Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 5

Answer

$y'=\dfrac{1-x}{e^{x}}$

Work Step by Step

$y=\dfrac{x}{e^{x}}$ Differentiate using the quotient rule: $y'=\dfrac{(e^{x})(x)'-(x)(e^{x})'}{(e^{x})^{2}}=\dfrac{e^{x}-xe^{x}}{e^{2x}}$ Take out common factor $e^{x}$ $y'=\dfrac{e^{x}(1-x)}{e^{2x}}=\dfrac{1-x}{e^{x}}$ $y'=\dfrac{1-x}{e^{x}}$
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