Section 14.3 Partial Derivatives

Section 14.3 Partial Derivatives

SECTION 14.3 PARTIAL DERIVATIVES ¤ 405

5. (a) If we start at (1 2) and move in the positive -direction, the graph of  increases. Thus (1 2)is positive.

(b) If we start at (1 2) and move in the positive -direction, the graph of  decreases. Thus (1 2)is negative.

6. (a) The graph of  decreases if we start at (−1 2) and move in the positive -direction, so ^(−1 2) is negative.

(b) The graph of  decreases if we start at (−1 2) and move in the positive -direction, so ^(−1 2) is negative.

7. (a) = _^(), so is the rate of change of in the -direction. is negative at (−1 2) and if we move in the positive -direction, the surface becomes less steep. Thus the values of are increasing and (−1 2) is positive.

(b) is the rate of change of in the -direction. is negative at (−1 2) and if we move in the positive -direction, the surface becomes steeper. Thus the values of are decreasing, and (−1 2) is negative.

8. (a) =_^ (), so is the rate of change of in the -direction. is positive at (1 2) and if we move in the positive

-direction, the surface becomes steeper, looking in the positive -direction. Thus the values of are increasing and

(1 2)is positive.

(b) is negative at (−1 2) and if we move in the positive -direction, the surface gets steeper (with negative slope), looking in the positive -direction. This means that the values of are decreasing as  increases, so (−1 2) is negative.

9.First of all, if we start at the point (3 −3) and move in the positive -direction, we see that both  and  decrease, while  increases. Both  and  have a low point at about (3 −15), while  is 0 at this point. So  is definitely the graph of ^, and one of  and  is the graph of . To see which is which, we start at the point (−3 −15) and move in the positive -direction.

traces out a line with negative slope, while  traces out a parabola opening downward. This tells us that  is the -derivative of . So  is the graph of ,  is the graph of , and  is the graph of .

10.(2 1)is the rate of change of  at (2 1) in the -direction. If we start at (2 1), where (2 1) = 10, and move in the positive -direction, we reach the next contour line [where ( ) = 12] after approximately 06 units. This represents an average rate of change of about_06² . If we approach the point (2 1) from the left (moving in the positive -direction) the output values increase from 8 to 10 with an increase in  of approximately 09 units, corresponding to an average rate of change of_09² . A good estimate for (2 1)would be the average of these two, so (2 1) ≈ 28. Similarly, ^(2 1)is the rate of change of  at (2 1) in the -direction. If we approach (2 1) from below, the output values decrease from 12 to 10 with a change in  of approximately 1 unit, corresponding to an average rate of change of −2. If we start at (2 1) and move in the positive -direction, the output values decrease from 10 to 8 after approximately 0.9 units, a rate of change of⁻²_09. Averaging these two results, we estimate (2 1) ≈ −21.

11.  ( ) = 16 − 4²− ² ⇒ ^( ) = −8 and ^( ) = −2 ⇒ ^(1 2) = −8 and ^(1 2) = −4. The graph of  is the paraboloid  = 16 − 4²− ²and the vertical plane  = 2 intersects it in the parabola  = 12 − 4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

408 ¤ CHAPTER 14 PARTIAL DERIVATIVES 24.  = ^

 + ² ⇒ 

 =0( + ²) − ^(1)

( + ²)² = − ^

( + ²)², 

 =^( + ²) − ^(2)

( + ²)² =^( + ²− 2) ( + ²)² 25. ( ) = (² − ³)⁵ ⇒ ^( ) = 5(² − ³)⁴· 2 = 10(² − ³)⁴,

( ) = 5(² − ³)⁴(²− 3²) = 5(²− 3²)(² − ³)⁴

26. ( ) = sin( cos ) ⇒ ^( ) = cos( cos ) · cos  = cos  cos( cos ),

( ) = cos( cos )(− sin ) = − sin  cos( cos )

