Section 14.3 Partial Derivatives 32. Find the first partial derivatives of the function.

Section 14.3 Partial Derivatives

32. Find the first partial derivatives of the function. u = x^yz. Solution:

408 ¤ CHAPTER 14 PARTIAL DERIVATIVES 26. ( ) = arctan

√



⇒ ^( ) = 1 1 +

√

² ·√

 =

√ 1 + ²,

( ) = 1 1 +

√

² · 

1 2^−12



= 

2√

 (1 + ²)

27.  = sin  cos  ⇒ 

 = cos  cos , 

 = − sin  sin 

28. ( ) = ^ ⇒ ^( ) = ^−1, ( ) = ^ln 

29.  ( ) =

 



cos(^)  ⇒ ^( ) = 



 



cos

^

 = cos(^)by the Fundamental Theorem of Calculus, Part 1;

( ) = 



 



cos

^

 = 





−

 



cos

^





= − 



 



cos

^

 = − cos(^).

30.  ( ) =

 



³+ 1  ⇒

( ) = 



 



³+ 1  = 





−

 



³+ 1 



= − 



 



³+ 1  = −

³+ 1by the Fundamental

Theorem of Calculus, Part 1; ( ) = 



 



³+ 1  =



³+ 1.

31. (  ) = ³²+ 2 ⇒ ^(  ) = 3²², (  ) = ³²+ 2, (  ) = 2³ + 2

32. (  ) = ²^− ⇒ (  ) = ²

 · ^−(−) + ^−· 1

= (1 − )²^−, (  ) = 2^−,

(  ) = ²^−(−) = −²²^−

33.  = ln( + 2 + 3) ⇒ 

 = 1

 + 2 + 3, 

 = 2

 + 2 + 3, 

 = 3

 + 2 + 3

34.  =  tan( + 2) ⇒ 

 =  [sec²( + 2)](1) =  sec²( + 2), 

 = tan( + 2),



 =  [sec²( + 2)](2) = 2 sec²( + 2)

35.  =  sin⁻¹() ⇒ 

=  sin⁻¹(), 

 =  · 1

1 − ()²() + sin⁻¹() ·  = 

1 − ²²+  sin⁻¹(),



 =  · 1

1 − ()²() = ²

1 − ²²

36.  = ^ ⇒ ^= 

^{()−1}, = ^ln  ·1

 = ^

 ln , = ^ln  · −

² = −^

² ln 

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39. Find the indicated partial derivative. f (x, y, z) = ln¹⁻

√

x²+y²+z² 1+√

x²+y²+z²; f_y(1, 2, 2).

Solution:

SECTION 14.3 PARTIAL DERIVATIVES ¤ 409

37. (   ) = ² cos() ⇒ ^(   ) = 2 cos(), (   ) = ²cos(),

(   ) = −² sin()(1) = (−²) sin(), (   ) = −² sin()(−⁻²) = (²²) sin()

38. (   ) = + ²

 + ² ⇒ (   ) = 1

 + ²() = 

 + ²,

_(   ) = 1

 + ²(2) = 2

 + ², _(   ) = ( + ²)(0) − ( + ²)()

( + ²)² = −( + ²) ( + ²)² ,

_(   ) = ( + ²)(0) − ( + ²)(2)

( + ²)² = −2( + ²) ( + ²)²

39.  =

²1+ ²2+ · · · + ^2. For each  = 1,   , , _ = ¹₂

²1+ ²2+ · · · + ^2−12

(2) = 

²₁+ ²₂+ · · · + ^2.

40.  = sin(1+ 22+ · · · + ^). For each  = 1,   , , _ =  cos(1+ 22+ · · · + ^).

41. ( ) = ^ ⇒ ^( ) =  · ^(−²) + ^· 1 = 1 −





^, so (0 1) =

 1 −0

1



^01 = 1.

42. ( ) =  sin⁻¹() ⇒ ^( ) =  · 1

1 − ()²() + sin⁻¹() · 1 = 

1 − ²² + sin⁻¹(),

so  1¹₂

= 1 · ¹2



1 − 1²1 2

2 + sin⁻¹ 1 · ¹2

=

1

2 3 4

+ sin^{−1 1}₂ = √¹ 3 +^₆.

