Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 419: 47


$g'(u)=ue^{\sqrt secu^{2}}(\sqrt secu^{2})(tanu^{2})$

Work Step by Step

$g'(u)=\frac{d}{du}(e^{\sqrt secu^{2}})$ $=e^{\sqrt secu^{2}}\frac{d}{du}({\sqrt secu^{2}})$ $=e^{\sqrt secu^{2}}\frac{1}{2(\sqrt secu^{2})}\frac{d}{du}({secu^{2}})$ $=e^{\sqrt secu^{2}}\frac{1}{2(\sqrt secu^{2})}({secu^{2}})({tanu^{2}}).2u$ Hence, $g'(u)=ue^{\sqrt secu^{2}}(\sqrt secu^{2})(tanu^{2})$
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