Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - Review - Exercises: 23

Answer

$\dfrac {1-2xy-y^{4}}{4xy^{3}+x^{2}-3} $

Work Step by Step

$xy^{4}+x^{2}y=x+3y\Rightarrow \dfrac {d}{dx}\left( xy^{4}\right) +\dfrac {d}{dx}\left( x^{2}y\right) =\dfrac {d}{dx}\left( x\right) +\dfrac {d}{dx}\left( 3y\right) \Rightarrow y^{4}+4xy^{3}\dfrac {dy}{dx}+2xy+x^{2}\dfrac {dy}{dx}=1+3\dfrac {dy}{dx}\Rightarrow \dfrac {dy}{dx}=\dfrac {1-2xy-y^{4}}{4xy^{3}+x^{2}-3} $
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