Section 14.5 The Chain Rule

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Section 14.5 The Chain Rule

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+ ²sin 





 =



+



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

√²+ ²

= ^



 cos  − 

√²+ ² sin 



12.  = tan(),  = 2 + 3,  = 3 − 2 ⇒



= 



+



 = sec²()(1) · 2 + sec²()(−⁻²) · 3

= 2

sec² 



−3

² sec² 



=2 − 3

² sec² 





26.  = ^,  = ²,  = ²,  = ² ⇒



 =



+



+



 = ^(2) + ^(0) + ^(²) = ^(2 + ²),



 =



+



 +



 = ^(²) + ^(2) + ^(0) = ^(²+ 2),



 =



 +



 +



 = ^(0) + ^(²) + ^(2) = ^(²+ 2).

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.

26.  = ^,  = ²,  = ²,  = ² ⇒



 =



+



+



 = ^(2) + ^(0) + ^(²) = ^(2 + ²),



 =



+



 +



 = ^(²) + ^(2) + ^(0) = ^(²+ 2),



 =



 +



 +



 = ^(0) + ^(²) + ^(2) = ^(²+ 2).

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.

438 ¤ CHAPTER 14 PARTIAL DERIVATIVES

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

 = −

 = − −

^−  = 

^−  and



 = −

 = − −

^−  = 

^− .

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

 = −

 = − ln 

 − 2 = ln  2 −  and



 = −

 = − + ()

 − 2 =  + 

2 − ².

35. Since  and  are each functions of ,  ( ) is a function of , so by the Chain Rule, 

 = 





 +





. After 3seconds,  =√

1 +  =√1 + 3 = 2,  = 2 +¹₃ = 2 + ¹₃(3) = 3, 

 = 1

2√

1 + = 1 2√

1 + 3 =1 4, and

 = 1 3. Then

 = (2 3)

 + (2 3)

 = 4₁

4

+ 3₁

3

= 2. Thus the temperature is rising at a rate of 2^◦Cs.

36. (a) Since  is negative, a rise in average temperature (while annual rainfall remains constant) causes a decrease in wheat production at the current production levels. Since  is positive, an increase in annual rainfall (while the average temperature remains constant) causes an increase in wheat production.

(b) Since the average temperature is rising at a rate of 015^◦Cyear, we know that  = 015. Since rainfall is decreasing at a rate of 01 cmyear, we know  = −01. Then, by the Chain Rule,



 =





 +





 = (−2)(015) + (8)(−01) = −11. Thus we estimate that wheat production will decrease at a rate of 11 unitsyear.

37.  = 14492 + 46 − 0055²+ 000029³+ 0016, so

 = 46 − 011 + 000087²and 

 = 0016.

According to the graph, the diver is experiencing a temperature of approximately 125^◦Cat  = 20 minutes, so



 = 46 − 011(125) + 000087(125)²≈ 336. By sketching tangent lines at  = 20 to the graphs given, we estimate



 ≈ 1 2and

 ≈ −1

10. Then, by the Chain Rule,

 =





 +





 ≈ (336)

−10¹

+ (0016)1 2

≈ −033.

Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 033 ms per minute.

38.  = ²3, so

 = 





 +





 = 2

3 18 +²

3 (−25) = 20,160 − 12,000 = 8160 in³s.

39. (a)  = , so by the Chain Rule,



 =





 +





 +





 = 

+ 

 + 

 = 2 · 2 · 2 + 1 · 2 · 2 + 1 · 2 · (−3) = 6 m³s.

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

 = −

 = − −

^−  = 

^−  and



 = −

 = − −

^−  = 

^− .

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

 = −

 = − ln 

 − 2 = ln  2 −  and



 = −

 = − + ()

 − 2 =  + 

2 − ².

35. Since  and  are each functions of ,  ( ) is a function of , so by the Chain Rule, 

 = 





 +





. After 3seconds,  =√

1 +  =√1 + 3 = 2,  = 2 +¹₃ = 2 + ¹₃(3) = 3, 

 = 1

2√

1 + = 1 2√

1 + 3 =1 4, and

 = 1 3. Then

 = (2 3)

 + (2 3)

 = 41 4

+ 31 3

= 2. Thus the temperature is rising at a rate of 2^◦Cs.

