Answer
$$f'\left( t \right) = \frac{{\left( {2{t^{3/2}} + 1} \right)\left( {3{t^{1/2}}} \right)\left( {\ln \left( {2{t^{3/2}} + 1} \right)} \right) - 6{t^2}}}{{\left( {2{t^{3/2}} + 1} \right){{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( t \right) = \frac{{2{t^{3/2}}}}{{\ln \left( {2{t^{3/2}} + 1} \right)}} \cr
& {\text{differentiate with respect to }}t \cr
& f'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{2{t^{3/2}}}}{{\ln \left( {2{t^{3/2}} + 1} \right)}}} \right) \cr
& {\text{use quotient rule}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {2{t^{3/2}} + 1} \right)} \right)\frac{d}{{dt}}\left( {2{t^{3/2}}} \right) - 2{t^{3/2}}\frac{d}{{dt}}\left( {\ln \left( {2{t^{3/2}} + 1} \right)} \right)}}{{{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr
& {\text{use }}\frac{d}{{dt}}\ln g\left( t \right) = \frac{{g'\left( t \right)}}{{g\left( t \right)}}{\text{ and the power rule for derivatives}} \cr
& f'\left( t \right) = \frac{{\left( {\ln \left( {2{t^{3/2}} + 1} \right)} \right)\left( {3{t^{1/2}}} \right) - 2{t^{3/2}}\left( {\frac{{3{t^{1/2}}}}{{2{t^{3/2}} + 1}}} \right)}}{{{{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& f'\left( t \right) = \frac{{\left( {2{t^{3/2}} + 1} \right)\left( {3{t^{1/2}}} \right)\left( {\ln \left( {2{t^{3/2}} + 1} \right)} \right) - 6{t^2}}}{{\left( {2{t^{3/2}} + 1} \right){{\left( {\ln \left( {{t^2} + 1} \right) + 1} \right)}^2}}} \cr} $$
