University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 32



Work Step by Step

We evaluate the function: $y=xe^{-{x}}$ on differentiating the above: $\frac{dy}{dx}=\frac{d{(xe^{(-x)})}}{dx}$ on applying the product rule: $\frac{dy}{dx}=e^{-x}{\frac{dx}{dx}}+x{\frac{de^{(-x)}}{dx}}$ or $\frac{dy}{dx}=e^{(-x)}-xe^{(-x)}$ or ${dy}=(e^{(-x)}-xe^{(-x)}){dx}$ or${dy}=(1-x)e^{(-x)}{dx}$ The final answer is: ${dy}=(1-x)e^{(-x)}{dx}$
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