University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 204: 117


The slope of the curve is $-1/2$ at $(1,1)$ and not defined at $(1,-1)$.

Work Step by Step

$$x^3y^3+y^2=x+y$$ a) Find the derivative of the function using implicit differentiation: $$(3x^2y^3+3y^2x^3y')+2yy'=1+y'$$ $$3y^2x^3y'+2yy'-y'=1-3x^2y^3$$ $$y'(3x^3y^2+2y-1)=1-3x^2y^3$$ $$y'=\frac{1-3x^2y^3}{3x^3y^2+2y-1}$$ b) The slope of the curve at $(1,1)$ is $$y'=\frac{1-3\times1^2\times1^3}{3\times1^3\times1^2+2\times1-1}=\frac{1-3}{3+2-1}$$ $$y'=-\frac{2}{4}=-\frac{1}{2}$$ The slope of the curve at $(1,-1)$ is $$y'=\frac{1-3\times1^2\times(-1)^3}{3\times1^3\times(-1)^2+2\times(-1)-1}=\frac{1-3(-1)}{3\times1-2-1}$$ $$y'=\frac{1+3}{3-2-1}=\frac{4}{0}$$ This means the slope is not defined here.
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