   Chapter 7.4, Problem 55E ### 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 ) = y 3 e x 2 ;     ( 1 ,   − 1 )

To determine

To calculate: The four second partial derivatives and four second partial derivatives at the point (1,1) for the function f(x,y)=y3ex2.

Explanation

Given information:

The provided function is f(x,y)=y3ex2.

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)=y3ex2

Partially derivative of the function f(x,y)=y3ex2 with respect to x.

fx(x,y)=x(y3ex2)=y3x(ex2)=y3ex2(2x)=2xy3ex2

Partially derivative of the function f(x,y)=y3ex2 with respect to y.

fy(x,y)=y(y3ex2)=ex2y(y3)=ex2(3y2)=3y2ex2

Again, partially derivative of the function fx(x,y)=2xy3ex2 with respect to x.

fxx(x,y)=x(2xy3ex2)=2y3(xx(ex2)+ex2x(x))=2y3(xex22x+ex21)=ex2(4x2y3+2y3)

Substitute (x,y)=(1,1) for the derivative fxx(x,y)=ex2(4x2y3+2y3).

fxx(1,1)=e(1)2(4(1)2(1)3+2(1)3)=e(4(1)+2(1))=6e

Again, partially derivative of the function fx(x,y)=2xy3ex2 with respect to y

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