Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 4



Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(x^n)'=nx^{n-1}$ Using the quotient and chain rules, we have $$\frac{d}{dx}\frac{x^3}{y^2}=\frac{y^2(3x^2)-x^3(2y\frac{dy}{dx})}{y^4}$$ $$=\frac{3x^2}{y^2}-\frac{2x^3}{y^3}\frac{dy}{dx}$$
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