Section 3.6 Derivatives of Logarithmic Functions

226 ¤ CHAPTER 3 DIFFERENTIATION RULES

3.6 Derivatives of Logarithmic Functions

1. The differentiation formula for logarithmic functions,

(log ) =

1

ln , is simplest when = because ln = 1.

2. () = ln− ⇒ 0() = · 1

+ (ln) · 1− 1 = 1 + ln− 1 = ln

3. () = sin(ln) ⇒ 0() = cos(ln) ·

ln = cos(ln) · 1

=

cos(ln)

4. () = ln(sin2 ) = ln(sin)2 = 2 ln |sin| ⇒ 0() = 2 · 1

sin· cos = 2 cot

5. () =5√

ln = (ln)15 ⇒ 0() = 15(ln)−45

(ln) =

1

5(ln)45· 1

=

1

5 5

(ln)4

6. () = ln5√ = ln15 = 1

5ln ⇒ 0() =

1

5· 1

=

1

5

7. () = log10 (1 + cos) ⇒ 0() =1

(1 + cos) ln 10

(1 + cos) =

− sin

(1 + cos) ln 10

8. () = log10

√ ⇒ 0() =

1√ ln 10

√ =

1√ ln 10

1

2√

=1

2(ln 10)

Or: () = log10

√ = log10

12 = 12

log10 ⇒ 0() =1

2

1

ln 10=

1

2 (ln 10)

9. () = ln(−2) = ln+ ln −2 = ln− 2 ⇒ 0() =1

− 2

10. () =√

1 + ln ⇒ 0() = 12(1 + ln )−12

(1 + ln ) =

1

2√

1 + ln · 1

=

1

2√

1 + ln

11. () = (ln )2 sin ⇒ 0() = (ln )2cos + sin · 2 ln · 1

= ln

ln cos +

2 sin

12. () = ln+

√2 − 1

⇒ 0() =1

+√2 − 1

1 +

√2 − 1

=

1

+√2 − 1

·√2 − 1 + √2 − 1

=1√

2 − 1

13. () = ln√2 − 1

= ln + ln(2 − 1)12 = ln + 1

2ln(2 − 1) ⇒

0() =1

+

1

2· 1

2 − 1· 2 =

1

+

2 − 1=

2 − 1 + · (2 − 1)

=22 − 1

(2 − 1)

14. () =ln

1− ⇒ 0() =

(1− )(1)− (ln )(−1)

(1− )2·

=1− + ln

(1− )2

15. () = ln ln ⇒ 0() =1

ln

ln =

1

ln · 1

=

1

ln

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SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 227

16. = ln1 + − 3

⇒ 0 =1

1 + − 3

(1 + − 3) =

1− 32

1 + − 3

17. () = 2 log2 ⇒ 0() = 21

ln 2+ log2 · 2 ln 2 = 2

1

ln 2+ log2 (ln 2)

.

Note that log2 (ln 2) =ln

ln 2(ln 2) = ln by the change of base formula. Thus, 0() = 2

1

ln 2+ ln

.

18. = ln(csc − cot) ⇒

0 =1

csc− cot

(csc− cot) =

1

csc− cot(− csc cot+ csc2 ) =

csc(csc− cot)

csc− cot= csc

19. = ln(− + −) = ln(−(1 + )) = ln(−) + ln(1 + ) = − + ln(1 + ) ⇒

0 = −1 +1

1 + =−1− + 1

1 + = −

1 +

20. () = ln

2 − 2

2 + 2= ln

2 − 2

2 + 2

12

=1

2ln

2 − 2

2 + 2

= 1

2ln(2 − 2)− 1

2ln(2 + 2) ⇒

0() =1

2· 1

2 − 2· (−2)− 1

2· 1

2 + 2· (2) =

2 − 2−

2 + 2=

(2 + 2)− (2 − 2)

(2 − 2)(2 + 2)

=3 + 2 − 3 + 2

(2 − 2)(2 + 2)=

22

4 − 4

21. = tan [ln(+ )] ⇒ 0 = sec2[ln(+ )] · 1

+ · = sec2[ln(+ )]

+

22. = log2( log5 ) ⇒

0 =1

( log5 )(ln 2)

( log5 ) =

1

( log5 )(ln 2)

· 1

ln 5+ log5

=

1

( log5 )(ln 5)(ln 2)+

1

(ln 2).

