Answer
$$4t\left( {2{t^3} - 1} \right)\left( {{t^3} - 2} \right)\left\langle {3t\left( {{t^3} - 2} \right),1,0} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{u}}\left( t \right) = 2{t^3}{\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}} - 8{\bf{k}} \cr
& {\text{Calculate }}\frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^4} - 2t} \right)} \right].{\text{ Use }}\frac{d}{{dt}}\left[ {{\bf{u}}\left( {f\left( t \right)} \right)} \right] = {\bf{u}}'\left( {f\left( t \right)} \right)f'\left( t \right) \cr
& {\bf{u}}'\left( t \right) = 6{t^2}{\bf{i}} + 2t{\bf{j}} \cr
& {\bf{u}}'\left( {{t^4} - 2t} \right) = 6{\left( {{t^4} - 2t} \right)^2}{\bf{i}} + 2\left( {{t^4} - 2t} \right){\bf{j}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {{t^4} - 2t} \right] = 4{t^3} - 2 \cr
& {\text{Thus}}{\text{,}} \cr
& \frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^4} - 2t} \right)} \right] = \left[ {6{{\left( {{t^4} - 2t} \right)}^2}{\bf{i}} + 2\left( {{t^4} - 2t} \right){\bf{j}}} \right]\left( {4{t^3} - 2} \right) \cr
& {\text{Factoring}} \cr
& \frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^4} - 2t} \right)} \right] = 2\left( {4{t^3} - 2} \right)\left( {{t^4} - 2t} \right)\left[ {3\left( {{t^4} - 2t} \right){\bf{i}} + {\bf{j}}} \right] \cr
& \frac{d}{{dt}}\left[ {{\bf{u}}\left( {{t^4} - 2t} \right)} \right] = 4t\left( {2{t^3} - 1} \right)\left( {{t^3} - 2} \right)\left[ {3t\left( {{t^3} - 2} \right){\bf{i}} + {\bf{j}}} \right] \cr
& or \cr
& 4t\left( {2{t^3} - 1} \right)\left( {{t^3} - 2} \right)\left\langle {3t\left( {{t^3} - 2} \right),1,0} \right\rangle \cr} $$
