Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 14

Answer

$\frac{\frac{1}{v}-1+\ln v}{(1-v)^2}$

Work Step by Step

$\frac{d}{dv}\frac{\ln v}{1-v}$ $=\frac{(1-v)\frac{d}{dx}\ln v-\ln v\frac{d}{dx}(1-v)}{(1-v)^2}$ $=\frac{(1-v)*\frac{1}{v}-\ln v*(-1)}{(1-v)^2}$ $=\frac{\frac{1-v}{v}+\ln v}{(1-v)^2}$ $=\frac{\frac{1-v}{v}+\ln v}{(1-v)^2}$ $=\frac{\frac{1}{v}-1+\ln v}{(1-v)^2}$
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