Note that log5 (ln 5) =ln

ln 5(ln 5) = ln by the change of base formula. Thus, 0 =

1

ln ln 2+

1

ln 2=

1 + ln

ln ln 2.

23. =√ ln ⇒ 0 =

√ · 1

+ (ln)

1

2√

=2 + ln

2√

⇒

00 =2√ (1)− (2 + ln)(1

√ )

(2√ )2

=2√− (2 + ln)(1

√ )

4=

2− (2 + ln)√(4)

= − ln

4√

24. =ln

1 + ln⇒ 0 =

(1 + ln)(1)− (ln)(1)

(1 + ln)2=

1

(1 + ln)2⇒

00 = −

[(1 + ln)2]

[(1 + ln)2]2[Reciprocal Rule] = − · 2(1 + ln) · (1) + (1 + ln)2

2(1 + ln)4

= − (1 + ln)[2 + (1 + ln)]

2(1 + ln)4= − 3 + ln

2(1 + ln)3

25. = ln |sec| ⇒ 0 =1

sec

sec =

1

secsec tan = tan ⇒ 00 = sec2

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SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 227

16. = ln1 + − 3

⇒ 0 =1

1 + − 3

(1 + − 3) =

1− 32

1 + − 3

17. () = 2 log2 ⇒ 0() = 21

ln 2+ log2 · 2 ln 2 = 2

1

ln 2+ log2 (ln 2)

.

Note that log2 (ln 2) =ln

ln 2(ln 2) = ln by the change of base formula. Thus, 0() = 2

1

ln 2+ ln

.

18. = ln(csc − cot) ⇒

0 =1

csc− cot

(csc− cot) =

1

csc− cot(− csc cot+ csc2 ) =

csc(csc− cot)

csc− cot= csc

19. = ln(− + −) = ln(−(1 + )) = ln(−) + ln(1 + ) = − + ln(1 + ) ⇒

0 = −1 +1

1 + =−1− + 1

1 + = −

1 +

20. () = ln

2 − 2

2 + 2= ln

2 − 2

2 + 2

12

=1

2ln

2 − 2

2 + 2

= 1

2ln(2 − 2)− 1

2ln(2 + 2) ⇒

0() =1

2· 1

2 − 2· (−2)− 1

2· 1

2 + 2· (2) =

2 − 2−

2 + 2=

(2 + 2)− (2 − 2)

(2 − 2)(2 + 2)

=3 + 2 − 3 + 2

(2 − 2)(2 + 2)=

22

4 − 4

21. = tan [ln(+ )] ⇒ 0 = sec2[ln(+ )] · 1

+ · = sec2[ln(+ )]

+

22. = log2( log5 ) ⇒

0 =1

( log5 )(ln 2)

( log5 ) =

1

( log5 )(ln 2)

· 1

ln 5+ log5

=

1

( log5 )(ln 5)(ln 2)+

1

(ln 2).

Note that log5 (ln 5) =ln

ln 5(ln 5) = ln by the change of base formula. Thus, 0 =

1

ln ln 2+

1

ln 2=

1 + ln

ln ln 2.

23. =√ ln ⇒ 0 =

√ · 1

+ (ln)

1

2√

=2 + ln

2√

⇒

00 =2√ (1)− (2 + ln)(1

√ )

(2√ )2

=2√− (2 + ln)(1

√ )

4=

2− (2 + ln)√(4)

= − ln

4√

24. =ln

1 + ln⇒ 0 =

(1 + ln)(1)− (ln)(1)

(1 + ln)2=

1

(1 + ln)2⇒

00 = −

[(1 + ln)2]

[(1 + ln)2]2[Reciprocal Rule] = − · 2(1 + ln) · (1) + (1 + ln)2

2(1 + ln)4

= − (1 + ln)[2 + (1 + ln)]

2(1 + ln)4= − 3 + ln

2(1 + ln)3

25. = ln |sec| ⇒ 0 =1

sec

sec =

1

secsec tan = tan ⇒ 00 = sec2

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228 ¤ CHAPTER 3 DIFFERENTIATION RULES

26. = ln(1 + ln) ⇒ 0 =1

1 + ln· 1

=

1

(1 + ln)⇒

00 = −

[(1 + ln)]

[(1 + ln)]2[Reciprocal Rule] = −(1) + (1 + ln)(1)

