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: 7

Answer

$y'=-\dfrac{4x^{3}+2xy^{2}}{2x^{2}y+3y^{2}}$

Work Step by Step

$x^{4}+x^{2}y^{2}+y^{3}=5$ Differentiate each term: $(x^{4})'+(x^{2}y^{2})'+(y^{3})'=(5)'$ Use the product rule to find $(x^{2}y^{2})'$: $4x^{3}+(x^{2})(y^{2})'+(y^{2})(x^{2})'+(3y^{2})(y')=0$ $4x^{3}+(x^{2})(2y)(y')+(y^{2})(2x)+(3y^{2})(y')=0$ Solve for $y'$: $2x^{2}yy'+3y^{2}y'=-4x^{3}-2xy^{2}$ $y'(2x^{2}y+3y^{2})=-4x^{3}-2xy^{2}$ $y'=\dfrac{-4x^{3}-2xy^{2}}{2x^{2}y+3y^{2}}=-\dfrac{4x^{3}+2xy^{2}}{2x^{2}y+3y^{2}}$
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