27. ( ) = tan⁻¹(²) ⇒ ^( ) = 1

1 + (²)² · ²= ²

1 + ²⁴, ( ) = 1

1 + (²)² · 2 = 2

1 + ²⁴ 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.  =√

⁴+ ²cos  ⇒ 

 = ¹₂(⁴+ ²cos )^−12(4³) = 2³

√⁴+ ²cos ,



= ¹₂(⁴+ ²cos )^−12(2 cos ) =  cos 

√⁴+ ²cos , 

 =¹₂(⁴+ ²cos )^−12[²(− sin )] = − ²sin  2√

⁴+ ²cos 

36.  = ^ ⇒ ^=

 ^{()−1}, = ^ln  ·1

 =^

 ln , = ^ln  ·−

² = −^

² ln 

° 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

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.

° 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

410 ¤ CHAPTER 14 PARTIAL DERIVATIVES

45.  ( ) = ²− ³ ⇒

( ) = lim

→0

 ( +  ) − ( )

 = lim

→0

( + )²− ( + )³ − (²− ³)



= lim

→0

(²− 3² − 3 − ²)

 = lim

→0(²− 3² − 3 − ²) = ²− 3²

( ) = lim

→0

 (  + ) − ( )

 = lim

→0

( + )²− ³( + ) − (²− ³)

 = lim

→0

(2 +  − ³)



= lim

→0(2 +  − ³) = 2 − ³ 46.  ( ) = 

 + ² ⇒

( ) = lim

→0

 ( +  ) − ( )

 = lim

→0

+

++² −+^2

 ·( +  + ²)( + ²) ( +  + ²)( + ²)

= lim

→0

( + )( + ²) − ( +  + ²)

( +  + ²)( + ²) = lim

→0

²

( +  + ²)( + ²)

= lim

→0

²

( +  + ²)( + ²) = ² ( + ²)²

( ) = lim

→0

 (  + ) − ( )

 = lim

→0



+(+)² −+^2

 ·

 + ( + )² 

 + ²

 + ( + )²

( + ²)

= lim

→0

( + ²) − 

 + ( + )²

[ + ( + )²]( + ²) = lim

→0

(−2 − )

[ + ( + )²]( + ²)

= lim

→0

−2 − 

[ + ( + )²]( + ²) = −2

( + ²)²

47. ²+ 2²+ 3²= 1 ⇒ 

(²+ 2²+ 3²) = 

(1) ⇒ 2 + 0 + 6

 = 0 ⇒ 6

 = −2 ⇒



= −2

6 = − 3, and 

(²+ 2²+ 3²) = 

(1) ⇒ 0 + 4 + 6

 = 0 ⇒ 6

 = −4 ⇒



 = −4

6 = −2

3.

48. ²− ²+ ²− 2 = 4 ⇒ 

(²− ²+ ²− 2) = 

(4) ⇒ 2 − 0 + 2 

− 2

 = 0 ⇒ (2 − 2)

 = −2 ⇒ 

 = −2

2 − 2 = 

1 − , and 

(²− ²+ ²− 2) = 

(4) ⇒ 0 − 2 + 2

− 2

 = 0 ⇒ (2 − 2)

 = 2 ⇒ 

 = 2

2 − 2= 

 − 1.

49. ^ =  ⇒ 

(^) = 

() ⇒ ^ 

= 





+  · 1



⇒ ^ 

− 

 =  ⇒ (^− )

= , so

 = 

^− .

° 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

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.

° 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

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.

° 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

1

414 ¤ CHAPTER 14 PARTIAL DERIVATIVES

70.  = ^^^. If  = 0, or if  = 0 or 1, or if  = 0, 1, or 2, then ⁶

 ²³ = 0. Otherwise

 = ^^^−1,

²

² = ( − 1)^^^−2, ³

³ = ( − 1)( − 2)^^^−3, ⁴

 ³ = ( − 1)( − 2)^^−1^−3,

⁵

²³ = ( − 1)( − 1)( − 2)^^−2^−3, and ⁶

 ²³ = ( − 1)( − 1)( − 2)^−1^−2^−3. 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

416 ¤ CHAPTER 14 PARTIAL DERIVATIVES By symmetry, = 2²− ²− ²

(²+ ²+ ²)^52 and = 2²− ²− ² (²+ ²+ ²)^52. Thus + + =2²− ²− ²+ 2²− ²− ²+ 2²− ²− ²

(²+ ²+ ²)^52 = 0.