36. (a) Since  is negative, a rise in average temperature (while annual rainfall remains constant) causes a decrease in wheat production at the current production levels. Since  is positive, an increase in annual rainfall (while the average temperature remains constant) causes an increase in wheat production.

(b) Since the average temperature is rising at a rate of 015^◦Cyear, we know that  = 015. Since rainfall is decreasing at a rate of 01 cmyear, we know  = −01. Then, by the Chain Rule,



 =





 +





 = (−2)(015) + (8)(−01) = −11. Thus we estimate that wheat production will decrease at a rate of 11 unitsyear.

37.  = 14492 + 46 − 0055²+ 000029³+ 0016, so

 = 46 − 011 + 000087²and 

 = 0016.

According to the graph, the diver is experiencing a temperature of approximately 125^◦Cat  = 20 minutes, so



 = 46 − 011(125) + 000087(125)²≈ 336. By sketching tangent lines at  = 20 to the graphs given, we estimate



 ≈ 1 2and

 ≈ −1

10. Then, by the Chain Rule,

 =





 +





 ≈ (336)

−10¹

+ (0016)1 2

≈ −033.

Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 033 ms per minute.

38.  = ²3, so

 = 





 +





 = 2

3 18 +²

3 (−25) = 20,160 − 12,000 = 8160 in³s.

39. (a)  = , so by the Chain Rule,



 =





 +





 +





 = 

+ 

 + 

 = 2 · 2 · 2 + 1 · 2 · 2 + 1 · 2 · (−3) = 6 m³s.

(b)  = 2( +  + ), so by the Chain Rule,



 =





+





 +





 = 2( + )

+ 2( + )

 + 2( + )



= 2(2 + 2)2 + 2(1 + 2)2 + 2(1 + 2)(−3) = 10 m²s

(c) ²= ²+ ²+ ² ⇒ 2

 = 2

 + 2

 + 2

 = 2(1)(2) + 2(2)(2) + 2(2)(−3) = 0 ⇒

 = 0ms.

40.  =

 ⇒



 = 





 + 





 = 1





 − 

²



 = 1





 − 





 = 1

400(−001) −008

400(003) = −0000031 As

41. 

 = 005,

 = 015,  = 831

 and 

 = 831





 − 831

²



. Thus when  = 20 and  = 320,



 = 831

015

20 −(005)(320) 400



≈ −027 Ls.

42.  = 147^065^035and considering  , , and  as functions of time  we have



 =





 + 





 = 147(065)^−035^035

 + 147(035)^065^−065

 . We are given that

 = −2 and

 = 05, so when  = 30 and  = 8, the rate of change of production

 is 147(065)(30)^−035(8)^035(−2) + 147(035)(30)^065(8)^−065(05) ≈ −0596. Thus production at that time is decreasing at a rate of about $596,000 per year.

43. Let  be the length of the ﬁrst side of the triangle and  the length of the second side. The area  of the triangle is given by

 =¹₂ sin where  is the angle between the two sides. Thus  is a function of , , and , and , , and  are each in turn functions of time . We are given that

 = 3, 

 = −2, and because  is constant,

 = 0. By the Chain Rule,



 =





 +





 +





 ⇒ 

 = ¹₂ sin  ·

 +¹₂ sin  ·

 +¹₂ cos  ·

. When  = 20,  = 30, and  = 6 we have

0 =¹₂(30) sin^₆

(3) +¹₂(20) sin^₆

(−2) +¹2(20)(30)

cos^₆ 



= 45 · ¹2− 20 ·¹2 + 300 ·

√3 2 ·

 =²⁵₂ + 150√ 3



Solving for

 gives 

 = −252 150√

3 = − 1 12√

3, so the angle between the sides is decreasing at a rate of 1

12√ 3

≈ 0048 rads.