2(1 + ln)2= − 1 + 1 + ln

2(1 + ln)2= − 2 + ln

2(1 + ln)2

27. () =

1− ln(− 1)⇒

0() =

[1− ln(− 1)] · 1− · −1

− 1

[1− ln(− 1)]2

=

(− 1)[1− ln(− 1)] +

− 1

[1− ln(− 1)]2=

− 1− (− 1) ln(− 1) +

(− 1)[1− ln(− 1)]2

=2− 1− (− 1) ln(− 1)

(− 1)[1− ln(− 1)]2

Dom() = { | − 1 0 and 1− ln(− 1) 6= 0} = { | 1 and ln(− 1) 6= 1}= | 1 and − 1 6= 1

= { | 1 and 6= 1 + } = (1 1 + ) ∪ (1 + ∞)

28. () =√

2 + ln = (2 + ln)12 ⇒ 0() =1

2(2 + ln)−12 · 1

=

1

2√

2 + ln

Dom() = { | 2 + ln ≥ 0} = { | ln ≥ −2} = { | ≥ −2} = [−2∞).

29. () = ln(2 − 2) ⇒ 0() =1

2 − 2(2− 2) =

2(− 1)

(− 2).

Dom() = { | (− 2) 0} = (−∞ 0) ∪ (2∞).

30. () = ln ln ln ⇒ 0() =1

ln ln· 1

ln· 1

.

Dom() = { | ln ln 0} = { | ln 1} = { | } = (∞).

31. () = ln(+ ln) ⇒ 0() =1

+ ln

( + ln) =

1

+ ln

1 +

1

.

Substitute 1 for to get 0(1) =1

1 + ln 1

1 +

1

1

=

1

1 + 0(1 + 1) = 1 · 2 = 2.

32. () = cos(ln2) ⇒ 0() = − sin(ln2)

ln2 = − sin(ln2)

1

2(2) = −2 sin(ln2)

.

Substitute 1 for to get 0(1) = −2 sin(ln 12)

1= −2 sin 0 = 0.

33. = ln(2 − 3 + 1) ⇒ 0 =1

2 − 3+ 1· (2− 3) ⇒ 0(3) = 1

1· 3 = 3, so an equation of a tangent line at

(3 0) is − 0 = 3(− 3), or = 3− 9.

34. = 2 ln ⇒ 0 = 2 · 1

+ (ln)(2) ⇒ 0(1) = 1 + 0 = 1 , so an equation of a tangent line at (1 0) is

− 0 = 1(− 1), or = − 1.

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SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 229

35. () = sin + ln ⇒ 0() = cos + 1.

This is reasonable, because the graph shows that increases when 0 is

positive, and 0() = 0 when has a horizontal tangent.

36. =ln

⇒ 0 =

(1)− ln

2=

1− ln

2.

0(1) =1− 0

12= 1 and 0() =

1− 1

2= 0 ⇒ equations of tangent

lines are − 0 = 1(− 1) or = − 1 and − 1 = 0(− )

or = 1.

37. () = + ln(cos) ⇒ 0() = +1

cos· (− sin) = − tan.

0(4) = 6 ⇒ − tan

4= 6 ⇒ − 1 = 6 ⇒ = 7.

38. () = log(32 − 2) ⇒ 0() =

1

(32 − 2) ln· 6.

0(1) = 3 ⇒ 1

ln · 6 = 3 ⇒ 2 = ln ⇒ = 2.

39. = (2 + 1)5(4 − 3)6 ⇒ ln = ln(2+ 1)5(4 − 3)6

⇒ ln = 5 ln(2 + 1) + 6 ln(4 − 3) ⇒1

0 = 5 · 1

2 + 1· 2 + 6 · 1

4 − 3· 43 ⇒

0 =

10

2 + 1+

243

4 − 3

= (2+ 1)5(4 − 3)6

10

2+ 1+

243

4 − 3

.

[The answer could be simplified to 0 = 2(2+ 1)4(4 − 3)5(294 + 123 − 15), but this is unnecessary.]

