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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
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.
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
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.
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
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.
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.
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.
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−
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.
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
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.
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
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.
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