University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 198: 22


${dy}=(\frac{1}{3\sqrt x(1+\sqrt x)^2}){dx}$

Work Step by Step

We evaluate the function: $y=\frac{2\sqrt x}{3{(1+\sqrt x)}}$ on differentiating the above: $\frac{dy}{dx}=\frac{d\frac{2\sqrt x}{3{(1+\sqrt x)}}}{dx}$ or ${dy}=(\frac{{3{(1+\sqrt x)}}\frac{2}{2\sqrt {x}})-{\frac{6\sqrt x}{2\sqrt x}}}{9(1+\sqrt x)^2}){dx}$ or ${dy}=(\frac{1+\sqrt x-\sqrt x}{3\sqrt x(1+\sqrt x)^2}){dx}$ or ${dy}=(\frac{1}{3\sqrt x(1+\sqrt x)^2}){dx}$ The final answer is: ${dy}=(\frac{1}{3\sqrt x(1+\sqrt x)^2}){dx}$
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