40. =√

22 + 1

10 ⇒ ln = ln√+ ln

2

+ ln(2 + 1)10 ⇒ ln = 12

ln + 2 + 10 ln(2 + 1) ⇒

1

0 =

1

2· 1

+ 2+ 10 · 1

2 + 1· 2 ⇒ 0 =

√

2

(2 + 1)10

1

2+ 2+

20

2 + 1

41. =

− 1

4 + 1⇒ ln = ln

− 1

4 + 1

12

⇒ ln =1

2ln(− 1)− 1

2ln(4 + 1) ⇒

1

0 =

1

2

1

− 1− 1

2

1

4 + 1· 43 ⇒ 0 =

1

2(− 1)− 23

4 + 1

⇒ 0 =

− 1

4 + 1

1

2− 2− 23

4 + 1

42. =√

2−(+ 1)23 ⇒ ln = ln12

2−(+ 1)23⇒

ln = 12

ln + (2 − ) + 23

ln( + 1) ⇒ 1

0 =

1

2· 1

+ 2− 1 +

2

3· 1

+ 1⇒

0 =

1

2+ 2− 1 +

2

3+ 3

⇒ 0 =

√

2−( + 1)23

1

2+ 2− 1 +

2

3+ 3

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

230 ¤ CHAPTER 3 DIFFERENTIATION RULES

43. = ⇒ ln = ln ⇒ ln = ln ⇒ 0 = (1) + (ln) · 1 ⇒ 0 = (1 + ln) ⇒0 = (1 + ln)

44. = cos ⇒ ln = lncos ⇒ ln = cos ln ⇒ 1

0 = cos · 1

+ ln · (− sin) ⇒

0 = cos

− ln sin

⇒ 0 = cos

cos

− ln sin

45. = sin ⇒ ln = ln sin ⇒ ln = sin ln ⇒ 0

= (sin) · 1

+ (ln)(cos) ⇒

0 =

sin

+ ln cos

⇒ 0 = sin

sin

+ ln cos

46. =√ ⇒ ln = ln

√ ⇒ ln = ln12 ⇒ ln = 1

2 ln ⇒ 1

0 =

1

2 · 1

+ ln · 1

2⇒

0 =

12

+ 12

ln ⇒ 0 = 1

2

√ (1 + ln)

47. = (cos) ⇒ ln = ln(cos) ⇒ ln = ln cos ⇒ 1

0 = · 1

cos· (− sin) + ln cos · 1 ⇒

0 =

ln cos− sin

cos

⇒ 0 = (cos)(ln cos− tan)

48. = (sin)ln ⇒ ln = ln(sin)ln ⇒ ln = ln · ln sin ⇒ 1

0 = ln · 1

sin· cos+ ln sin · 1

⇒

0 =

ln · cos

sin+

ln sin

⇒ 0 = (sin)ln

ln cot+

ln sin

49. = (tan)1 ⇒ ln = ln(tan)1 ⇒ ln =1

ln tan ⇒

1

0 =

1

· 1

tan· sec2 + ln tan ·

− 1

2

⇒ 0 =

sec2

tan− ln tan

2

⇒

0 = (tan)1

sec2

tan− ln tan

2

or 0 = (tan)1 · 1

csc sec− ln tan

50. = (ln)cos ⇒ ln = cos ln(ln) ⇒ 0

= cos · 1

ln· 1

+ (ln ln)(− sin) ⇒

0 = (ln)cos cos

ln− sin ln ln

51. = ln(2 + 2) ⇒ 0 =1

2 + 2

(2 + 2) ⇒ 0 =

2+ 20

2 + 2⇒ 20 + 20 = 2 + 20 ⇒

20 + 20 − 20 = 2 ⇒ (2 + 2 − 2)0 = 2 ⇒ 0 =2

2 + 2 − 2

52. = ⇒ ln = ln ⇒ · 1

+ (ln) · 0 = · 1

· 0 + ln ⇒ 0 ln−

0 = ln −

⇒

0 =ln −

ln−

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SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 231

53. () = ln(− 1) ⇒ 0() =1

(− 1)= (− 1)−1 ⇒ 00() = −(− 1)−2 ⇒ 000 () = 2(− 1)−3 ⇒

(4)() = −2 · 3(− 1)−4 ⇒ · · · ⇒ ()() = (−1)−1 · 2 · 3 · 4 · · · · · (− 1)(− 1)− = (−1)−1 (− 1)!

(− 1)

54. = 8 ln, so9 = 80 = 8(87 ln+ 7). But the eighth derivative of 7 is 0, so we now have

8(87 ln) = 7(8 · 76 ln+ 86) = 7(8 · 76 ln) = 6(8 · 7 · 65 ln) = · · · = (8!0 ln) = 8!

55. If () = ln (1 + ), then 0() =1

1 + , so 0(0) = 1.

Thus, lim→0

ln(1 + )

= lim

→0

()

= lim

→0

()− (0)

− 0= 0(0) = 1.

