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: 36

Answer

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

Work Step by Step

$y=x^{2}e^{-1/x}$ Differentiate using the product rule: $y'=(x^{2})(e^{-1/x})'+(e^{-1/x})(x^{2})'=...$ Use the chain rule to find $(e^{-1/x})'$: $...=(x^{2})[(e^{-1/x})(-\dfrac{1}{x})']+(e^{-1/x})(2x)=...$ Write $-\dfrac{1}{x}$ as $-x^{-1}$ and continue with the differentiation process: $...=(x^{2})[(e^{-1/x})(-x^{-1})']+(e^{-1/x})(2x)=...$ $...=(x^{2})[(e^{-1/x})(x^{-2})]+(e^{-1/x})(2x)=...$ Simplify: $...=e^{-1/x}+2xe^{-1/x}=e^{-1/x}(1+2x)$
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