   Chapter 7.5, Problem 9SWU ### 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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# In Exercises 9-14, find all first and second partial derivatives. z = 4 x 3 − 3 y 2

To determine

To calculate: The first and the second partial derivatives of z=4x33y2.

Explanation

Given Information:

The provided equation is,

z=4x33y2

Formula used:

For a function in two variables f(x,y),

Partial derivative can be found with respect to one variable keeping the other variable a constant,

fx, keeping y as constant

And fy, keeping x as constant

Similarly, one could take the partial derivatives of the partial derivatives. These are called second partial derivatives.

x(fx)=2fx2x(fy)=v2fxyy(fx)=2fyxy(fy)=2fy2

And

x(x)n=n(x)n1

Calculation:

Consider the given equation,

z=4x33y2

For the first partial derivative, differentiate above equation partially with respect to x:

zx=x(4x33y2)=x(4x3)x(3y2)=4x(x3)0=43x2

On further simplification:

zx=12x2

Differentiate z=4x33y2 partially with respect to y:

zy=y(4x33y2)=y

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