Answer
$\frac{d}{{dt}}{{\rm{e}}^t}\left( {1,t,{t^2}} \right) = {{\rm{e}}^t}\left( {1,t + 1,{t^2} + 2t} \right)$
Work Step by Step
$\frac{d}{{dt}}{{\rm{e}}^t}\left( {1,t,{t^2}} \right) = \left( {\frac{d}{{dt}}{{\rm{e}}^t}} \right)\left( {1,t,{t^2}} \right) + {{\rm{e}}^t}\frac{d}{{dt}}\left( {1,t,{t^2}} \right)$
$ = {{\rm{e}}^t}\left( {1,t,{t^2}} \right) + {{\rm{e}}^t}\left( {0,1,2t} \right)$
$\frac{d}{{dt}}{{\rm{e}}^t}\left( {1,t,{t^2}} \right) = {{\rm{e}}^t}\left( {1,t + 1,{t^2} + 2t} \right)$
