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: 46

Answer

$y'=2[cos(2x)e^{sin2x}+cos(e^{2x})(e^{2x})]$

Work Step by Step

$y'=\frac{d}{dx}[e^{sin2x}+sine^{(2x)}]$ $=\frac{d}{dx}[e^{sin2x}]+\frac{d}{dx}[sine^{(2x)}]$ $=e^{sin2x}\frac{d}{dx}[sin2x]+cos(e^{2x})\frac{d}{dx}[e^{2x}]$ $=e^{sin2x}cos2x.2+cos(e^{2x})(2e^{2x})$ Hence, $y'=2[cos(2x)e^{sin2x}+cos(e^{2x})(e^{2x})]$
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