Answer
$$\frac{1}{{2\sqrt t }}\left\langle {{e^t}, - 2{e^{ - t}}, - 2{e^{2t}}} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{v}}\left( t \right) = {e^t}{\bf{i}} + 2{e^{ - t}}{\bf{j}} - {e^{2t}}{\bf{k}} \cr
& {\text{Calculate }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {\sqrt t } \right)} \right].{\text{ Use }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {f\left( t \right)} \right)} \right] = {\bf{v}}'\left( {f\left( t \right)} \right)f'\left( t \right) \cr
& {\bf{v}}'\left( t \right) = {e^t}{\bf{i}} - 2{e^{ - t}}{\bf{j}} - 2{e^{2t}}{\bf{k}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {\sqrt t } \right] = \frac{1}{{2\sqrt t }} \cr
& {\text{Thus}}{\text{,}} \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {\sqrt t } \right)} \right] = \left( {{e^t}{\bf{i}} - 2{e^{ - t}}{\bf{j}} - 2{e^{2t}}{\bf{k}}} \right)\left( {\frac{1}{{2\sqrt t }}} \right) \cr
& or \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {\sqrt t } \right)} \right] = \frac{1}{{2\sqrt t }}\left\langle {{e^t}, - 2{e^{ - t}}, - 2{e^{2t}}} \right\rangle \cr} $$