1

(2)

440 ¤ CHAPTER 14 PARTIAL DERIVATIVES 44. =

 + 

 − ^



=

332+34 332−40

460 ≈ 5766 Hz. ^and are functions of time , so



 = 





 +





 =

 1

 − ^



·

 +  + 

( − ^)²·



=

1 332−40



(460) (12) +₍₃₃₂³³²⁺³⁴

−40)²(460) (14) ≈ 465 Hzs 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. By the Chain Rule,

 = 

^cos  +

^sin , 

 = 

(−^sin ) +

^cos . Then





2

=





2

^2cos² + 2





^2cos  sin  +





2

^2sin²and





2

=





2

^2sin² − 2





^2cos  sin  +





2

^2sin². Thus





2

+





2

^−2=





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

Similarly

²

² = −^cos 

− ^sin  









− ^sin 

 + ^cos  









= −^cos 

− ^sin 



−^sin ²

² + ^cos  ²

 



−^sin 

 + ^cos 



^cos ²

² − ^sin  ²

 



= −^cos 

− ^sin 

 + ^2sin²²

² − 2^2cos  sin  ²

  + ^2cos²²

² Thus ^−2

²

² +²

²



= (cos² + sin²)

²

² +²

²



=²

² +²

², as desired.

51. 

= 

2 +

2. Then

²

 = 





2

 + 





2



=²

²



 2 + 









2 +





2 + ²

²



2 + 









2 +

2

= 4²

² + ²

 4²+ 0 + 4²

² + ²

 4²+ 2



By the continuity of the partials, ²

 = 4²

² + 4²

² + (4²+ 4²) ²

 + 2

. 52. By the Chain Rule,

(a) 

 =

cos  +

sin  (b) 

= −

 sin  +

 cos 

(c) ²

  = ²

  = 





cos  +

sin 



= − sin 

+ cos  









+ cos 

+ sin  









= − sin 

+ cos 

²

²



+ ²

 







+ cos 

+ sin ²

²



+ ²

 





= − sin 

+ cos 



− sin ²

² +  cos  ²

 



+ cos 

+ sin 



 cos ²

² −  sin  ²

 



= − sin 

−  cos  sin ²

² +  cos² ²

 + cos 

+  cos  sin ²

² −  sin² ²

 

= cos 

 − sin 

+  cos  sin 

²

² −²

²



+ (cos² − sin²) ²

 

53. 

 = 

cos  +

sin and 

= −

 sin  +

 cos . Then

²

² = cos 

²

²cos  + ²

 sin 

 + sin 

²

²sin  + ²

 cos 



= cos²²

² + 2 cos  sin  ²

  + sin²²

²

SECTION 14.5 THE CHAIN RULE ¤ 443 and

²

² = − cos 

+ (− sin )

²

²(− sin ) + ²

  cos 



− sin 

 +  cos 

²

² cos  + ²

 (− sin )



= − cos 

−  sin 

+ ²sin²²

² − 2²cos  sin  ²

 + ²cos²²

² Thus

²

² + 1

²

²

² +1





 = (cos² + sin²)²

² +

sin² + cos² ²

²

−1

cos 

−1

sin 

+1



 cos  

+ sin 





= ²

² +²

² as desired.

54. (a) 

 = 



 +



. Then

²

² = 









 + 











= 









 +²

²



+ 









 +²

²





= ²

²





2

+ ²

 







 +²

²



+²

²





2

+ ²

 







 +²

²





= ²

²





2

+ 2 ²

 







 +²

²





2

+²

²



+²

²





(b) ²

 = 







 +







=

²

²



+ ²

 







 +



²

 +

²

²



+ ²

 







 +



²

 

= ²

²







 + ²

 







 +







 +



²

 +



²

 +²

²









55. (a) Since  is a polynomial, it has continuous second-order partial derivatives, and

 ( ) = ()²() + 2()()²+ 5()³= ³² + 2³²+ 5³³= ³(² + 2²+ 5³) = ³ ( ).

Thus,  is homogeneous of degree 3.

