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: 31


$$ y'= \frac{1}{t(1-(\ln t)^2)}.$$

Work Step by Step

Since $ y=\tanh^{-1} (\ln t) $, then the derivative, by using the chain rule, is given by $$ y'= \frac{1}{1-(\ln t))^2}(\ln t )'=\frac{1/t }{1-(\ln t))^2}=\frac{1}{t(1-(\ln t)^2)}.$$
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