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: 5


$y' = \dfrac{1-2xy-3y^3}{x^2+9xy^2}$

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 $x^2y+3xy^3-x-3=0$ Then derivate the whole equation. Rember the put y' every time you derivate y $2xy+x^2y'+3y^3+9xy^2y'-1=0$ *Note: Here you have to apply the product rule twice Solve for y' and you have the answer $y'(x^2+9xy^2) = 1-2xy-3y^3$ $y' = \dfrac{1-2xy-3y^3}{x^2+9xy^2}$
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