Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 178: 43

Answer

$\frac{dy}{dx}=\frac{2-3x^2-4y}{4x-4y^{1/3}}$

Work Step by Step

Take the derivative of the equation on each side separately. Apply chain rule when differentiating the "y" variables since we are differentiating with respect to x: $3x^2+4x\frac{dy}{dx}+4y-\frac{4}{3}\times3y^{1/3}\frac{dy}{dx}=2$ Move all terms with dy/dx to one side of the equation, and isolate dy/dx: $4x\frac{dy}{dx}-4y^{1/3}\frac{dy}{dx}=2-3x^2-4y$ $\frac{dy}{dx}(4x-4y^{1/3})=2-3x^2-4y$ $\frac{dy}{dx}=\frac{2-3x^2-4y}{4x-4y^{1/3}}$
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