 − x4

+ 6x2 arcsin (x2) =4x4

√1 − x4

+ 6x2 arcsin (x2)

21. f(x) =1

ln 3ln

1 + x

1 − x; f ′(x) =

1ln 3

(1 − x

1 + x

)(2

(1 − x)2

)=

2ln 3(1 − x2)

22. f ′(x) =1√

x2 + 4√x2 + 4 − 1

· 12

2x√x2 + 4

==x

(x2 + 4)√x2 + 3

23. Substitution: Set u = ex, du = ex dx.∫ex√

1 − e2xdx =

∫1√

1 − u2dx = arcsin u + C = arcsin ex + C

24. Substitution: let u = lnx. Then du =1xdx, u(1) = 0, u(e) = 1.

∫ e

1

√lnx

xdx =

∫ 1

0

u1/2 du = 23

[u3/2

]10

= 23

25. Substitution: let u = sinx, du = cosx dx∫cosx

4 + sin2 xdx =

∫1

4 + u2du =

12

arctan(u

2

)+ C =

12

arctan(

sinx

2

)+ C

26. Substitution: let u = ln cosx, du = − tanx dx∫tanx ln cosx dx = −

∫u du = −1

2u2 + C = −1

2ln2 cosx + C

27. Substitution: let u =√x, du =

12

1√xdx

∫sec

√x√

xdx = 2

∫secu du = 2 ln | secu + tanu| + C = 2 ln | sec

√x + tan

√x| + C

28. Substitution: let u = x2, du = 2x dx∫1

x√x4 − 9

dx =12

∫1

u√u2 − 32

du =16arcsec

x 2

3+ C

29. Substitution: let u = lnx, du =1xdx

∫5lnx

xdx =

5lnx

ln 5+ C




30. Substitution: let u = x3, du = 3x2 dx∫ 2

0

x2ex3dx =

13ex

3 |20 =13(e8 − 1)

31. Substitution: u = x4/3 + 1, du = 43x

1/3 dx, u(1) = 2, u(8) = 17∫ 8

1

x1/3

x4/3 + 1dx =

34

∫ 17

2

1udu =

34[lnu]172 =

34

(ln 17 − ln 2) =34

ln(

172

)

32. Substitution: let u = secx, du = secx tanx dx∫secx tanx

1 + sec2 xdx =

∫1

1 + u2du = arctanu + C = arctan secx + C

33. Substitution: u = 2x, du = 2x ln 2 dx∫2x sinh 2x dx =

1ln 2

∫sinhu du =

1ln 2

coshu + C =1

ln 2cosh 2x + C

34. Substitution: let u = e2x, du = 2e2x dx∫ex

ex + e−xdx =

∫e2x

1 + e2xdx =

12

ln(1 + e2x) + C

35.∫ 5

2

1x2 − 4x + 13

dx =∫ 5

2

1(x− 2)2 + 32

dx =13

[arctan

x− 23

]52

=π

12

36. Substitution: let u = x− 1, du = dx∫1√

15 + 2x− x2dx =

∫1

42 − (x− 1)2dx = arcsin

x− 14

+ C

37.∫ 2

0

sech2(x

2) dx = 2

[tanh

x

2

]20

= 2 tanh 1 = 2e2 − 1e2 + 1

38.∫

tanh22x dx = x− 12tanh 2x + C

39. A =∫ 1

0

x

1 + x2dx =

12

[ln(x2 + 1)

]10

=ln 22

40. A =∫ 1

0

11 + x2

dx = [arctanx]10 =π

4

41. A =∫ 1/2

0

1√1 − x2

dx = [arcsinx]1/20 =π

6

42. A =∫ 1/2

0

x√1 − x2

dx = −[√

1 − x2]1/20

=2 −

√3

2




43. Fix x > 0. Then, by the mean-value theorem,

ln(1 + x) − ln 1 =1

1 + c(x− 0) =

x

1 + cfor some c ∈ (0, x)

Sincex

1 + x<

x

1 + c< x, it follows that

x

1 + x< ln(1 + x) < x

Now fix x ∈ (−1, 0). Then,

ln 1 − ln(1 + x) =1

1 + c(0 − x) =

−x

1 + cfor some c ∈ (x, 0)

Since −x <−x

1 + c<

−x

1 + x, it follows that

x

1 + x< ln(1 + x) < x

Thusx

1 + x< ln(1 + x) < x for all x > −1 (the result is obvious if x = 0).

(b) The result follows from part (a) and the pinching theorem (Theorem 2.5.1)

44. Using the hint,

lnn + 1m

=∫ n+1

m

dx

x=∫ m+1

m

dx

x+ · · · +

∫ n+1

n

dx

x<

1m

+ · · · + 1n

lnn

m− 1=∫ n

m−1

dx

x=∫ m

m−1

dx

x+ · · · +

∫ n

n−1

dx

x>

1m

+ · · · + 1n.

