   Chapter 7.4, Problem 53E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding and Evaluating Second Partial Derivatives In Exercises 53-56, find the four second partial derivatives and evaluate each at the given point. f ( x , y ) = x 4 − 3 x 2 y 2 + y 2 ;   ( 1 ,   0 )

To determine

To calculate: The four second partial derivatives and value of four second partial derivatives at the point (1,0) for the function f(x,y)=x43x2y2+y2.

Explanation

Given information:

The provided function is f(x,y)=x43x2y2+y2.

Formula used:

Consider the function z=f(x,y) then for the value of zx consider y to be constant and differentiate with respect to x and the value of zy consider x to be constant and differentiate with respect to y.

According to Higher-Order Partial Derivatives,

x(fx)=2fx2=fxxy(fy)=2fy2=fyyy(fx)=2fyx=fxyx(fy)=2fxy=fyx

Calculation:

Consider the provided function is,

f(x,y)=x43x2y2+y2

Partially derivative of the function f(x,y)=x43x2y2+y2 with respect to x.

fx(x,y)=x(x43x2y2+y2)=x(x4)3y2x(x2)+y2x(1)=4x36xy2

Partially derivative of the function f(x,y)=x43x2y2+y2 with respect to y.

fy(x,y)=y(x43x2y2+y2)=x4y(1)3x2y(y2)+y(y2)=6x2y+2y

Again, partially derivative of the function fx(x,y)=4x36xy2 with respect to x.

fxx(x,y)=x(4x36xy2)=4x(x3)6y2x(x)=12x26y2

Substitute (x,y)=(1,0) for the derivative fxx(x,y)=12x26y2

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