Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 21

Answer

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

Work Step by Step

$y=\sqrt{\dfrac{x}{x+1}}$ Let's rewrite the function like this: $y=(\dfrac{x}{x+1})^{1/2}$ Differentiate using the chain rule: $y'=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}(\dfrac{x}{x+1})'=...$ Use the quotient rule to find $(\dfrac{x}{x+1})'$: $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(x)'-(x)(x+1)'}{(x+1)^{2}}]=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}[\dfrac{(x+1)(1)-(x)(1)}{(x+1)^{2}}]=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{x-x+1}{(x+1)^{2}}=...$ $...=\dfrac{1}{2}(\dfrac{x}{x+1})^{-1/2}\dfrac{1}{(x+1)^{2}}=\dfrac{1}{2}\dfrac{1}{(x+1)^{2}}\sqrt{\dfrac{x+1}{x}}$
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