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: 9


$y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$

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 $\sin(x^2y^2)-x=0$ Then derivate the whole equation. Rember the put y' every time you derivate y $(2xy^2+2x^2yy')\cos(x^2y^2)-1=0$ *Note: In this one, you have to apply the chain rule Solve for y' and you have the answer $y'(2x^2y\cos (x^2y^2))=1-2xy^2 \cos(x^2y^2)$ $y' = \dfrac{1-2xy^2 \cos(x^2y^2)}{2x^2y\cos (x^2y^2)}$
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