Answer
\[\left\langle {30{t^{14}} + 24{t^3} + 3,14{t^{13}} - 12{t^{11}} + 9{t^2} - 3, - 96{t^{11}} - 24} \right\rangle \]
Work Step by Step
$$\eqalign{
& {\text{Let }}{\bf{u}}\left( t \right) = 2{t^3}{\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}} - 8{\bf{k}} \cr
& \left( {{t^{12}} + 3t} \right){\bf{u}}\left( t \right) \cr
& {\text{Calculate the derivative}} \cr
& \frac{d}{{dt}}\left[ {\left( {{t^{12}} + 3t} \right){\bf{u}}\left( t \right)} \right] \cr
& {\text{Use the product rule}} \cr
& = {\bf{u}}\left( t \right)\left( {12{t^{11}} + 3} \right) + \left( {{t^{12}} + 3t} \right){\bf{u}}'\left( t \right) \cr
& = \left( {2{t^3}{\bf{i}} + \left( {{t^2} - 1} \right){\bf{j}} - 8{\bf{k}}} \right)\left( {12{t^{11}} + 3} \right) + \left( {{t^{12}} + 3t} \right)\left( {6{t^2}{\bf{i}} + 2t{\bf{j}}} \right) \cr
& = \left( {24{t^{14}} + 6{t^3}} \right){\bf{i}} + \left( {12{t^{13}} + 3{t^2} - 12{t^{11}} - 3} \right){\bf{j}} - \left( {96{t^{11}} - 24} \right){\bf{k}} \cr
& \,\,\,\, + \left( {6{t^{14}} + 18{t^3}} \right){\bf{i}} + \left( {2{t^{13}} + 6{t^2}} \right){\bf{j}} \cr
& {\text{Combining like terms}} \cr
& = \left( {24{t^{14}} + 6{t^3} + 6{t^{14}} + 18{t^3}} \right){\bf{i}} \cr
& + \left( {12{t^{13}} + 3{t^2} - 12{t^{11}} - 3 + 2{t^{13}} + 6{t^2}} \right){\bf{j}} - \left( {96{t^{11}} + 24} \right){\bf{k}} \cr
& = \left( {30{t^{14}} + 24{t^3}} \right){\bf{i}} + \left( {14{t^{13}} - 12{t^{11}} + 9{t^2} - 3} \right){\bf{j}} - \left( {96{t^{11}} + 24} \right){\bf{k}} \cr
& or \cr
& \left\langle {30{t^{14}} + 24{t^3} + 3,14{t^{13}} - 12{t^{11}} + 9{t^2} - 3, - 96{t^{11}} - 24} \right\rangle \cr} $$
