University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 79



Work Step by Step

$$y=\log_{3}\Big(\Big(\frac{x+1}{x-1}\Big)^{\ln3}\Big)$$ We have $\log_ax^b=b\log_ax$. Therefore, $$y=\ln3\log_3\Big(\frac{x+1}{x-1}\Big)$$ The derivative of $y$ is $$y'=\ln3\Big(\log_3\Big(\frac{x+1}{x-1}\Big)\Big)'$$ Recall Theorem 7: $$\frac{d}{dr}\log_au=\frac{1}{u\ln a}\frac{du}{dr}$$ That means: $$y'=\ln3\times\frac{1}{\frac{x+1}{x-1}\ln3}\Big(\frac{x+1}{x-1}\Big)'=\frac{1}{\frac{x+1}{x-1}}\times\frac{(x-1)-(x+1)}{(x-1)^2}$$ $$y'=\frac{x-1}{x+1}\times\frac{x-1-x-1}{(x-1)^2}=\frac{x-1}{x+1}\times\frac{-2}{(x-1)^2}$$ $$y'=\frac{-2}{(x+1)(x-1)}$$
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