Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 9

Answer

$g'(x)=\frac{1-2x}{x}$

Work Step by Step

Use the derivative of natural log and apply the chain rule: $g'(x)=\frac{1}{xe^{-2x}}\times(-2xe^{-2x}+e^{-2x})$ After simplifying: $g'(x)=\frac{1-2x}{x}$
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