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: 5



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: $x^2(x-y)^2=x^2-\frac{y^2}{d/dx}$ $(x^2)'(x-y)^2+x^2((x-y)^2)'=2x-2yy'$ $2x(x-y)^2+2x^2(x-y)(x-y)'=2x-2yy'$ This implies, $2yy'-2x^2(x-y)y'=2x-2x(x-y)^2-2x^2(x-y)$ $y'(2y-2x^2(x-y))=2x-2x(x-y)^2-2x^2(x-y)$ $y'=\dfrac{2x-2x(x-y)^2-2x^2(x-y}{2y-2x^2(x-y)}$ Hence, $y'=\dfrac{-2x^3+3x^2y-xy^2+x}{y-x^3+x^2y}$
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