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 164: 1



Work Step by Step

Apply rule of differentiation: $y'=a'(x)+b'(x)$ Here, $a'(x)=m'(x)n(x)+m(x)n'(x)$ Given: $x^2y+xy^2=\frac{6}{d/dx}$ $(x^2)'y+x^2.y'+x'y^2+x(y^2)'=0$ $2xy+2xyy'+y^2+x^2y'=0$ $2xyy'+x^2y'=-2xy-y^2$ $y'(2xy+x^2)=-2xy-y^2$ Hence, $y'=\dfrac{-2xy-y^2}{2xy+x^2}$
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