56. Let = . Then = , and as → ∞, → ∞.

Therefore, lim→∞

1 +

= lim

→∞

1 +

1

=

lim

→∞

1 +

1

= by Equation 6.

3.7 Rates of Change in the Natural and Social Sciences

1. (a) = () = 3 − 82 + 24 (in meters) ⇒ () = 0() = 32 − 16+ 24 (in ms)

(b) (1) = 3(1)2 − 16(1) + 24 = 11 ms

(c) The particle is at rest when () = 0. 32 − 16+ 24 = 0 ⇒ −(−16)±

(−16)2 − 4(3)(24)

2(3)=

16±√−32

6.

The negative discriminant indicates that is never 0 and that the particle never rests.

(d) From parts (b) and (c), we see that () 0 for all , so the particle is always moving in the positive direction.

(e) The total distance traveled during the first 6 seconds

(since the particle doesn’t change direction) is

(6)− (0) = 72− 0 = 72 m.

(f )

(g) () = 32 − 16+ 24 ⇒() = 0() = 6− 16 (in (ms)s or ms

2).

(1) = 6(1)− 16 = −10 ms2

(h)

(i) The particle is speeding up when and have the same sign. is always positive and is positive when 6− 16 0 ⇒ 8

3, so the particle is speeding up when 8

3. It is slowing down when and have opposite signs; that is, when

0 ≤ 83.

2. (a) = () =9

2 + 9(in meters) ⇒ () = 0() =

(2 + 9)(9)− 9(2)

(2 + 9)2=−92 + 81

(2 + 9)2=−9(2 − 9)

(2 + 9)2(in ms)

(b) (1) =−9(1− 9)

(1 + 9)2=

72

100= 072 ms

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SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 231

53. () = ln(− 1) ⇒ 0() =1

(− 1)= (− 1)−1 ⇒ 00() = −(− 1)−2 ⇒ 000 () = 2(− 1)−3 ⇒

(4)() = −2 · 3(− 1)−4 ⇒ · · · ⇒ ()() = (−1)−1 · 2 · 3 · 4 · · · · · (− 1)(− 1)− = (−1)−1 (− 1)!

(− 1)

54. = 8 ln, so9 = 80 = 8(87 ln+ 7). But the eighth derivative of 7 is 0, so we now have

8(87 ln) = 7(8 · 76 ln+ 86) = 7(8 · 76 ln) = 6(8 · 7 · 65 ln) = · · · = (8!0 ln) = 8!

55. If () = ln (1 + ), then 0() =1

1 + , so 0(0) = 1.

Thus, lim→0

ln(1 + )

= lim

→0

()

= lim

→0

()− (0)

− 0= 0(0) = 1.

56. Let = . Then = , and as → ∞, → ∞.

Therefore, lim→∞

1 +

= lim

→∞

1 +

1

=

lim

→∞

1 +

1

= by Equation 6.

3.7 Rates of Change in the Natural and Social Sciences

1. (a) = () = 3 − 82 + 24 (in meters) ⇒ () = 0() = 32 − 16+ 24 (in ms)

(b) (1) = 3(1)2 − 16(1) + 24 = 11 ms

(c) The particle is at rest when () = 0. 32 − 16+ 24 = 0 ⇒ −(−16)±

(−16)2 − 4(3)(24)

2(3)=

16±√−32

6.

The negative discriminant indicates that is never 0 and that the particle never rests.

(d) From parts (b) and (c), we see that () 0 for all , so the particle is always moving in the positive direction.

(e) The total distance traveled during the first 6 seconds

(since the particle doesn’t change direction) is

(6)− (0) = 72− 0 = 72 m.

(f )

(g) () = 32 − 16+ 24 ⇒() = 0() = 6− 16 (in (ms)s or ms

2).

(1) = 6(1)− 16 = −10 ms2

(h)

(i) The particle is speeding up when and have the same sign. is always positive and is positive when 6− 16 0 ⇒ 8

3, so the particle is speeding up when 8

3. It is slowing down when and have opposite signs; that is, when

0 ≤ 83.

2. (a) = () =9

2 + 9(in meters) ⇒ () = 0() =

(2 + 9)(9)− 9(2)

(2 + 9)2=−92 + 81

(2 + 9)2=−9(2 − 9)

(2 + 9)2(in ms)

(b) (1) =−9(1− 9)

(1 + 9)2=

72

100= 072 ms

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1