Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises: 14

Answer

$\frac{dy}{dx} = \frac{\sin y^2 - 2yx\cos x^2}{\sin x^2-2xy\cos y^2}$

Work Step by Step

$y\sin x^2 = x \sin y^2$ Differentiate both sides with respect to x $\frac{dy}{dx}\sin x^2 + 2yx\cos x^2 =\sin y^2 + 2xy\cos y^2 \frac{dy}{dx} $ Isolate $\frac{dy}{dx}$ $\frac{dy}{dx}\sin x^2 - 2xy\cos y^2 \frac{dy}{dx} = \sin y^2 - 2yx\cos x^2$ $\frac{dy}{dx}(\sin x^2-2xy\cos y^2) = \sin y^2 - 2yx\cos x^2$ $\frac{dy}{dx} = \frac{\sin y^2 - 2yx\cos x^2}{\sin x^2-2xy\cos y^2}$
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