Answer
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {{{\rm{e}}^t}{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = \left( {{{\rm{e}}^3},3{{\rm{e}}^3},4{{\rm{e}}^3}} \right)$
Work Step by Step
$\frac{d}{{dt}}\left( {{{\rm{e}}^t}{{\bf{r}}_2}\left( t \right)} \right) = {{\rm{e}}^t}{{\bf{r}}_2}\left( t \right) + {{\rm{e}}^t}{{\bf{r}}_2}'\left( t \right)$
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {{{\rm{e}}^t}{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = {{\rm{e}}^3}{{\bf{r}}_2}\left( 3 \right) + {{\rm{e}}^3}{{\bf{r}}_2}'\left( 3 \right)$
$ = {{\rm{e}}^3}\left( {1,1,0} \right) + {{\rm{e}}^3}\left( {0,2,4} \right)$
$\frac{d}{{dt}}\left( {{{\rm{e}}^t}{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = \left( {{{\rm{e}}^3},3{{\rm{e}}^3},4{{\rm{e}}^3}} \right)$
