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 - Page 158: 41


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

Work Step by Step

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