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 199: 23



Work Step by Step

We evaluate the function: $2\sqrt (y^3)+xy-x=0$ on differentiating the above: $\frac{d{(2\sqrt (y^3)+xy-x})}{dx}=0$ or $2\times\frac{3}{2}\times\sqrt{y}\frac{dy}{dx}+x\frac{dy}{dx}+y-1=0$ or ${(3\sqrt{y}+x)}\frac{dy}{dx}=(1-y)$ or${dy}=\frac{1-y}{(3\sqrt{y}+x)}{dx}$ The final answer is: ${dy}=\frac{1-y}{(3\sqrt{y}+x)}{dx}$
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