Answer
$\dfrac {1-t^{2}}{\left( 1+t^{2}\right) ^{2}}sec^{2}\dfrac {t}{1+t^{2}}$
Work Step by Step
$\dfrac {d}{dt}\tan \left( \dfrac {t}{1+t^{2}}\right) =\dfrac {1}{\cos ^{2}\left( \dfrac {t}{1+t^{2}}\right) }\times \dfrac {d}{dt}\left( \dfrac {t}{1+t^{2}}\right) =sec^2\dfrac {t}{1+t^{2}}\times \dfrac {\left( \dfrac {d}{dt}\left( t\right) \right) \times \left( 1+t^{2}\right) -\dfrac {d}{dt}\left( 1+t^{2}\right) \times t}{\left( 1+t^{2}\right) ^{2}}= \dfrac {1-t^{2}}{\left( 1+t^{2}\right) ^{2}}sec^{2}\dfrac {t}{1+t^{2}}$
