Answer
$g'\left( t \right) = \frac{{4{t^3}}}{{\left( {1 + {t^8}} \right)\arctan \left( {{t^4}} \right)}}$
Work Step by Step
$$\eqalign{
& g\left( t \right) = \ln \left( {\arctan \left( {{t^4}} \right)} \right) \cr
& {\text{Differentiate}}{\text{, apply }}\frac{d}{{dt}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dt}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {\ln \left( {\arctan \left( {{t^4}} \right)} \right)} \right] \cr
& g'\left( t \right) = \frac{1}{{\arctan \left( {{t^4}} \right)}}\frac{d}{{dt}}\left[ {\left( {\arctan \left( {{t^4}} \right)} \right)} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr
& g'\left( t \right) = \frac{1}{{\arctan \left( {{t^4}} \right)}}\left( {\frac{1}{{1 + {{\left( {{t^4}} \right)}^2}}}} \right)\frac{d}{{dt}}\left[ {{t^4}} \right] \cr
& {\text{Compute derivative}} \cr
& g'\left( t \right) = \frac{1}{{\arctan \left( {{t^4}} \right)}}\left( {\frac{1}{{1 + {t^8}}}} \right)\left( {4{t^3}} \right) \cr
& {\text{Simplify}} \cr
& g'\left( t \right) = \frac{{4{t^3}}}{{\left( {1 + {t^8}} \right)\arctan \left( {{t^4}} \right)}} \cr} $$
