14.5 The Chain Rule

Section 14.5 The Chain Rule

3. Use the Chain Rule to find dz/dt. z = xy³− x²y, x = t²+ 1, y = t²− 1.

Solution:

432 ¤ CHAPTER 14 PARTIAL DERIVATIVES 4. The level curves of ( ) =

1 +  1 + 

13

− 1 are

1 +  1 + 

13

− 1 =  ⇒ 1 + 

1 +  = (1 + )³ ⇒

 = 1 +  (1 + )³ − 1.

From the level curves, we see that increasing  (from 0) by a small amount has a similar effect on the value of  as decreasing  by a small amount. However, for larger changes, a decrease in  gives greater values of  than a similar increase in .

14.5 The Chain Rule

1.  = ³− ²,  = ²+ 1,  = ²− 1 ⇒



 = 





 +





 = (³− 2)(2) + (3²− ²)(2) = 2(³− 2 + 3²− ²) 2.  =  − 

 + 2,  = ^,  = ^− ⇒



 = 





 +





 = ( + 2)(1) − ( − )(1)

( + 2)² (^) + ( + 2)(−1) − ( − )(2)

( + 2)² (−^−)

= 3

( + 2)²(^) + −3

( + 2)²(−^−) = 3

( + 2)²

^+ ^−

3.  = sin  cos ,  =√

,  = 1 ⇒



 = 





 +





 = (cos  cos )

1 2^−12

+ (− sin  sin )

−⁻²

= 1 2√

cos  cos  + 1

² sin  sin  4.  = √1 + ,  = tan ,  = arctan  ⇒



 = 





 +





 = ¹₂(1 + )^−12() · sec² +¹₂(1 + )^−12() · 1 1 + ²

= 1

2√ 1 + 



 sec² +  1 + ²



5.  = ^,  = ²,  = 1 − ,  = 1 + 2 ⇒



 = 





 +





 + 





 = ^· 2 + ^

1





· (−1) + ^

−

²

· 2 = ^

 2 − 

 −2

²



6.  = tan⁻¹(),  = ^,  = 1 − ^− ⇒



 = 





 +





 = 1

1 + ()²(−⁻²) · ^+ 1

1 + ()²(1) · (−^−)(−1)

= − 

²+ ² · ^+ 1

 + ² · ^−= ^−− ^

²+ ²

° 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

14. Use the Chain Rule to find ∂z/∂s and ∂z/∂t.

z =√

xe^xy, x = 1 + st, y = s²− t² Solution:

SECTION 14.5 THE CHAIN RULE ¤ 433

7.  = ( − )⁵,  = ²,  = ² ⇒



 = 





 +





 = 5( − )⁴(1) · 2 + 5( − )⁴(−1) · ²= 5( − )⁴

2 − ²



 = 





 +





 = 5( − )⁴(1) · ²+ 5( − )⁴(−1) · 2 = 5( − )⁴

²− 2

8.  = tan⁻¹(²+ ²),  =  ln ,  = ^ ⇒



= 





 +





 = 2

1 + (²+ ²)² · ln  + 2

1 + (²+ ²)² · ^

= 2

1 + (²+ ²)² ( ln  + ^)



 = 





 +





 = 2

1 + (²+ ²)² ·

 + 2

1 + (²+ ²)² · ^

= 2

1 + (²+ ²)²

 

