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: 6



Work Step by Step

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