Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 39


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

Work Step by Step

Since $ y=arcsin (e^x)=\sin^{-1}e^x $, then the derivative $ y'$ is given by $$ y'= \frac{1}{\sqrt{1-(e^x)^2}} (e^x)'= \frac{e^x}{\sqrt{1-(e^x)^2}} .$$
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