Answer
$$f'\left( t \right) = \frac{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)\left( {2t + {t^2} + 1} \right) - 2t\left( {\ln \left( {{t^2} + 1} \right) + t} \right)}}{{\left( {{t^2} + 1} \right){{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( t \right) = \frac{{\ln \left( {{t^2} + 1} \right) + t}}{{\ln \left( {{t^2} + 1} \right) + 1}} \cr
& {\text{differentiate with respect to }}t \cr
& f'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{\ln \left( {{t^2} + 1} \right) + t}}{{\ln \left( {{t^2} + 1} \right) + 1}}} \right) \cr
& {\text{use quotient rule}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)\frac{d}{{dt}}\left( {\ln \left( {{t^2} + 1} \right) + t} \right) - \left( {\ln \left( {{t^2} + 1} \right) + t} \right)\frac{d}{{dt}}\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}}{{{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr
& {\text{use }}\frac{d}{{dt}}\ln g\left( t \right) = \frac{{g'\left( t \right)}}{{g\left( t \right)}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)\left( {\frac{{2t}}{{{t^2} + 1}} + 1} \right) - \left( {\ln \left( {{t^2} + 1} \right) + t} \right)\left( {\frac{{2t}}{{{t^2} + 1}}} \right)}}{{{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)\left( {\frac{{2t + {t^2} + 1}}{{{t^2} + 1}}} \right) - \left( {\ln \left( {{t^2} + 1} \right) + t} \right)\left( {\frac{{2t}}{{{t^2} + 1}}} \right)}}{{{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)\left( {2t + {t^2} + 1} \right) - 2t\left( {\ln \left( {{t^2} + 1} \right) + t} \right)}}{{\left( {{t^2} + 1} \right){{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr} $$
