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: 35

Answer

$y'=\dfrac{4e^{2x}}{(1+e^{2x})^{2}}\sin(\dfrac{1-e^{2x}}{1+e^{2x}})$

Work Step by Step

$y=\cos(\dfrac{1-e^{2x}}{1+e^{2x}})$ Differentiate using the chain rule: $y'=-(\dfrac{1-e^{2x}}{1+e^{2x}})'\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$ Use the quotient rule to find $(\dfrac{1-e^{2x}}{1+e^{2x}})'$: $...=-[\dfrac{(1+e^{2x})(1-e^{2x})'-(1-e^{2x})(1+e^{2x})'}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=$ $...=-[\dfrac{-(1+e^{2x})(e^{2x})'-(1-e^{2x})(e^{2x})'}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$ Use the chain rule to find $(e^{2x})'$: $...=-[\dfrac{-(1+e^{2x})(2e^{2x})-(1-e^{2x})(2e^{2x})}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$ Simplify: $...=-[\dfrac{-2e^{2x}-2e^{4x}-2e^{2x}+2e^{4x}}{(1+e^{2x})^{2}}]\sin(\dfrac{1-e^{2x}}{1+e^{2x}})=...$ $...=\dfrac{4e^{2x}}{(1+e^{2x})^{2}}\sin(\dfrac{1-e^{2x}}{1+e^{2x}})$
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