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: 12


$$\frac{dy}{dx}=\frac{3x^2y^2-4x^3}{\cos y-2x^3y}$$

Work Step by Step

$$x^4+\sin y=x^3y^2$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(x^4)+\frac{d}{dx}(\sin y)=\frac{d}{dx}(x^3y^2)$$ $$4x^3+\cos y\frac{dy}{dx}=3x^2y^2+x^3(2y)\frac{dy}{dx}$$ $$4x^3+\cos y\frac{dy}{dx}=3x^2y^2+2x^3y\frac{dy}{dx}$$ 2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$\cos y\frac{dy}{dx}-2x^3y\frac{dy}{dx}=3x^2y^2-4x^3$$ $$\frac{dy}{dx}=\frac{3x^2y^2-4x^3}{\cos y-2x^3y}$$
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