Answer
$P'(v)=\frac{v\ln v-v+1}{v(1-v)^2}$
Work Step by Step
$P(v)=\frac{\ln v}{1-v}\\
P'(v)=\frac{(1-v)(\frac{1}{v})-(\ln v)(-1)}{(1-v)^2}\\
P'(v)=\frac{\frac{1-v+v\ln v}{v}}{(1-v)^2}\\
P'(v)=\frac{v\ln v-v+1}{v(1-v)^2}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 14
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