45. Assume a > 0: A =∫ 2a

a

a2

xdx =

[a2 lnx

]2aa

= a2(ln 2a− ln a) = a2 ln 2

46. A =∫ 1/2

−1/2

sec 12πx dx =

2π

[ln∣∣sec 1

2πx + tan 12πx∣∣ ]1/2

−1/2=

2π

ln√

2 + 1√2 − 1

47. (a) V =∫ √

3

0

π1

1 + x2dx = π[arctanx]

√3

0 =π2

3

(b) V =∫ √

3

0

2πx(1 + x2)−1/2 dx = 2π[√

1 + x2]√3

0= 2π

48. (a) V =∫ 1/2

0

π√1 + x2

dx = π ln1 +

√5

2

(b) V =∫ 1/2

0

2πx1

(1 + x2)1/4dx =

4π3[(1 + x2)3/4

]1/20

=4π3

[(54

)3/4

− 1

]




49. f(x) =lnx

x, domain: (0,∞)

f ′(x) =1 − lnx

x2

critical pt. x = e

f ′′(x) =2 lnx− 3

x3

f ′(x) > 0 on (0, e),

f ′(x) < 0 on (e,∞)

f ′′(x) > 0 on(e3/2,∞

)f ′′(x) < 0 on

(0, e3/2

)

2 4 6 8x

-2

-1

0.5y

50. f(x) = x2e−x2, domain: (−∞, ∞)

f ′(x) = 2(x− x3

)e−x2

= 2x(1 − x)(1 + x)e−x2

critical pts.. x = 0, x = 1, x = −1

f ′′(x) = 2(2x4 − 5x2 + 1)e−x2

f ′(x) > 0 on (−∞,−1) ∪ (0, 1),

f ′(x) < 0 on (−1, 0) ∪ (1, ∞)

f ′′(x) > 0 on(−∞, − 1

2

√5 +

√17)∪(− 1

2

√5 −

√17, 1

2

√5 −

√17)∪(

12

√5 +

√17, ∞

)f ′′(x) < 0 on

(− 1

2

√5 +

√17, − 1

2

√5 −

√17)∪(

12

√5 −

√17, 1

2

√5 +

√17)

-1-2 1 2x

0.5

y

51.∫ 1

0

b√1 − b2x2

dx: substitution: u = bx, du = b dx, u(0) = 0, u(1) = b

∫ 1

0

b√1 − b2x2

dx =∫ b

0

1√1 − u2

du = [arcsin u]b0 = arcsin b.

∫ a

0

1√1 − x2

dx = arcsin a; =⇒ a = b

52.∫ 1

0

a

1 + a2x2dx = arctan a =

∫ a

0

11 + x2

dx

53. Let 6 a.m. correspond to t = 0; measure time in hours.

P (t) = 20ekt; P (2) = 40 = 20e2k =⇒ k =ln 22

=⇒ P (t) = 20et2 ln 2 = 20 (2)t/2

(a) P (6) = 20(2)3 = 20(8) = 160 grams

(b) 20(2)t/2 = 200 =⇒ t

2ln 2 = ln 10 =⇒ t =

2 ln 10ln 2

∼= 6.64 hours




54. A(t) = A(0)ekt . A(1) = 0.8A(0) =⇒ ek = 0.8 =⇒ k = ln 0.8 =⇒ T = − ln 2ln 0.8

∼= 3.1

The half-life of the substance is approximately 3.1 years.

55. (a) A(t) = 100ekt. From the relation k =− ln 2T

, where T is the half-life, A(t) = 100e−t

140 ln 2

(b) 100e−t

140 ln 2 = 75 =⇒ −t

140ln 2 = ln 0.75 =⇒ t =

−140 ln 0.75ln 2

∼= 58.11 days

56. PUS = 227ekt; PUS(10) = 249 = 227e10k =⇒ k =ln(249/227)

10∼= 0.00925

PM = 62emt;PM (10) = 79 = 62e10m =⇒ m =ln(79/62)

10∼= 0.02423

To find when the populations will be equal, solve 227ekt = 62emt for t:

227ekt = 62emt =⇒ e(m−k)t =22762

=⇒ t =ln(227/62)m− k

∼= 86.64

If the populations continue to grow at the given rates, the two populations will be equal in the year2067.

57. P (t) = P0ekt. Since the doubling time is 2 years, k = 1

2 ln 2 and P (t) = P0et2 ln 2 = P0(2)t/2.

(a) 40, 000 = P (4) = P0(2)2 =⇒ 4P0 = 25, 000 =⇒ P0 = 6250.

(b) P (t) = 6250(2)t/2 = 40, 000 =⇒ (2)t/2 = 6.4 =⇒ t =2 ln 6.4

ln 2∼= 5.36; it will take

approximately 5.36 years for the population to reach 40,000.

58.1p

+1q

= 1 =⇒ q =p

p− 1. Assume that a ≤ b. See the figure: clearly ab ≤ area Ω1 + area Ω2.

area Ω1 =∫ a

0

xp−1dx =[xp

p

]a0

=ap

p.

y = xp−1 =⇒ x = y1/(p−1);

area Ω2 =∫ b

0

y1/(p−1)dy =[

y1+1/(p−1)

1 + 1/(p− 1)

]b0

=b1+1/(p−1)

1 + 1/(p− 1)=

bq

q

The same argument applies if b < a.

calculus one and several variables 10e salas solutions manual ch07

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