Answer
$g'\left( t \right) = \frac{{2t}}{{3 + {t^2}}}$
Work Step by Step
$$\eqalign{
& g\left( t \right) = \ln \left( {3 + {t^2}} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {\ln \left( {3 + {t^2}} \right)} \right] \cr
& {\text{Use the formula }}\frac{d}{{dt}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dt}},{\text{ }}{\text{ }} \cr
& {\text{Let }}u = 3 + {t^2},{\text{ then}} \cr
& g'\left( t \right) = \frac{1}{{3 + {t^2}}}\frac{d}{{dt}}\left[ {3 + {t^2}} \right] \cr
& g'\left( t \right) = \frac{1}{{3 + {t^2}}}\left( {2t} \right) \cr
& {\text{simplify}} \cr
& g'\left( t \right) = \frac{{2t}}{{3 + {t^2}}} \cr} $$
