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: 13

Answer

$y'=-xe^{-3x}(3x-2)$

Work Step by Step

$y=x^{2}e^{-3x}$ Differentiate using the product rule: $y'=(x^{2})(e^{-3x})'+(e^{-3x})(x^{2})'=...$ Now, apply the chain rule to find $(e^{-3x})'$: $...=(x^{2})[(e^{-3x})(-3x)']+(e^{-3x})(2x)=(x^{2})(-3e^{-3x})+2xe^{-3x}=...$ $...=-3x^{2}e^{-3x}+2xe^{-3x}=...$ Take out common factor $-x^{2}e^{-3x}$ to present a better looking answer: $...=-xe^{-3x}(3x-2)$
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