Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 28


$$\frac{2 u 2^{u}}{\sqrt{u^{2}+1}} \left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right)$$

Work Step by Step

Given $$ y=\frac{2 u 2^{u}}{\sqrt{u^{2}+1}} $$ Take $\ln $ for both sides \begin{align*} \ln y&= \ln \frac{2 u 2^{u}}{\sqrt{u^{2}+1}}\\ &=\ln 2 u 2^{u}- \ln \sqrt{u^{2}+1}\\ &=\ln 2u +\ln 2^u -\frac{1}{2}\ln (u^2+1)\\ &= \frac{1}{u}+ \ln 2 -\frac{1}{2} \frac{2u}{u^2+1} \end{align*} Then \begin{align*} \frac{y'}{y} &= \frac{1}{u}+ \ln 2 -\frac{1}{2} \frac{2u}{u^2+1}\\ &= \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\\ y'&= y\left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right)\\ &=\frac{2 u 2^{u}}{\sqrt{u^{2}+1}} \left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right) \end{align*}
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