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: 8



Work Step by Step

$$x^3=\frac{2x-y}{x+3y}$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(x^3)=\frac{d}{dx}\Big(\frac{2x-y}{x+3y}\Big)=\frac{(2-\frac{dy}{dx})(x+3y)-(2x-y)(1+3\frac{dy}{dx})}{(x+3y)^2}$$ $$3x^2=\frac{(2x+6y)-\frac{dy}{dx}(x+3y)-(2x-y)-\frac{dy}{dx}(6x-3y)}{(x+3y)^2}$$ $$3x^2=\frac{7y-\frac{dy}{dx}(x+3y)-\frac{dy}{dx}(6x-3y)}{(x+3y)^2}$$ $$7y-\frac{dy}{dx}(x+3y)-\frac{dy}{dx}(6x-3y)=3x^2(x+3y)^2$$ 2) Collect all the terms with $dy/dx$ on one side and solve for $dy/dx$: $$\frac{dy}{dx}(x+3y)+\frac{dy}{dx}(6x-3y)=7y-3x^2(x+3y)^2$$ $$7x\frac{dy}{dx}=7y-3x^2(x+3y)^2$$ $$\frac{dy}{dx}=\frac{7y-3x^2(x+3y)^2}{7x}$$
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