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: 10

Answer

$y'=\dfrac{-y^2\sin(xy^2)}{2xy\sin(xy^2)+1}$

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