Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 7


$\dfrac {8t^{3}}{(t^{4}+1)^2}$

Work Step by Step

$\dfrac {d}{dx}\dfrac {t^{4}-1}{t^{4}+1}=\dfrac {d}{dx}\left( 1-\dfrac {2}{t^{4}+1}\right) =\dfrac {2}{\left( t^{4}+1\right) ^{2}}\times \dfrac {d}{dx}\left( t^{4}+1\right) =\dfrac {8t^{3}}{(t^{4}+1)^2}$
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