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: 4



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: $x^3-xy+y^3=\frac{1}{d/dx}$ $3x^2-(x'y+xy')+3y^2y'=0$ This implies, $3x^2-y-xy'+3y^2y'=0$ $y'(3y^2-x)=y-3x^2$ Hence, $y'=\dfrac{(y-3x^2)}{(3y^2-x)}$
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