43. (  ) = ln1 −

²+ ²+ ² 1 +

²+ ²+ ² ⇒

(  ) = 1 1 −

²+ ²+ ² 1 +

²+ ²+ ²

·

1 +

²+ ²+ ²

−¹2(²+ ²+ ²)^−12· 2

− 1 −

²+ ²+ ²

1

2(²+ ²+ ²)^−12· 2

 1 +

²+ ²+ ²2

= 1 +

²+ ²+ ² 1 −

²+ ²+ ² ·−(²+ ²+ ²)^−12 1 +

²+ ²+ ²+ 1 −

²+ ²+ ²

 1 +

²+ ²+ ²2

= −(²+ ²+ ²)^−12(2)

1 −

²+ ²+ ²

1 +

²+ ²+ ² =  −2

²+ ²+ ²[1 − (²+ ²+ ²)]

so (1 2 2) = √ −2(2)

1²+ 2²+ 2²[1 − (1²+ 2²+ 2²)] = √ −4

9 (1 − 9) = 1 6.

44. (  ) = ^ ⇒ ^(  ) = (^ln )() = ^ln , so ( 1 0) = 1⁽¹⁾⁽⁰⁾ln  = 1.

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44. Use implicit differentiation to find ∂z/∂x and ∂z/∂y.

yz + x ln y = z² Solution:

SECTION 14.3 PARTIAL DERIVATIVES ¤ 411



 (^) = 

() ⇒ ^ 

 = 





+  · 1



⇒ ^ 

 − 

 =  ⇒ (^− )

 = , so



 = 

^− .

50.  +  ln  = ² ⇒ 

( +  ln ) = 

(²) ⇒ 

+ ln  = 2

 ⇒ ln  = 2 

− 

 ⇒

ln  = (2 − )

, so

= ln  2 − .



 ( +  ln ) = 

 (²) ⇒ 

 +  · 1 +  · 1

 = 2 

 ⇒  + 

 = 2

 − 

 ⇒

 + 

 = (2 − )

, so 

 =  + ()

2 −  =  + 

(2 − ).

51. (a)  = () + () ⇒ 

 = ⁰(), 

 = ⁰() (b)  = ( + ). Let  =  + . Then 

 = 





 = 

(1) = ⁰() = ⁰( + ),



 = 





 = 

(1) = ⁰() = ⁰( + ).

52. (a)  = ()() ⇒ 

 = ⁰()(), 

 =  ()⁰() (b)  = (). Let  = . Then 

 = and 

 = . Hence 

= 





 = 

·  = ⁰() = ⁰() and 

 = 





 = 

·  = ⁰() = ⁰().

(c)  = 







. Let  = 

. Then

 = 1

 and 

 = −

². Hence 

 = 





 = ⁰()1

 = ⁰()

 and 

 = 





 = ⁰()



−

²



= −⁰()

² .

53. ( ) = ⁴ − 2³² ⇒ ( ) = 4³ − 6²², ( ) = ⁴− 4³. Then ( ) = 12² − 12²,

( ) = 4³− 12², ( ) = 4³− 12², and ( ) = −4³.

54. ( ) = ln( + ) ⇒ ( ) = 

 +  = ( + )⁻¹, ( ) = 

 +  = ( + )⁻¹. Then

( ) = −( + )⁻²() = − ²

( + )², ( ) = −( + )⁻²() = − 

( + )²,

( ) = −( + )⁻²() = − 

( + )², and ( ) = −( + )⁻²() = − ² ( + )².

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62. Find the indicated partial derivative(s). V = ln(r + s²+ t³); _∂r∂s∂t^∂3V Solution:

1

SECTION 14.3 PARTIAL DERIVATIVES ¤ 413

60.  = ^sin  ⇒ ^= ^sin ,  = ^cos  + (sin )( · ^+ ^· 1) = ^( cos  +  sin  + sin ),

 = ^cos  + (sin )(^) = ^(cos  +  sin ),

= ^· sin  + (cos  +  sin ) · ^= ^(sin  +  cos  +  sin ). Thus  = . 61.  = cos(²) ⇒ ^ = − sin(²) · 2 = −2 sin(²),

 = −2 · cos(²) · ²+ sin(²) · (−2) = −2³ cos(²) − 2 sin(²)and

 = − sin(²) · ²= −²sin(²), = −²· cos(²) · 2 + sin(²) · (−2) = −2³ cos(²) − 2 sin(²).

Thus = .

62.  = ln( + 2) ⇒ ^= 1

 + 2 = ( + 2)⁻¹,  = (−1)( + 2)⁻²(2) = − 2

( + 2)² and

 = 1

 + 2 · 2 = 2( + 2)⁻¹, = (−2)( + 2)⁻²= − 2

( + 2)². Thus  = . 63. ( ) = ⁴²− ³ ⇒ ^= 4³²− 3², = 12²²− 6, ^= 24²− 6 and

= 8³ − 3², = 24² − 6.

