Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 8



Work Step by Step

$x^{3}-xy^{2}+y^{3}=1$ Differentiate each term: $(x^{3})'-(xy^{2})'+(y^{3})'=(1)'$ Use the product rule to find $(xy^{2})'$: $3x^{2}-[(x)(y^{2})'+(y^{2})(x)']+(3y^{2})(y')=0$ $3x^{2}-[(x)(2y)(y')+(y^{2})(1)]+(3y^{2})(y')=0$ $3x^{2}-2xyy'-y^{2}+3y^{2}y'=0$ Solve for $y'$: $-2xyy'+3y^{2}y'=-3x^{2}+y^{2}$ $y'(-2xy+3y^{2})=-3x^{2}+y^{2}$ $y'=\dfrac{-3x^{2}+y^{2}}{-2xy+3y^{2}}=\dfrac{y^{2}-3x^{2}}{3y^{2}-2xy}$
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