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: 41

Answer

$\frac{xe^x(x^3+2ex)}{(x^2+e^x)^2}$

Work Step by Step

$\frac{d}{dx}\frac{x^2e^x}{x^2+e^x}$ $=\frac{(x^2+e^x)\frac{d}{dx}(x^2e^x)-(x^2e^x)\frac{d}{dx}(x^2+e^x)}{(x^2+e^x)^2}$ $=\frac{(x^2+e^x)(x^2\frac{d}{dx}e^x+e^x\frac{d}{dx}x^2)-(x^2e^x)(2x+e^x)}{(x^2+e^x)^2}$ $=\frac{(x^2+e^x)(x^2e^x+e^x*2x)-(x^2e^x)(2x+e^x)}{(x^2+e^x)^2}$ $=\frac{xe^x(x^2+e^x)(x+2)-(x^2e^x)(2x+e^x)}{(x^2+e^x)^2}$ $=\frac{xe^x[(x^2+e^x)(x+2)-x(2x+e^x)]}{(x^2+e^x)^2}$ $=\frac{xe^x[(x^3+2x^2+xe^x+2e^x)-(2x^2+xe^x)]}{(x^2+e^x)^2}$ $=\frac{xe^x(x^3+2x^2+xe^x+2e^x-2x^2-xe^x)}{(x^2+e^x)^2}$ $=\frac{xe^x(x^3+2ex)}{(x^2+e^x)^2}$
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.