(b) Differentiating both sides of ( ) = ^ ( )with respect to  using the Chain Rule, we get



 ( ) = 

[^ ( )] ⇔



() ( ) ·()

 + 

() ( ) ·()

 =  

() ( ) +  

() ( ) = ^⁻¹ ( ).

Setting  = 1:  

 ( ) +  

 ( ) =  ( ).

SECTION 14.5 THE CHAIN RULE ¤ 443 and

²

² = − cos 

+ (− sin )

²

²(− sin ) + ²

  cos 



− sin 

 +  cos 

²

² cos  + ²

 (− sin )



= − cos 

−  sin 

+ ²sin²²

² − 2²cos  sin  ²

 + ²cos²²

² Thus

²

² + 1

²

²

² +1





 = (cos² + sin²)²

² +

sin² + cos² ²

²

−1

cos 

−1

sin 

+1



 cos  

+ sin 





= ²

² +²

² as desired.

54. (a) 

 = 



 +



. Then

²

² = 









 + 











= 









 +²

²



+ 









 +²

²





= ²

²





2

+ ²

 







 +²

²



+²

²





2

+ ²

 







 +²

²





= ²

²





2

+ 2 ²

 







 +²

²





2

+²

²



+²

²





(b) ²

 = 







 +







=

²

²



+ ²

 







 +



²

 +

²

²



+ ²

 







 +



²

 

= ²

²







 + ²

 







 +







 +



²

 +



²

 +²

²









55. (a) Since  is a polynomial, it has continuous second-order partial derivatives, and

 ( ) = ()²() + 2()()²+ 5()³= ³² + 2³²+ 5³³= ³(² + 2²+ 5³) = ³ ( ).

Thus,  is homogeneous of degree 3.

(b) Differentiating both sides of ( ) = ^ ( )with respect to  using the Chain Rule, we get



 ( ) = 

[^ ( )] ⇔



() ( ) ·()

 + 

() ( ) ·()

 =  

() ( ) +  

() ( ) = ^⁻¹ ( ).

Setting  = 1:  

 ( ) +  

 ( ) =  ( ).

56. Differentiating both sides of ( ) = ^ ( )with respect to  using the Chain Rule, we get



() ( ) · ()

 + 

() ( ) ·()

 =  

() ( ) +  

() ( ) = ^⁻¹ ( )and differentiating again with respect to  gives



 ²

 ()² ( ) · ()

 + ²

 ()  () ( ) · ()





+ 

 ²

 ()  () ( ) · ()

 + ²

 ()²  ( ) · ()





= ( − 1)^⁻¹ ( ).

Setting  = 1 and using the fact that = , we have ²+ 2+ ²= ( − 1)( ).

57. Differentiating both sides of ( ) = ^ ( )with respect to  using the Chain Rule, we get



 ( ) = 

[^ ( )] ⇔



 () ( ) · ()

 + 

 () ( ) · ()

 = ^ 

 ( ) ⇔ ^( ) = ^( ).

Thus ( ) = ^−1( ).

58.  (  ) = 0is assumed to deﬁne  as a function of  and , that is,  = ( ). So by (7),

= −



since 6= 0.

Similarly, it is assumed that  (  ) = 0 deﬁnes  as a function of  and , that is  = ( ). Then  (( )  ) = 0 and by the Chain Rule, 



 + 



 + 



 = 0. But 

= 0and 

 = 1, so 



+ = 0 ⇒ 

 = −



.

A similar calculation shows that

 = −



. Thus





 =



−





−





−





= −1.

59. Given a function deﬁned implicitly by  ( ) = 0, where  is differentiable and 6= 0, we know that

 = −



. Let

( ) = −



so 

= ( ). Differentiating both sides with respect to  and using the Chain Rule gives

²

² =





+





where

 = 





−





= −− ^

_² , 

 = 





−





= −− ^

_² . Thus

²

²=



−− ^

_²

 (1) +



−− ^

_²

 

−





= −_²− ^− ^+ _²

³

But  has continuous second derivatives, so by Clauraut’s Theorem, = and we have

²

² = −_²− 2^+ _²

_³ as desired.