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