Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises - Page 446: 30


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.