   Chapter 7.4, Problem 54E ### 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 3 + 2 x y 3 − 3 y ;     ( 3 ,   2 )

To determine

To calculate: The four second partial derivatives and four second partial derivatives at the point (3,2) for the function f(x,y)=x3+2xy33y.

Explanation

Given information:

The provided function is f(x,y)=x3+2xy33y.

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)=x3+2xy33y

Partially derivative of the function f(x,y)=x3+2xy33y with respect to x.

fx(x,y)=x(x3+2xy33y)=x(x3)+2y3x(x)3yx(1)=3x2+2y3

Partially derivative of the function f(x,y)=x3+2xy33y with respect to y.

fy(x,y)=y(x3+2xy33y)=x3y(1)+2xy(y3)3y(y)=6xy23

Again, partially derivative of the function fx(x,y)=3x2+2y3 with respect to x.

fxx(x,y)=x(3x2+2y3)=3x(x2)+2y3x(1)=6x

Substitute (x,y)=(3,2) for the derivative fxx(x,y)=6x.

fxx(3,2)=63=18

Again, partially derivative of the function fx(x,y)=3x2+2y3 with respect to y

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