Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 8

Answer

$\dfrac{dy}{dx}=\dfrac{y-4\sqrt{(xy)^3}}{2x^2\sqrt{xy}-x}.$

Work Step by Step

$\dfrac{d}{dx}(\sqrt{xy})=\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(1)\rightarrow$ Using the Chain Rule with $u=xy\rightarrow =\dfrac{dy}{dx}=(y+\dfrac{dy}{dx}(x))\rightarrow$ $\dfrac{d}{dx}(\sqrt{xy})=\dfrac{y+\dfrac{dy}{dx}(x)}{2\sqrt{xy}}\rightarrow$ $\dfrac{y+\dfrac{dy}{dx}(x)}{2\sqrt{xy}}=(2xy+\dfrac{dy}{dx}(x^2))\rightarrow$ $y-4\sqrt{(xy)^3}=\dfrac{dy}{dx}(2x^2\sqrt{xy}-x)\rightarrow$ $\dfrac{dy}{dx}=\dfrac{y-4\sqrt{(xy)^3}}{2x^2\sqrt{xy}-x}.$
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