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: 31


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

Work Step by Step

$\frac{d}{du} \frac{\ln u}{1+\ln(2u)}$ $=\frac{(1+\ln 2u)\frac{d}{du}\ln u-\ln u\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^2}$ $=\frac{(1+\ln(2u))\frac{1}{u}-\ln u*\frac{1}{2u}*\frac{d}{du}(2u)}{(1+\ln(2u))^2}$ $=\frac{\frac{1+\ln(2u)}{u}-\frac{\ln u}{u}}{(1+\ln(2u))^2}$ $=\frac{1+\ln(2u)-\ln u}{u(1+\ln(2u))^2}$ $=\frac{1+\ln 2+\ln u-\ln u}{u(1+\ln(2u))^2}$ $=\frac{1+\ln 2}{u(1+\ln(2u))^2}$
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