University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises: 20

Answer

$\frac{dy}{dx} = \frac{2}{3}e^{\frac{2x}{3}}$

Work Step by Step

$y = u^{\frac{2}{3}}$ where, $u = e^{x}$ Now, $\frac{dy}{du} = \frac{2}{3}u^{-\frac{1}{3}}$ and, $\frac{du}{dx} = e^{x}$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = \frac{2}{3}u^{-\frac{1}{3}} \times e^{x}$ $\frac{dy}{dx} = \frac{2}{3}e^{-\frac{1x}{3}} \times e^{x}$ $\frac{dy}{dx} = \frac{2}{3}e^{\frac{2x}{3}}$
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.