Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 197: 19


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

Work Step by Step

$y=\dfrac {t^{4}-1}{t^{4}+1}=\dfrac {t^{4}+1-2}{t^{4}+1}=1-\dfrac {2}{\left( t^{4}+1\right) }\Rightarrow y'=\dfrac {2\times \left( 4t^{3}\right) }{\left( t^{4}+1\right) ^{2}}=\dfrac {8t^{3}}{\left( t^{4}+1\right) ^{2}}$
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