Calculus 8th Edition

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

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 48


$$\frac{d}{d x}\left(\frac{1}{2} \tan ^{-1} x+\frac{1}{4} \ln \frac{(x+1)^{2}}{x^{2}+1}\right) =\frac{1}{\left(x+1\right)\left(x^2+1\right)}$$

Work Step by Step

Since \begin{align*} \frac{d}{d x}\left(\frac{1}{2} \tan ^{-1} x+\frac{1}{4} \ln \frac{(x+1)^{2}}{x^{2}+1}\right)&= \frac{d}{dx}\left(\frac{1}{2}\tan^{-1} \left(x\right)\right)+\frac{d}{dx}\left(\frac{1}{4}[2\ln (x+1)-\ln(x^2+1) \right)\\ &=\frac{1}{2(1+x^2)}+\frac{1}{2(x+1)}-\frac{x}{2(x^2+1)}\\ &=\frac{1}{\left(x+1\right)\left(x^2+1\right)} \end{align*}
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