   Chapter 7.4, Problem 39E ### 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 ( 5 x + 2 y 3 − 3 z ) ;   ( 4 ,   1 ,   − 1 )

To determine

To calculate: The first partial derivatives with respect to x,y and z for the function w=ln(5x+2y33z) at point (4,1,1).

Explanation

Given information:

The provided function is w=ln(5x+2y33z) and the point is (4,1,1).

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(5x+2y33z)

Partially derivative of the function w=ln(5x+2y33z) with respect to x.

wx=x(ln(5x+2y33z))=1(5x+2y33z)x(5x+2y33z)=1(5x+2y33z)(5)=5(5x+2y33z)

Substitute (x,y,z)=(4,1,1) into the function wx.

wx|(4,1,1)=5(54+2(1)33(1))=5(20+2+3)=15

Partially derivative of the function w=ln(5x+2y33z) with respect to y.

wy=y(ln(5x+2y33z))=1(5x+2y33z)y(5x+2y33z)=1(5x+2y33z)(6y)=6y(5x+2y33z)

Substitute (x,y,z)=(4,1,1) into the function wy

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