Section 3.6 Derivatives of Logarithmic Functions

(1)

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 )^15 ⇒ ⁰() = ¹₅(ln )^−45 

(ln ) = 1 5(ln )^45· 1

 = 1

5⁵ (ln )⁴

6.  () = ln√⁵

 = ln ^15=¹₅ln  ⇒ 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^12=¹₂log10 ⇒ 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 )^−12 

(1 + ln ) = 1 2√

1 + ln ·1

 = 1

2√ 1 + ln 

11.  () = (ln )²sin  ⇒ 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)^12= ln  +¹₂ln(²− 1) ⇒

⁰() = 1

+1 2· 1

²− 1· 2 = 1

+ 

²− 1=²− 1 +  · 

(²− 1) = 2²− 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 

SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 227 16.  = ln

1 +  − ³

 ⇒ ⁰= 1

1 +  − ³



(1 +  − ³) = 1 − 3² 1 +  − ³

17.  () = 2^log₂ ⇒ ⁰() = 2^ 1

 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

12

= 1 2ln

²− ²

²+ ²



=¹₂ln(²− ²) −¹2ln(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) = 2²

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

sec sec  tan  = tan  ⇒ ⁰⁰= sec²

SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 227 16.  = ln

1 +  − ³

 ⇒ ⁰= 1

1 +  − ³



(1 +  − ³) = 1 − 3² 1 +  − ³

17.  () = 2^log₂ ⇒ ⁰() = 2^ 1

 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

12

= 1 2ln

²− ²

²+ ²



=¹₂ln(²− ²) −¹2ln(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) = 2²

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

sec sec  tan  = tan  ⇒ ⁰⁰= sec²

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 )^12 ⇒ ⁰() = 1

2(2 + ln )^−12·1

= 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(ln ²) ⇒ ⁰() = − sin(ln ²) 

 ln ²= − sin(ln ²) 1

²(2) = −2 sin(ln ²)

 .

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) = ¹₁ · 3 = 3, so an equation of a tangent line at (3 0)is  − 0 = 3( − 3), or  = 3 − 9.

34.  = ²ln  ⇒ 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.

SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 229

35.  () = sin  + ln  ⇒ ⁰() = cos  + 1.

This is reasonable, because the graph shows that  increases when ⁰is positive, and ⁰() = 0when  has a horizontal tangent.

36.  =ln 

 ⇒ ⁰= (1) − ln 

² = 1 − ln 

² .

⁰(1) = 1 − 0

1² = 1and 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(3²− 2) ⇒ ⁰() = 1

(3²− 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· 4³ ⇒

⁰= 

 10

2 + 1+ 24³

⁴− 3



= (2 + 1)⁵(⁴− 3)⁶

 10

2 + 1+ 24³

⁴− 3

 .

[The answer could be simpliﬁed to ⁰= 2(2 + 1)⁴(⁴− 3)⁵(29⁴+ 12³− 15), but this is unnecessary.]

40.  =√

 ^²

²+ 110

⇒ ln  = ln√

 + ln ^²+ ln(²+ 1)¹⁰ ⇒ ln  = ¹2ln  + 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

12

⇒ ln  = 1

2ln( − 1) −1

2ln(⁴+ 1) ⇒ 1

⁰=1 2

1

 − 1−1 2

1

⁴+ 1· 4³ ⇒ ⁰= 

 1

2( − 1)− 2³

⁴+ 1



⇒ ⁰=

 − 1

⁴+ 1

 1

2 − 2− 2³

⁴+ 1



42.  =√

 ^²^−( + 1)^23 ⇒ ln  = ln

^12^²^−( + 1)^23

⇒

ln  =¹₂ln  + (²− ) +²3ln( + 1) ⇒ 1

⁰=1 2·1

+ 2 − 1 +2 3· 1

 + 1 ⇒

⁰= 

 1

2+ 2 − 1 + 2 3 + 3



⇒ ⁰=√

 ^²^−( + 1)^23

 1

2+ 2 − 1 + 2 3 + 3



230 ¤ CHAPTER 3 DIFFERENTIATION RULES

43.  = ^ ⇒ ln  = ln ^ ⇒ ln  =  ln  ⇒ ⁰ = (1) + (ln ) · 1 ⇒ ⁰= (1 + ln ) ⇒

⁰= ^(1 + ln )

44.  = ^{cos } ⇒ ln  = ln ^{cos } ⇒ 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  =  ln ^12 ⇒ ln  = ¹2 ln  ⇒ 1

⁰= 1 2 ·1

+ ln  ·1

2 ⇒

⁰= 1

2+¹₂ln 

⇒ ⁰=¹₂√

^(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 + 2⁰

²+ ² ⇒ ²⁰+ ²⁰= 2 + 2⁰ ⇒

²⁰+ ²⁰− 2⁰= 2 ⇒ (²+ ²− 2)⁰= 2 ⇒ ⁰= 2

²+ ²− 2

52. ^= ^ ⇒  ln  =  ln  ⇒  ·1

+ (ln ) · ⁰=  ·1

 · ⁰+ ln  ⇒ ⁰ln  −

⁰= ln  −

 ⇒

⁰= ln  − 

ln  − 

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.  = ⁸ln , so ⁹ = ⁸⁰= ⁸(8⁷ln  + ⁷). But the eighth derivative of ⁷is 0, so we now have

⁸(8⁷ln ) = ⁷(8 · 7⁶ln  + 8⁶) = ⁷(8 · 7⁶ln ) = ⁶(8 · 7 · 6⁵ln ) = · · · = (8! ⁰ln ) = 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− 8²+ 24(in meters) ⇒ () = ⁰() = 3²− 16 + 24 (in ms) (b) (1) = 3(1)²− 16(1) + 24 = 11 ms

(c) The particle is at rest when () = 0. 3²− 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) () = 3²− 16 + 24 ⇒

() = ⁰() = 6 − 16 (in (ms)s or ms²).

(1) = 6(1) − 16 = −10 ms²

(h)

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

 ⁸₃, so the particle is speeding up when   ⁸₃. 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 = −9²+ 81

(²+ 9)² = −9(²− 9)

(²+ 9)² (in ms) (b) (1) = −9(1 − 9)

(1 + 9)² = 72

100= 072 ms

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.  = ⁸ln , so ⁹ = ⁸⁰= ⁸(8⁷ln  + ⁷). But the eighth derivative of ⁷is 0, so we now have

⁸(8⁷ln ) = ⁷(8 · 7⁶ln  + 8⁶) = ⁷(8 · 7⁶ln ) = ⁶(8 · 7 · 6⁵ln ) = · · · = (8! ⁰ln ) = 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− 8²+ 24(in meters) ⇒ () = ⁰() = 3²− 16 + 24 (in ms) (b) (1) = 3(1)²− 16(1) + 24 = 11 ms

(c) The particle is at rest when () = 0. 3²− 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) () = 3²− 16 + 24 ⇒

() = ⁰() = 6 − 16 (in (ms)s or ms²).

(1) = 6(1) − 16 = −10 ms²

(h)

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

 ⁸₃, so the particle is speeding up when   ⁸₃. 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 = −9²+ 81

(²+ 9)² = −9(²− 9)

(²+ 9)² (in ms) (b) (1) = −9(1 − 9)

(1 + 9)² = 72

100= 072 ms