Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 32


$$ y'= \frac{1}{(1-x^2)\tanh^{-1}x}.$$

Work Step by Step

Since $ y=\ln (\tanh^{-1}x)$, then the derivative, by using the chain rule, is given by $$ y'= \frac{1}{\tanh^{-1}x}(\tanh^{-1}x )'=\frac{1}{\tanh^{-1}x}\frac{1 }{1-x^2}=\frac{1}{(1-x^2)\tanh^{-1}x}.$$
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