   Chapter 7.4, Problem 48E ### 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 Second Partial Derivatives In Exercises 45-52, find the four second partial derivatives. See Example 6. z = y 3 − 6 x 2 y 2 − 1

To determine

To calculate: The second partial derivatives for the function z=y36x2y21.

Explanation

Given information:

The provided function is z=y36x2y21.

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,

z=y36x2y21

Partially derivative of the function z=y36x2y21 with respect to x.

zx=x(y36x2y21)=y3x(1)6y2x(x2)x(1)=12xy2

Partially derivative of the function z=y36x2y21 with respect to y.

zy=y(y36x2y21)=y(y3)6x2x(y2)x(1)=3y212x2y

Again, partially derivative of the function zx=12xy2 with respect to x

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