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: 23

Answer

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

Work Step by Step

$$y'=\left(\sqrt{\frac{x}{x+1}}\right)'=\frac{1}{2\sqrt{\frac{x}{x+1}}}\left(\frac{x}{x+1}\right)'=\frac{1}{2\sqrt{\frac{x}{x+1}}}\frac{x'(x+1)-x(x+1)'}{(x+1)^2}=\frac{1}{2\sqrt{\frac{x}{x+1}}}\frac{1\cdot(x+1)-x\cdot1}{(x+1)^2}=\frac{1}{2\sqrt{\frac{x}{x+1}}(x+1)^2}$$
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