Answer
${\bf{r}}'\left( 0 \right) = \left( {2,0,6} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^{2t}},{{\rm{e}}^{ - 4{t^2}}},{{\rm{e}}^{6t}}} \right)$. So,
${\bf{r}}'\left( t \right) = \left( {2{{\rm{e}}^{2t}}, - 8t{{\rm{e}}^{ - 4{t^2}}},6{{\rm{e}}^{6t}}} \right)$
${\bf{r}}'\left( 0 \right) = \left( {2,0,6} \right)$
