   Chapter 3.6, Problem 42E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use logarithmic differentiation to find the derivative of the function. y = x e x 2 − x ( x + 1 ) 2 / 3

To determine

To find: The derivative of y by using logarithmic differentiation.

Explanation

Given:

The function is y=xex2x(x+1)23.

Result used: Chain Rule

If y=f(u) and u=g(x) are both differentiable functions, then dydx=dydududx.

Calculation:

Consider y=xex2x(x+1)23,

Take natural logarithm on both sides,

lny=ln[xex2x(x+1)23]=ln(x)12+lnex2x+ln(x+1)23             (Qlnab=lna+lnb)=12ln(x)+(x2x)lne+23ln(x+1)   (Qlnxa=alnx)=12ln(x)+(x2x)(1)+23ln(x+1)     (Qlnea=a)

Differentiate both sides with respect to x.

ddx(lny)=ddx(12ln(x)+(x2x)+23ln(x+1))=ddx(12lnx)+ddx(x2x)+ddx(23ln(x+1))

Let u=x+1

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 