Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 9

Answer

$\frac{1}{x}-2$

Work Step by Step

$\frac{d}{dx}\ln(xe^{-2x})$ $=\frac{1}{xe^{-2x}}\frac{d}{dx}(xe^{-2x})$ $=\frac{1}{xe^{-2x}}(x\frac{d}{dx}e^{-2x}+e^{-2x}\frac{d}{dx}x)$ $=\frac{1}{xe^{-2x}}(xe^{-2x}\frac{d}{dx}(-2x)+e^{-2x}*1)$ $=\frac{1}{xe^{-2x}}(xe^{-2x}*(-2)+e^{-2x})$ $=\frac{-2xe^{-2x}+e^{-2x}}{xe^{-2x}}$ $=\frac{-2xe^{-2x}}{xe^{-2x}}+\frac{e^{-2x}}{xe^{-2x}}$ $=-2+\frac{1}{x}$ $=\frac{1}{x}-2$
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