Explanation Given: The provide function is f ( x , y ) = 3 x 2 y 3 x 2 + y 2 . Approach: To find f x ( x , y ) differentiate the f ( x , y ) with respect to x by holding y . f x ( x , y ) = ∂ ∂ x f ( x , y ) To find f y ( x , y ) differentiate the f ( x , y ) with respect to y by holding x . f y ( x , y ) = ∂ ∂ y f ( x , y ) The quotient rule of partial derivative with respect to y , ∂ ∂ y [ f ( x , y ) h ( x , y ) ] = h ( x , y ) ∂ ∂ y f ( x , y ) − f ( x , y ) ∂ ∂ y h ( x , y ) [ h ( x , y ) ] 2 The quotient rule of partial derivative with respect to x , ∂ ∂ x [ f ( x , y ) h ( x , y ) ] = h ( x , y ) ∂ ∂ x f ( x , y ) − f ( x , y ) ∂ ∂ x h ( x , y ) [ h ( x , y ) ] 2 Calculation: Consider the given function, f ( x , y ) = 3 x 2 y 3 x 2 + y 2 . To find f x ( x , y ) differentiate the f ( x , y ) with respect to x by holding y . f x ( x , y ) = ∂ ∂ x ( 3 x 2 y 3 x 2 + y 2 ) Apply the quotient rule of partial derivative, ∂ ∂ x ( 3 x 2 y 3 x 2 + y 2 ) = ( x 2 + y 2 ) ∂ ∂ x ( 3 x 2 y 3 ) − ( 3 x 2 y 3 ) ∂ ∂ x ( x 2 + y 2 ) [ x 2 + y 2 ] 2 = ( x 2 + y 2 ) ( 6 x y 3 ) − ( 3 x 2 y 3 ) ( 2 x ) [ x 2 + y 2 ] 2 = 6 x 3 y 3 + 6 x y 5 − 6 x 3 y 3 [ x 2 + y 2 ] 2 = 6 x y 5 ( x 2 + y 2 ) 2 Thus, f x ( x , y ) = 6 x y 5 ( x 2 + y 2 ) 2 Substitute 2 for x and − 1 for y in the function f x ( x , y ) = 6 x y 5 ( x 2 + y 2 ) 2 . f x ( 2 , − 1 ) = 6 ( 2 ) ( − 1 ) 5 [ ( 2 ) 2 + ( − 1 ) 2 ] 2 = − 12 [ 4 + 1 ] 2 = − 12 25 To find f y ( x , y ) differentiate the f ( x , y ) with respect to y by holding x

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

2nd Edition

ISBN: 9780137553457

Author: Raymond Greenwell, Nathan Ritchey

Publisher: PEARSON+

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Question

Chapter 9.2, Problem 12E

To determine

To find:

The fx(x,y) and fy(x,y) then find fx(2,−1) and fy(−4,3) for f(x,y)=3x2y3x2+y2.

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Chapter 9.2, Problem 11E

Chapter 9.2, Problem 13E

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Chapter 9 Solutions

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

Ch. 9.1 - YOUR TURN For the function in Example 1, find...Ch. 9.1 - Prob. 2YT Ch. 9.1 - Prob. 3YT Ch. 9.1 - EXERCISES Let f(x,y)=2x3y+5.Find the following. a....Ch. 9.1 - Prob. 2E Ch. 9.1 - EXERCISES Let h(x,y)=x2+2y2. Find the following....Ch. 9.1 - EXERCISES Let f(x,y)=9x+5ylogx. Find the...Ch. 9.1 - EXERCISES Let f(x,y)=ex+ln(x+y). Find the...Ch. 9.1 - EXERCISES Let f(x,y)=xex+y. Find the following. a....Ch. 9.1 - EXERCISES Let f(x,y)=xsin(x2y). Find the...

Knowledge Booster

The f x ( x , y ) and f y ( x , y ) then find f x ( 2 , − 1 ) and f y ( − 4 , 3 ) for f ( x , y ) = 3 x 2 y 3 x 2 + y 2 .

Solution Summary: The author explains that the provide function is f_x(x,y), and the partial derivative function

To find:

The fx(x,y) and fy(x,y) then find fx(2,−1) and fy(−4,3) for f(x,y)=3x2y3x2+y2.

Want to see the full answer?

Chapter 9 Solutions

The f x ( x , y ) and f y ( x , y ) then find f x ( 2 , − 1 ) and f y ( − 4 , 3 ) for f ( x , y ) = 3 x 2 y 3 x 2 + y 2 .

Solution Summary: The author explains that the provide function is f_x(x,y), and the partial derivative function

To find: The fx(x,y) and fy(x,y) then find fx(2,−1) and fy(−4,3) for f(x,y)=3x2y3x2+y2.

Want to see the full answer?

Chapter 9 Solutions

To find:

The fx(x,y) and fy(x,y) then find fx(2,−1) and fy(−4,3) for f(x,y)=3x2y3x2+y2.