   Chapter 7.4, Problem 40E ### 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 Partial Derivatives In Exercises 33-40, find the first partial derivatives with respect to x, y, and z, and evaluate each at the given point. w = ln x 2 + y 2 + z 2 ;   ( 3 ,   0 ,   4 )

To determine

To calculate: The first partial derivatives with respect to x,y and z for the function w=ln(x2+y2+z2) at point (3,0,4).

Explanation

Given information:

The provided function is w=ln(x2+y2+z2) and the point is (3,0,4).

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.

Calculation:

Consider the provided function is,

w=ln(x2+y2+z2)

Partially derivative of the function w=ln(x2+y2+z2) with respect to x.

wx=x(ln(x2+y2+z2))=1(x2+y2+z2)x(x2+y2+z2)=1(x2+y2+z2)12x2+y2+z2x(x2+y2+z2)=x(x2+y2+z2)

Substitute (x,y,z)=(3,0,4) into the function wx.

wx|(3,0,4)=3((3)2+(0)2+(4)2)=3(9+0+16)=325

Partially derivative of the function w=ln(x2+y2+z2) with respect to y.

wy=y(ln(x2+y2+z2))=1(x2+y2+z2)y(x2+y2+z2)=1(x2+y2+z2)12x2+y2+z2y(x2+y2+z2)=y(x2+y2+z2)

Substitute (x,y,z)=(3,0,4) into the function wy

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