78. (a)  = sin() sin() ⇒ ^=  sin() cos(), = −²²sin() sin(), =  cos() sin(),

= −²sin() sin(). Thus = ². (b)  = 

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²− ²)(2²)

(²²− ²)⁴ = 2⁴³+ 6²² (²²− ²)³ ,

 = (−1)(²²− ²)⁻²(−2) = 2

(²²− ²)²,

= 2(²²− ²)²− 2 (2)(²²− ²)(−2)

(²²− ²)⁴ = 2²³− 2²+ 8²

(²²− ²)³ =2²³+ 6² (²²− ²)³. Thus = ².

(c)  = ( − )⁶+ ( + )⁶ ⇒ ^= −6( − )⁵+ 6( + )⁵, = 30²( − )⁴+ 30²( + )⁴,

= 6( − )⁵+ 6( + )⁵, = 30( − )⁴+ 30( + )⁴. Thus = ². (d)  = sin( − ) + ln( + ) ⇒ ^= − cos( − ) + 

 + , = −²sin( − ) − ² ( + )²,

= cos( − ) + 1

 + , = − sin( − ) − 1

( + )². Thus = ². 79. Let  =  + ,  =  − . Then ^= [ () + ()]

 = ()





 +()





 = ⁰() − ⁰()and

= [⁰() − ⁰()]

 = [⁰⁰() + ⁰⁰()] = ²[⁰⁰() + ⁰⁰()]. Similarly, by using the Chain Rule we have

= ⁰() + ⁰()and = ⁰⁰() + ⁰⁰(). Thus = ².

80. For each ,  = 1     , = ^{11+22+···+} and ²² = ²^{11+22+···+}.

Then²

²₁ +²

²₂ + · · · + ²

²_ =

²₁+ ²2+ · · · + ²

^{11+22+···+}= ^{11+22+···+}= 

since ²1+ ²2+ · · · + ²= 1.

81. ( ) = 1

√4^{−2(4)} ⇒



= 1

√4· ^{−2(4)}

−²(−1)(4)⁻²(4)

+ ^{−2(4)}·

−¹2

(4)^−32(4)

= (4)^−32



4 · ²

4² − 2



^{−2(4)}= 2

(4)^32

 ² 2− 1



^{−2(4)},



= 1

√4^{−2(4)}·−2

4 = −2

(4)^32^{−2(4)}, and

° 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

SECTION 14.3 PARTIAL DERIVATIVES ¤ 417

²

² = −2

(4)^32



 · ^{−2(4)}·−2

4+ ^{−2(4)}· 1



= −2

(4)^32



− ² 2+ 1



^{−2(4)}= 2

(4)^32

 ² 2− 1



^{−2(4)}.

Thus

 = 2

(4)^32

 ² 2− 1



^{−2(4)}= 

 2

(4)^32

 ² 2− 1



^{−2(4)}



=  ²

².

82. (a)  = −60(2)(1 + ²+ ²)², so at (2 1), = −240(1 + 4 + 1)²= −²⁰3.

(b)  = −60(2)(1 + ²+ ²)², so at (2 1), = −12036 = −¹⁰3. Thus from the point (2 1) the temperature is decreasing at a rate of ²⁰₃^◦Cm in the -direction and is decreasing at a rate of¹⁰₃^◦Cm in the -direction.

83.By the Chain Rule, taking the partial derivative of both sides with respect to 1gives

⁻¹





1

= [(11) + (12) + (13)]

1 or −⁻² 

1 = −⁻²1 . Thus 

1

= ²

²₁.

