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: 6

Answer

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

Work Step by Step

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