## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 1

#### Answer

$$2x+6y^2\frac{dy}{dx}=0.$$ $$\frac{dy}{dx}=-\frac{x}{3y^2}.$$ $$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$

#### Work Step by Step

Differentiating the equation $x^2+2y^3=6$ with respect to $x$ and using the chain rule, we get $$2x+6y^2\frac{dy}{dx}=0.$$ Hence, $$\frac{dy}{dx}=-\frac{x}{3y^2}.$$ Moreover, we have $$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$

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