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: 60


At $P(1,0)$, the graph has a vertical tangent $x=1$.

Work Step by Step

$$y^3+\cos xy=x^2$$ The graph of the curve is enclosed below. a) As the point $(1,0)$ lies in the curve, $P$ satisfies the equation. b) Find $dy/dx$ using implicit differentiation: $$3y^2+(-\sin xy)\frac{d}{dy}(xy)=2x$$ $$3y^2-\sin xy(y+x\frac{dy}{dx})=2x$$ $$3y^2-y\sin xy-x\sin xy\frac{dy}{dx}=2x$$ $$x\sin xy\frac{dy}{dx}=3y^2-y\sin xy-2x$$ $$\frac{dy}{dx}=\frac{3y^2-y\sin xy-2x}{x\sin xy}$$ - For $P(1,0)$: $$\frac{dy}{dx}=\frac{3\times0^2-0\times\sin(1\times0)-2\times1}{1\times\sin(1\times0)}=\frac{0-0-2}{1\times\sin0}=-\frac{2}{0}$$ $dy/dx$ is not defined for $P(1,0)$. c) Since $dy/dx$ is not defined for $P(1,0)$ and from the graph of the curve, we see that at $(1,0)$, the curve has a vertical tangent $x=1$.
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