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: 34



Work Step by Step

$$y=\frac{1}{2}\ln\frac{1+x}{1-x}$$ We have $$y'=\frac{1}{2}\frac{1}{\frac{1+x}{1-x}}\Big(\frac{1+x}{1-x}\Big)'=\frac{1}{2}\frac{1-x}{1+x}\frac{(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{1-x-(1+x)(-1)}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{1-x+1+x}{(1-x)^2}$$ $$y'=\frac{1-x}{2(1+x)}\times\frac{2}{(1-x)^2}$$ $$y'=\frac{1}{(1+x)(1-x)}$$ $$y'=\frac{1}{1-x^2}$$
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