Section 3.6 Derivatives of Logarithmic and Inverse Trigonometric Functions

(1)

Section 3.6 Derivatives of Logarithmic and Inverse Trigonometric Functions

49. Use logarithmic differentiation to find the derivative of the function. y = x^x Solution:

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  − 

62. Show that lim

n→∞ 1 + ^x_nⁿ

= e^xfor any x > 0.

Solution:

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

82. (a) One way of defining sec⁻¹x is to say that y = sec⁻¹x ⇔ sec y = x and 0 ≤ y < π/2 or π ≤ y < 3π/2. Show that, with this definition

d

dx(sec⁻¹x) = 1 x√

x²− 1

(b) Another way of defining sec⁻¹x that is sometimes used is to say that y = sec⁻¹x ⇔ sec y = x and 0 ≤ y ≤ π, y 6= π/2. Show that, with this definition,

d

dx(sec⁻¹x) = 1

|x|√ x²− 1 Solution:

SECTION 3.6 DERIVATIVES OF LOGARITHMIC AND INVERSE TRIGONOMETRIC FUNCTIONS ¤ 247 80. () = arctan(²− ) ⇒ ⁰() = 1

1 + (²− )² · 

(²− ) = 2 − 1 1 + (²− )²

Note that ⁰= 0where the graph of  has a horizontal tangent. Also note that ⁰is negative when  is decreasing and ⁰is positive when  is increasing.

81.Let  = cos⁻¹. Then cos  =  and 0 ≤  ≤  ⇒ − sin 

 = 1 ⇒



 = − 1

sin  = − 1

1 − cos² = − 1

√1 − ². [Note that sin  ≥ 0 for 0 ≤  ≤ .]

82. (a) Let  = sec⁻¹. Then sec  =  and  ∈ 0^₂

∪

^3₂ . Differentiate with respect to : sec  tan 







= 1 ⇒



 = 1

sec  tan  = 1 sec 

sec² − 1 = 1

√

²− 1. Note that tan² = sec² − 1 ⇒ tan  =

sec² − 1

since tan   0 when 0    ^₂ or     ^3₂ .

(b)  = sec⁻¹ ⇒ sec  =  ⇒ sec  tan 

= 1 ⇒ 

 = 1

sec  tan . Now tan² = sec² − 1 = ²− 1, so tan  = ±√

²− 1. For  ∈ 0^₂

,  ≥ 1, so sec  =  = || and tan  ≥ 0 ⇒



 = 1

√

²− 1= 1

||√

²− 1. For  ∈_

2 

,  ≤ −1, so || = − and tan  = −√

²− 1 ⇒



 = 1

sec  tan  = 1



−√

²− 1 = 1 (−)√

²− 1= 1

||√

²− 1.

83.If  = ⁻¹(), then () = . Differentiating implicitly with respect to  and remembering that  is a function of ,

we get ⁰()

 = 1, so 

 = 1

⁰() ⇒ 

⁻¹0

() = 1

⁰(⁻¹()).

84. (4) = 5 ⇒ ⁻¹(5) = 4. By Exercise 83,

⁻¹0

(5) = 1

⁰(⁻¹(5))= 1

⁰(4)= 1 23=3

2.

85. () =  + ^ ⇒ ⁰() = 1 + ^. Observe that (0) = 1, so that ⁻¹(1) = 0. By Exercise 83, we have (⁻¹)⁰(1) = 1

⁰(⁻¹(1))= 1

⁰(0)= 1

1 + ⁰ = 1 1 + 1= 1

2.

86. () = ³+ 3 sin  + 2 cos  ⇒ ⁰() = 3²+ 3 cos  − 2 sin . Observe that (0) = 2, so that ⁻¹(2) = 0.

By Exercise 83, we have (⁻¹)⁰(2) = 1

⁰(⁻¹(2))= 1

⁰(0)= 1

3(0)²+ 3 cos 0 − 2 sin 0 = 1 3(1) =1

3.

83. Derivatives of Inverse Functions Suppose that f is a one-to-one differentiable function and its inverse function f⁻¹ is also differentiable. Use implicit differentiation to show that

(f⁻¹)⁰(x) = 1 f⁰(f⁻¹(x)) provided that the denominator is not 0.

Solution:

SECTION 3.6 DERIVATIVES OF LOGARITHMIC AND INVERSE TRIGONOMETRIC FUNCTIONS ¤ 247 80.  () = arctan(²− ) ⇒ ⁰() = 1

1 + (²− )² · 

(²− ) = 2 − 1 1 + (²− )²

81. Let  = cos⁻¹. Then cos  =  and 0 ≤  ≤  ⇒ − sin 

 = 1 ⇒



= − 1

sin  = − 1

1 − cos² = − 1

√1 − ². [Note that sin  ≥ 0 for 0 ≤  ≤ .]

