#### Answer

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

#### Work Step by Step

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