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: 2



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: $x^3+y^3=\frac{18xy}{d/dx}$ $3x^2+3y^2y'=18(x'y+xy')$ $3x^2+3y^2y'=18(y+xy')$ $3x^2+3y^2y'=18y+18xy'$ $-18xy'+3y^2y'=18y-3x^2$ $y'=\dfrac{18y-3x^2}{3y^2-18x}$ Hence, $y'=\dfrac{6y-x^2}{y^2-6x}$
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