∪

^3₂ . Differentiate with respect to : sec  tan 







= 1 ⇒



= 1

sec² − 1 = 1

√

sec² − 1

since tan   0 when 0   ^₂ or    ^3₂ .

(b)  = sec⁻¹ ⇒ sec  =  ⇒ sec  tan 

 = 1 ⇒ 

= 1

²− 1. For  ∈ 0^₂

,  ≥ 1, so sec  =  = || and tan  ≥ 0 ⇒



 = 1

√

²− 1 = 1

||√

²− 1. For  ∈ 2 

,  ≤ −1, so || = − and tan  = −√

²− 1 ⇒



= 1

sec  tan  = 1



−√

²− 1 = 1 (−)√

²− 1= 1

||√

²− 1.

83. If  = ⁻¹(), then () = . Differentiating implicitly with respect to  and remembering that  is a function of ,

we get ⁰()

 = 1, so

= 1

⁰() ⇒ 

⁻¹0

() = 1

⁰(⁻¹()).

84.  (4) = 5 ⇒ ⁻¹(5) = 4. By Exercise 83,

⁻¹₀

(5) = 1

⁰(⁻¹(5)) = 1

⁰(4) = 1 23= 3

2.

85.  () =  + ^ ⇒ ⁰() = 1 + ^. Observe that (0) = 1, so that ⁻¹(1) = 0. By Exercise 83, we have (⁻¹)⁰(1) = 1

⁰(⁻¹(1)) = 1

⁰(0) = 1

1 + ⁰ = 1 1 + 1=1

2.

86.  () = ³+ 3 sin  + 2 cos  ⇒ ⁰() = 3²+ 3 cos  − 2 sin . Observe that (0) = 2, so that ⁻¹(2) = 0.

⁰(⁻¹(2)) = 1

⁰(0) = 1

3(0)²+ 3 cos 0 − 2 sin 0 = 1 3(1)= 1

3.

1

(2)

85. Use the formula in Exercise 83.

If f (x) = x + e^x, find (f⁻¹)⁰(1).

Solution:

SECTION 3.6 DERIVATIVES OF LOGARITHMIC AND INVERSE TRIGONOMETRIC FUNCTIONS ¤ 247 80. () = arctan(²− ) ⇒ ⁰() = 1

1 + (²− )² · 

(²− ) = 2 − 1 1 + (²− )²

81.Let  = cos⁻¹. Then cos  =  and 0 ≤  ≤  ⇒ − sin 

 = 1 ⇒



 = − 1

sin  = − 1

1 − cos² = − 1

√1 − ². [Note that sin  ≥ 0 for 0 ≤  ≤ .]

∪

^3₂ 

. Differentiate with respect to : sec  tan 







= 1 ⇒



 = 1

sec² − 1 = 1

√

sec² − 1

since tan   0 when 0    ^₂ or     ^3₂ .

(b)  = sec⁻¹ ⇒ sec  =  ⇒ sec  tan 

= 1 ⇒ 

 = 1

²− 1. For  ∈ 0^₂

,  ≥ 1, so sec  =  = || and tan  ≥ 0 ⇒



 = 1

√

²− 1= 1

||√

²− 1. For  ∈_

2 

,  ≤ −1, so || = − and tan  = −√

²− 1 ⇒



 = 1

sec  tan  = 1



−√

²− 1 = 1 (−)√

²− 1= 1

||√

²− 1.

83.If  = ⁻¹(), then () = . Differentiating implicitly with respect to  and remembering that  is a function of ,

we get ⁰()

 = 1, so

 = 1

⁰() ⇒ 

⁻¹0() = 1

⁰(⁻¹()).

84. (4) = 5 ⇒ ⁻¹(5) = 4. By Exercise 83,

⁻¹0

(5) = 1

⁰(⁻¹(5))= 1

⁰(4)= 1 23=3

2.

85. () =  + ^ ⇒ ⁰() = 1 + ^. Observe that (0) = 1, so that ⁻¹(1) = 0. By Exercise 83, we have (⁻¹)⁰(1) = 1

⁰(⁻¹(1))= 1

⁰(0)= 1

1 + ⁰ = 1 1 + 1= 1

2.

86. () = ³+ 3 sin  + 2 cos  ⇒ ⁰() = 3²+ 3 cos  − 2 sin . Observe that (0) = 2, so that ⁻¹(2) = 0.

⁰(⁻¹(2))= 1

⁰(0)= 1

3(0)²+ 3 cos 0 − 2 sin 0 = 1 3(1)=1

3.

2