Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 41


$$ g'(x)= \frac{e^x}{1-e^{2x}}.$$

Work Step by Step

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