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: 59

Answer

The tangent line to the curve at $P(0,1)$ is $y=x+1$.

Work Step by Step

$$y^2+y=\frac{2+x}{1-x}$$ The graph of the curve is enclosed below. a) As the point $(0,1)$ lies in the curve, $P$ satisfies the equation. b) Find $dy/dx$ using implicit differentiation: $$2y\frac{dy}{dx}+\frac{dy}{dx}=\frac{(2+x)'(1-x)-(1-x)'(2+x)}{(1-x)^2}=\frac{1(1-x)-(-1)(2+x)}{(1-x)^2}$$ $$2y\frac{dy}{dx}+\frac{dy}{dx}=\frac{1-x+(2+x)}{(1-x)^2}$$ $$\frac{dy}{dx}(2y+1)=\frac{3}{(1-x)^2}$$ $$\frac{dy}{dx}=\frac{3}{(1-x)^2(2y+1)}$$ For $P(0,1)$: $$\frac{dy}{dx}=\frac{3}{(1-0)^2(2\times1+1)}=\frac{3}{1^2\times3}=1$$ c) The tangent line to the curve at $P(0,1)$ is $$y-1=1(x-0)$$ $$y-1=x$$ $$y=x+1$$ The tangent line is also enclosed below.
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