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: 58

Answer

The tangent line to the curve at $P(1,1)$ is $y=-x+2$.
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Work Step by Step

$$x^5+y^3x+yx^2+y^4=4$$ 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: $$5x^4+(3y^2x\frac{dy}{dx}+y^3)+(x^2\frac{dy}{dx}+2yx)+4y^3\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(3y^2x+x^2+4y^3)=-(5x^4+y^3+2yx)$$ $$\frac{dy}{dx}=-\frac{5x^4+y^3+2yx}{3y^2x+x^2+4y^3}$$ For $P(1,1)$: $$\frac{dy}{dx}=-\frac{5\times1^4+1^3+2\times1\times1}{3\times1^2\times1+1^2+4\times1^3}=-\frac{5+1+2}{3+1+4}=-\frac{8}{8}=-1$$ c) The tangent line to the curve at $P(1,1)$ is $$y-1=-(x-1)$$ $$y-1=-x+1$$ $$y=-x+2$$ The tangent line is also enclosed below.
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