University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 164: 14


$$\frac{dy}{dx}=\frac{\cos(2x+3y)-2x\sin(2x+3y)-y\cos x}{\sin x+3x\sin(2x+3y)}$$

Work Step by Step

$$x\cos(2x+3y)=y\sin x$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(x\cos(2x+3y))=\frac{d}{dx}(y\sin x)$$ $$\cos(2x+3y)+x\frac{d}{dx}(\cos(2x+3y))=\frac{dy}{dx}\sin x+y\cos x$$ $$\cos(2x+3y)+x(-\sin(2x+3y))\frac{d}{dx}(2x+3y)=\frac{dy}{dx}\sin x+y\cos x$$ $$\cos(2x+3y)-x\sin(2x+3y)(2+3\frac{dy}{dx})=\frac{dy}{dx}\sin x+y\cos x$$ $$\cos(2x+3y)-2x\sin(2x+3y)-3x\sin(2x+3y)\frac{dy}{dx}=\frac{dy}{dx}\sin x+y\cos x$$ 2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$\frac{dy}{dx}\sin x+3x\sin(2x+3y)\frac{dy}{dx}=\cos(2x+3y)-2x\sin(2x+3y)-y\cos x$$ $$\frac{dy}{dx}=\frac{\cos(2x+3y)-2x\sin(2x+3y)-y\cos x}{\sin x+3x\sin(2x+3y)}$$
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