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: 10

Answer

$$\frac{dy}{dx}=-\frac{y}{x}$$

Work Step by Step

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