 + ^ 9.  = ln(3 + 2),  =  sin ,  =  cos  ⇒



 = 





 +





 = 3

3 + 2(sin ) + 2

3 + 2(− sin ) = 3 sin  − 2 sin  3 + 2



 = 





 +





 = 3

3 + 2( cos ) + 2

3 + 2(cos ) = 3 cos  + 2 cos  3 + 2

10.  = √ ^,  = 1 + ,  = ²− ² ⇒



= 





 +





 =√

 · ^() + ^·¹2^−12

() +√

 ^() (2) =



√

 +  2√

+ 2^32



^



 = 





 +





 =√

 · ^() + ^·¹2^−12

() +√

 ^() (−2) =



√

 + 

2√− 2^32



^

11.  = ^cos ,  = ,  =√

²+ ² ⇒



 = 





+ 





 = ^cos  ·  + ^(− sin ) ·¹2(²+ ²)^−12(2) = ^cos  − ^sin  · 

√²+ ²

= ^



 cos  − 

√²+ ² sin 





 = 





 + 





 = ^cos  ·  + ^(− sin ) ·¹2(²+ ²)^−12(2) = ^cos  − ^sin  · 

√²+ ²

= ^



 cos  − 

√²+ ²sin 



12.  = arcsin( − ),  = ²+ ²,  = 1 − 2 ⇒



 = 





 +





 = 1

1 − ( − )²(1) · 2 + 1

1 − ( − )²(−1) · (−2) = 2 + 2

1 − ( − )²



 = 





 +





 = 1

1 − ( − )²(1) · 2 + 1

1 − ( − )²(−1) · (−2) = 2 + 2

1 − ( − )²

° 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

33. Use Equation 5 to find dy/dx. tan⁻¹(x²y) = x + xy² Solution:

SECTION 14.5 THE CHAIN RULE ¤ 437 When  = −1,  = 2, and  = 1 we have  = 2,  = 4, and  = −1, so 

 = ⁻⁴(−4 + 8) = 4⁻⁴,



 = ⁻⁴(1 − 8) = −7⁻⁴, and 

 = ⁻⁴(−8 − 16) = −24⁻⁴.

27.  cos  = ²+ ², so let  ( ) =  cos  − ²− ²= 0. Then by Equation 6



= −

 = −− sin  − 2

cos  − 2 = 2 +  sin  cos  − 2 .

28. cos() = 1 + sin , so let  ( ) = cos() − 1 − sin  = 0. Then by Equation 6



= −

 = − − sin()()

− sin()() − cos  = −  sin() cos  +  sin().

29. tan⁻¹(²) =  + ², so let  ( ) = tan⁻¹(²) −  − ² = 0. Then

( ) = 1

1 + (²)²(2) − 1 − ²= 2

1 + ⁴² − 1 − ²= 2 − (1 + ²)(1 + ⁴²) 1 + ⁴² ,

( ) = 1

1 + (²)²(²) − 2 = ²

1 + ⁴² − 2 =²− 2(1 + ⁴²) 1 + ⁴²

and 

= −

 = −[2 − (1 + ²)(1 + ⁴²)](1 + ⁴²)

[²− 2(1 + ⁴²)](1 + ⁴²) = (1 + ²)(1 + ⁴²) − 2

²− 2(1 + ⁴²)

= 1 + ⁴²+ ²+ ⁴⁴− 2

²− 2 − 2⁵³

30. ^sin  =  + , so let  ( ) = ^sin  −  −  = 0. Then 

 = −

 = −^cos  − 1 − 

^sin  −  = 1 +  − ^cos 

^sin  −  .

31. ²+ 2²+ 3²= 1, so let  (  ) = ²+ 2²+ 3²− 1 = 0. Then by Equations 7



= −

 = −2

6 = −

3 and 

 = −

 = −4

6 = −2

3.

32. ²− ²+ ²− 2 = 4, so let  (  ) = ²− ²+ ²− 2 − 4 = 0. Then by Equations 7



= −

 = − 2

2 − 2 = 

1 −  and 

 = −

 = − −2

2 − 2 = 

 − 1.

33. ^ = , so let  (  ) = ^−  = 0. Then 

 = −

 = − −

^−  = 

^−  and



 = −

 = − −

^−  = 

^− .

34.  +  ln  = ², so let  (  ) =  +  ln  − ² = 0. Then 

= −

 = − ln 

 − 2 = ln  2 −  and



 = −

 = − + ()

 − 2 =  + 

2 − ².

° 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

38. Use Equations 6 to find ∂z/∂x and ∂z/∂y. yz + x ln y = z² Solution:

SECTION 14.5 THE CHAIN RULE ¤ 437 When  = −1,  = 2, and  = 1 we have  = 2,  = 4, and  = −1, so 

 = ⁻⁴(−4 + 8) = 4⁻⁴,



 = ⁻⁴(1 − 8) = −7⁻⁴, and 

 = ⁻⁴(−8 − 16) = −24⁻⁴.

27.  cos  = ²+ ², so let  ( ) =  cos  − ²− ²= 0. Then by Equation 6



 = −

 = −− sin  − 2

cos  − 2 = 2 +  sin  cos  − 2 .

28. cos() = 1 + sin , so let  ( ) = cos() − 1 − sin  = 0. Then by Equation 6



 = −

 = − − sin()()

− sin()() − cos  = −  sin() cos  +  sin().

