Answer
$$\left\langle {5t\sqrt t , - 6\sqrt t ,\frac{1}{{\sqrt t }}} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{v}}\left( t \right) = \left\langle {{t^2}, - 2t,1} \right\rangle {\text{ and }}g\left( t \right) = 2\sqrt t \cr
& {\text{Find the derivatives}} \cr
& {\bf{v}}'\left( t \right) = \left\langle {2t, - 2,0} \right\rangle {\text{ and }}g'\left( t \right) = \frac{1}{{\sqrt t }} \cr
& {\text{Calculate the derivative }}\frac{d}{{dt}}\left[ {g\left( t \right){\bf{v}}\left( t \right)} \right].{\text{ Using the product rule}} \cr
& \frac{d}{{dt}}\left[ {g\left( t \right){\bf{v}}\left( t \right)} \right] = g\left( t \right){\bf{v}}'\left( t \right) + {\bf{v}}\left( t \right)g'\left( t \right) \cr
& \cr
& {\text{Substitute }}g\left( t \right){\text{ and }}{\bf{v}}\left( t \right){\text{ and their derivatives}} \cr
& \frac{d}{{dt}}\left[ {g\left( t \right){\bf{v}}\left( t \right)} \right] = 2\sqrt t \left\langle {2t, - 2,0} \right\rangle + \frac{1}{{\sqrt t }}\left\langle {{t^2}, - 2t,1} \right\rangle \cr
& {\text{Simplify}} \cr
& \frac{d}{{dt}}\left[ {g\left( t \right){\bf{v}}\left( t \right)} \right] = \left\langle {4t\sqrt t , - 4\sqrt t ,0} \right\rangle + \left\langle {t\sqrt t , - 2\sqrt t ,\frac{1}{{\sqrt t }}} \right\rangle \cr
& \frac{d}{{dt}}\left[ {g\left( t \right){\bf{v}}\left( t \right)} \right] = \left\langle {5t\sqrt t , - 6\sqrt t ,\frac{1}{{\sqrt t }}} \right\rangle \cr} $$
