Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 42

Answer

$$y'=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt x}}}\left(1+\frac{1}{2\sqrt{x+\sqrt x}}\left(1+\frac{1}{2\sqrt x}\right)\right)$$

Work Step by Step

$$y'=\left(\sqrt{x+\sqrt{x+\sqrt x}}\right)'=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt x}}}\left(x+\sqrt{x+\sqrt x}\right)'=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt x}}}\left(1+\frac{1}{2\sqrt{x+\sqrt x}}(x+\sqrt x)'\right)=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt x}}}\left(1+\frac{1}{2\sqrt{x+\sqrt x}}\left(1+\frac{1}{2\sqrt x}\right)\right)$$
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.