64. ( ) = sin(2 + 5) ⇒ ^ = cos(2 + 5) · 5 = 5 cos(2 + 5), ^= −5 sin(2 + 5) · 2 = −10 sin(2 + 5),

= −10 cos(2 + 5) · 5 = −50 cos(2 + 5)

65. (  ) = ^2 ⇒ ^= ^2· ²= ²^2, = ²· ^2(²) + ^2· ²= (⁴+ ²)^2,

 = (⁴+ ²) · ^2(2) + ^2· (4³+ 2) = (2²²⁵+ 6³+ 2)^2. 66. (  ) = ^sin() ⇒ ^= ^sin(), = ^cos() ·  = ^cos(),

= ^(− sin() · ) + cos() · ^ = ^[cos() −  sin()].

67.  =√

 + ² ⇒ 

 = ¹₂( + ²)^−12(2) = ( + ²)^−12,

²

  = 

−¹2

( + ²)^−32(1) = −¹2( + ²)^−32, ³

² = −¹2

−³2

( + ²)^−52(1) = ³₄( + ²)^−52.

68.  = ln( + ²+ ³) ⇒ 

 = 3²

 + ²+ ³ = 3²( + ²+ ³)⁻¹,

²

  = 3²(−1)( + ²+ ³)⁻²(2) = −6²( + ²+ ³)⁻²,

³

   = −6²(−2)( + ²+ ³)⁻³(1) = 12²( + ²+ ³)⁻³= 12² ( + ²+ ³)³.

69.  = ^sin  ⇒ 

 = ^cos  + sin  · ^() = ^(cos  +  sin ),

²

  = ^(sin ) + (cos  +  sin ) ^() = ^(sin  +  cos  +  sin ),

³

² = ^( sin ) + (sin  +  cos  +  sin ) · ^ () = ^(2 sin  +  cos  +  sin ).

° 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.c

67. If f (x, y, z) = xy²z³+ arcsin(x√

z), find fxzy. [Hint:Which order of differentiation is easiest?]

Solution:

414 ¤ CHAPTER 14 PARTIAL DERIVATIVES 70.  = √

 −  = ( − )^12 ⇒ 

 = 

1

2( − )^−12(−1)

= −¹2( − )^−12,

²

  = −¹2

−¹2( − )^−32(1)

= ¹₄( − )^−32, ³

   = ¹₄( − )^−32.

71. Assuming that the third partial derivatives of  are continuous (easily verified), we can write  = . Then

 (  ) = ²³+ arcsin

√



⇒ ^ = 2³+ 0, = 2³, and  = 6²= .

72. Let (  ) =√

1 + and (  ) = √1 −  so that  =  + . Then ^ = 0 = = and

= 0 = = . But (since the partial derivatives are continous on their domains) = and  = , so

 = + = 0 + 0 = 0.

73. By Definition 4, (3 2) = lim

→0

 (3 +  2) − (3 2)

 which we can approximate by considering  = 05 and  = −05:

(3 2) ≈  (35 2) − (3 2)

05 = 224 − 175

05 = 98, (3 2) ≈  (25 2) − (3 2)

−05 = 102 − 175

−05 = 146. Averaging these values, we estimate (3 2)to be approximately 122. Similarly, (3 22) = lim

→0

 (3 +  22) − (3 22)

 which

we can approximate by considering  = 05 and  = −05: ^(3 22) ≈  (35 22) − (3 22)

05 = 261 − 159

05 = 204,

(3 22) ≈  (25 22) − (3 22)

−05 = 93 − 159

−05 = 132. Averaging these values, we have (3 22) ≈ 168.

To estimate (3 2), we first need an estimate for (3 18):

(3 18) ≈  (35 18) − (3 18)

05 = 200 − 181

05 = 38, (3 18) ≈  (25 18) − (3 18)

−05 = 125 − 181

−05 = 112.

Averaging these values, we get (3 18) ≈ 75. Now ^( ) = 

[( )]and ( )is itself a function of two variables, so Definition 4 says that ( ) = 

 [( )] = lim

→0

(  + ) − ^( )

 ⇒

(3 2) = lim

→0

(3 2 + ) − ^(3 2)

 . We can estimate this value using our previous work with  = 02 and  = −02:

(3 2) ≈ (3 22) − ^(3 2)

02 = 168 − 122

02 = 23, (3 2) ≈ (3 18) − ^(3 2)

−02 = 75 − 122

−02 = 235.

Averaging these values, we estimate (3 2)to be approximately 2325.

74. (a) If we fix  and allow  to vary, the level curves indicate that the value of  decreases as we move through  in the positive

-direction, so is negative at  .

(b) If we fix  and allow  to vary, the level curves indicate that the value of  increases as we move through  in the positive

-direction, so is positive at  .

° 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.c

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