Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 40

Answer

$y'=2e^{\sin2x}\cos2x+2e^{2x}\cos(e^{2x})$

Work Step by Step

$y=e^{\sin2x}+\sin(e^{2x})$ Differentiate each term: $y'=(e^{\sin2x})'+[\sin(e^{2x})]'=...$ Apply the chain rule to find $(e^{\sin2x})'$ and $[\sin(e^{2x})]'$: $...=(e^{\sin2x})(\sin2x)'+[\cos(e^{2x})](e^{2x})'=...$ Use the chain rule one more time to find $(\sin2x)'$ and $(e^{2x})'$: $...=(e^{\sin2x})(\cos2x)(2x)'+[\cos(e^{2x})](e^{2x})(2x)'=...$ $...=2e^{\sin2x}\cos2x+2e^{2x}\cos(e^{2x})$
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