29. tan⁻¹(²) =  + ², so let  ( ) = tan⁻¹(²) −  − ²= 0. Then

( ) = 1

1 + (²)² (2) − 1 − ²= 2

1 + ⁴² − 1 − ²= 2 − (1 + ²)(1 + ⁴²) 1 + ⁴² ,

( ) = 1

1 + (²)² (²) − 2 = ²

1 + ⁴² − 2 = ²− 2(1 + ⁴²) 1 + ⁴²

and 

= −

 = −[2 − (1 + ²)(1 + ⁴²)](1 + ⁴²)

[²− 2(1 + ⁴²)](1 + ⁴²) = (1 + ²)(1 + ⁴²) − 2

²− 2(1 + ⁴²)

= 1 + ⁴²+ ²+ ⁴⁴− 2

²− 2 − 2⁵³

30. ^sin  =  + , so let  ( ) = ^sin  −  −  = 0. Then 

= −

 = −^cos  − 1 − 

^sin  −  = 1 +  − ^cos 

^sin  −  .

31. ²+ 2²+ 3²= 1, so let  (  ) = ²+ 2²+ 3²− 1 = 0. Then by Equations 7



 = −

 = −2

6 = −

3 and 

 = −

 = −4

6 = −2

3.

32. ²− ²+ ²− 2 = 4, so let  (  ) = ²− ²+ ²− 2 − 4 = 0. Then by Equations 7



 = −

 = − 2

2 − 2 = 

1 −  and 

 = −

 = − −2

2 − 2 = 

 − 1.

33. ^ = , so let  (  ) = ^−  = 0. Then 

 = −

 = − −

^−  = 

^−  and



 = −

 = − −

^−  = 

^− .

34.  +  ln  = ², so let  (  ) =  +  ln  − ²= 0. Then 

 = −

 = − ln 

 − 2 = ln  2 −  and



 = −

 = − + ()

 − 2 =  + 

2 − ².

° 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

49. Assume that all the given functions are differentiable. If z = f (x, y), where x = r cos θ and y = r sin θ, (a) find

∂z/∂r and ∂z/∂θ and (b) show that

 ∂z

∂x

² + ∂z

∂y

²

= ∂z

∂r

² + 1

r²

 ∂z

∂θ

²

1

Solution:

440 ¤ CHAPTER 14 PARTIAL DERIVATIVES 45. (a) By the Chain Rule, 

 = 

cos  + 

 sin , 

 = 

(− sin ) +

  cos .

(b)





2

=





2

cos² + 2





cos  sin  +





2

sin²,





2

=





2

²sin² − 2 





²cos  sin  +





2

²cos². Thus





2

+ 1

²





2

=





2

+





2

(cos² + sin²) =





2

+





2

.

46. 

 = 

+

 and 

 = 

− 

. Thus 





 =





2

−





2

.

47. Let  =  −  and  =  + . Then  = 1

[ () + ()]and



= 1









+









+ [ () + ()]



−1

²



= 1

[⁰()(1) + ⁰()(1)] − 1

² [ () + ()] = 1

[⁰() + ⁰()] − 1

² [ () + ()]



 = 1









 + 









= 1

[⁰()(−1) + ⁰()(1)] = 1

[−⁰() + ⁰()]

²

² = 1



 

[−⁰()]

 + 

[⁰()]





= 1

[−⁰⁰()(−1) + ⁰⁰()(1)] = 1

[⁰⁰() + ⁰⁰()]

Thus







²





= 

( [⁰() + ⁰()] − [() + ()])

=  [⁰⁰()(1) + ⁰⁰()(1)] + [⁰() + ⁰()] (1) − [⁰()(1) + ⁰()(1)]

=  [⁰⁰() + ⁰⁰()] + ⁰() + ⁰() − ⁰() − ⁰() =  [⁰⁰() + ⁰⁰()]

= ²· 1

[⁰⁰() + ⁰⁰()] = ²²

²

48. Let  =  +  and  =  − . Then  = 1

[ () + ()]and



= 1









+ 









= 1

[⁰()() + ⁰()()] = 

[⁰() + ⁰()]

²

² = 



 

[⁰()]

+ 

[⁰()]





= 

[⁰⁰()() + ⁰⁰()()] = ²

 [⁰⁰() + ⁰⁰()]



 = 1









 + 









+ [ () + ()]



−1

²



= 1

[⁰()(1) + ⁰()(−1)] − 1

²[ () + ()] = 1

[⁰() − ⁰()] − 1

²[ () + ()]

[continued]

° 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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