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: 40

Answer

$(2x+1)e^{-1/x}$

Work Step by Step

$\frac{d}{dx}x^2e^{-1/x}$ Use chain rule: $=x^2\frac{d}{dx}e^{-1/x}+e^{-1/x}\frac{d}{dx}x^2$ Use chain rule again to evaluate $\frac{d}{dx}e^{-1/x}$: $=x^2e^{-1/x}\frac{d}{dx}(-\frac{1}{x})+e^{-1/x}*2x$ $=x^2e^{-1/x}\frac{d}{dx}(-x^{-1})+e^{-1/x}*2x$ $=x^2e^{-1/x}x^{-2}+e^{-1/x}*2x$ $=e^{-1/x}+e^{-1/x}*2x$ $=(2x+1)e^{-1/x}$
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