Explanation Given: The function is f ( x , y ) = 2 x + 5 y 2 3 x 2 + y 2 ... ( 1 ) . Approach: The partial derivative of f with respect to y is the derivative of f obtained by treating x as a constant and y as a variable, while the partial derivative of f with respect to x is the derivative of f obtained by treating x as a variable and y as a constant. Calculation: Take partial derivative of f with respect to x in the equation ( 1 ) , f x ( x , y ) = 2 ( 3 x 2 + y 2 ) − 6 x ( 2 x + 5 y 2 ) ( 3 x 2 + y 2 ) 2 = 6 x 2 + 2 y 2 − 12 x 2 − 30 x y 2 ( 3 x 2 + y 2 ) 2 = − 6 x 2 + 2 y 2 − 30 x y 2 ( 3 x 2 + y 2 ) 2 Take partial derivative of f with respect to y in the equation ( 1 ) ,

2nd Edition

ISBN: 9780321964458

Author: Lial

Publisher: PEARSON EDUCATION (COLLEGE)

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In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 9.CR, Problem 30CR

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To find:

The value of $fx(x,y)$ and $fy(x,y)$ for the function $f(x,y)=2x+5y23x2+y2$ .

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Chapter 9.CR, Problem 29CR

Chapter 9.CR, Problem 31CR

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Chapter 9 Solutions

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The value of f x ( x , y ) and f y ( x , y ) for the function f ( x , y ) = 2 x + 5 y 2 3 x 2 + y 2 .

Solution Summary: The author explains the function f(x,y)=2x+5y

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To find:

The value of $fx(x,y)$ and $fy(x,y)$ for the function $f(x,y)=2x+5y23x2+y2$ .

Want to see the full answer?

Chapter 9 Solutions