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 166: 63


The tangent line to the curve at $P(1,1)$ is $$y=\frac{5}{13}x+\frac{8}{13}$$

Work Step by Step

$$2y^2+(xy)^{1/3}=x^2+2$$ The graph of the curve is enclosed below. a) As the point $(1,1)$ lies in the curve, $P$ satisfies the equation. b) Find $dy/dx$ using implicit differentiation: $$4y\frac{dy}{dx}+\frac{1}{3}(xy)^{-2/3}\frac{d}{dx}(xy)=2x$$ $$4y\frac{dy}{dx}+\frac{1}{3}(xy)^{-2/3}(y+x\frac{dy}{dx})=2x$$ $$4y\frac{dy}{dx}+\frac{1}{3}(xy)^{-2/3}y+\frac{1}{3}(xy)^{-2/3}x\frac{dy}{dx}=2x$$ $$\frac{dy}{dx}\Big(4y+\frac{1}{3}(xy)^{-2/3}\Big)=2x-\frac{1}{3}(xy)^{-2/3}y$$ $$\frac{dy}{dx}=\frac{2x-\frac{1}{3}(xy)^{-2/3}y}{4y+\frac{1}{3}(xy)^{-2/3}}$$ - For $P(1,1)$: $$\frac{dy}{dx}=\frac{2\times1-\frac{1}{3}(1\times1)^{-2/3}\times1}{4\times1+\frac{1}{3}(1\times1)^{-2/3}}=\frac{2-\frac{1}{3}\times1}{4+\frac{1}{3}\times1}=\frac{\frac{5}{3}}{\frac{13}{3}}=\frac{5}{13}$$ c) The tangent line to the curve at $P(1,1)$ is $$y-1=\frac{5}{13}(x-1)$$ $$y-1=\frac{5}{13}x-\frac{5}{13}$$ $$y=\frac{5}{13}x+\frac{8}{13}$$
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