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: 26


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

Work Step by Step

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