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: 24



Work Step by Step

We evaluate the function: $xy^2-4\sqrt {(x^3)}-y=0$ on differentiating the above: $\frac{d(xy^2-4\sqrt {(x^3)}-y)}{dx}=0$ or $2xy\frac{dy}{dx}+y^2-4\times\frac{3}{2}\times\sqrt{x}-\frac{dy}{dx}=0$ or ${(2xy-1)}\frac{dy}{dx}=(6\sqrt{x}-y^2)$ or ${dy}=\frac{(6\sqrt{x})-y^2}{(2xy-1)}{dx}$ The final answer is: ${dy}=\frac{(6\sqrt{x})-y^2}{(2xy-1)}{dx}$
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