84. = ^^, so

 = ^−1^and 

 = ^^−1. Then



+  

 = (^−1^) + (^^−1) = ^1+−1^+ ^^1+−1= ( + )^^= ( + )

85.If we fix  = 0  ( 0)is a function of a single variable , and

 = 

 is a separable differential equation. Then



 = 

 ⇒

 

 =





 ⇒ ln | | =  ln || +  (⁰), where (0)can depend on 0. Then

| | = ^{ ln|| + (0)}, and since   0 and   0, we have  = ^{ ln }^(0)= ^(0)^{ln }= 1(0)^where

1(0) = ^(0).

86. (a)  ( ) = 101^075^025 ⇒ ( ) = 101(075^−025)^025= 07575^−025^025and

( ) = 101^075(025^−075) = 02525^075^−075.

(b) The marginal productivity of labor in 1920 is (194 407) = 07575(194)^−025(407)^025≈ 0912. Recall that  , , and

are expressed as percentages of the respective amounts in 1899, so this means that in 1920, if the amount of labor is increased, production increases at a rate of about 0.912 percentage points per percentage point increase in labor. The marginal productivity of capital in 1920 is (194 407) = 02525(194)^075(407)^−075≈ 0145, so an increase in capital investment would cause production to increase at a rate of about 0.145 percentage points per percentage point increase in capital.

(c) The value of (194 407)is greater than the value of (194 407), suggesting that increasing labor in 1920 would have increased production more than increasing capital.

° 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

2

SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 421

(c) Since  differentiations are to be performed with three choices of variable at each differentiation, there are 3^th-order partial derivatives of a function of three variables.

103. Let () = ( 0) = (²)^−32⁰=  ||⁻³. But we are using the point (1 0), so near (1 0), () = ⁻². Then

⁰() = −2⁻³and ⁰(1) = −2, so using (1) we have ^(1 0) = ⁰(1) = −2.

104. (0 0) = lim

→0

 (0 +  0) − (0 0)

 = lim

→0

(³+ 0)^13− 0

 = lim

→0



= 1.

Or: Let () = ( 0) =√³

³+ 0 = . Then ⁰() = 1and ⁰(0) = 1so, by (1), (0 0) = ⁰(0) = 1.

105. (a) (b) For ( ) 6= (0 0),

( ) =(3² − ³)(²+ ²) − (³ − ³)(2) (²+ ²)²

=⁴ + 4²³− ⁵ (²+ ²)²

and by symmetry ( ) =⁵− 4³²− ⁴ (²+ ²)² .

(c) (0 0) = lim

→0

 ( 0) − (0 0)

 = lim

→0

(0²) − 0

 = 0and (0 0) = lim

→0

 (0 ) − (0 0)

 = 0.

(d) By (3), (0 0) =

 = lim

→0

(0 ) − ^(0 0)

 = lim

→0

(−⁵− 0)⁴

 = −1 while by (2),

(0 0) = 

 = lim

→0

( 0) − ^(0 0)

 = lim

→0

⁵⁴

 = 1.

(e) For ( ) 6= (0 0), we use a CAS to compute

( ) = ⁶+ 9⁴²− 9²⁴− ⁶ (²+ ²)³

Now as ( ) → (0 0) along the -axis, ^( ) → 1 while as ( ) → (0 0) along the -axis, ^( ) → −1. Thus ^isn’t continuous at (0 0) and Clairaut’s Theorem doesn’t apply, so there is no contradiction. The graphs of and are identical except at the origin, where we observe the discontinuity.

14.4 Tangent Planes and Linear Approximations

1.  =  ( ) = 2²+ ²− 5 ⇒ ^( ) = 4, ( ) = 2 − 5, so ^(1 2) = 4, (1 2) = −1.

By Equation 2, an equation of the tangent plane is  − (−4) = ^(1 2)( − 1) + ^(1 2)( − 2) ⇒

 + 4 = 4( − 1) + (−1)( − 2) or  = 4 −  − 6.

° 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

3