Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 43



Work Step by Step

$y=x^2e^{-3x}$ Use the product rule: $y'=x^2\frac{d}{dx}e^{-3x}+e^{-3x}\frac{d}{dx}x^2$ Use the chain rule to evaluate $\frac{d}{dx}e^{-3x}$: $y'=x^2e^{-3x}\frac{d}{dx}(-3x)+e^{-3x}*2x$ $y'=x^2e^{-3x}*(-3)+2xe^{-3x}$ $y'=-3x^2e^{-3x}+2xe^{-3x}$ Factor out $xe^{-3x}$: $y'=xe^{-3x}(-3x+2)$ $y'=xe^{-3x}(2-3x)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.