Answer
$$\left\langle {4, - \frac{2}{{\sqrt t }},0} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{v}}\left( t \right) = \left\langle {{t^2}, - 2t,1} \right\rangle {\text{ and }}g\left( t \right) = 2\sqrt t \cr
& {\text{Calculate }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {g\left( t \right)} \right)} \right].{\text{ Use }}\frac{d}{{dt}}\left[ {{\bf{v}}\left( {g\left( t \right)} \right)} \right] = {\bf{v}}'\left( {g\left( t \right)} \right)g'\left( t \right) \cr
& {\bf{v}}'\left( t \right) = \left\langle {2t, - 2,0} \right\rangle \cr
& {\bf{v}}'\left( {g\left( t \right)} \right) = \left\langle {2\left( {2\sqrt t } \right), - 2,0} \right\rangle \cr
& {\bf{v}}'\left( {g\left( t \right)} \right) = \left\langle {4\sqrt t , - 2,0} \right\rangle \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {2\sqrt t } \right] = \frac{1}{{\sqrt t }} \cr
& {\text{Thus}}{\text{,}} \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {g\left( t \right)} \right)} \right] = \left\langle {4\sqrt t , - 2,0} \right\rangle \frac{1}{{\sqrt t }} \cr
& {\text{Multiplying}} \cr
& \frac{d}{{dt}}\left[ {{\bf{v}}\left( {g\left( t \right)} \right)} \right] = \left\langle {4, - \frac{2}{{\sqrt t }},0} \right\rangle \cr} $$
