Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.7 Implicit Differentiation - Exercises Set 2.7 - Page 166: 4


$y' = \dfrac{3y^2-3x^2}{3y^2-6xy}$

Work Step by Step

In order to derivate this function you have to apply implicit differentation method. First, take the function to it's $f(x) = 0$ form $x^3+y^3-3xy= 0$ Then derivate the whole equation. Rember the put y' every time you derivate y $3x^2+3y^2y' - (3y^2+6xyy')= 0$ *Note: Here you have to apply the product rule Solve for y' and you have the answer $y'(3y^2-6xy) = 3y^2-3x^2$ $y' = \dfrac{3y^2-3x^2}{3y^2-6xy}$
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