Answer
\[\left\langle {1 + 6{t^2},4{t^3}, - 2 - 3{t^2}} \right\rangle \]
Work Step by Step
\[\begin{gathered}
{\text{Let }}{\mathbf{u}}\left( t \right) = \left\langle {1,t,{t^2}} \right\rangle {\text{ and }}{\mathbf{v}}\left( t \right) = \left\langle {{t^2}, - 2t,1} \right\rangle \hfill \\
\hfill \\
{\text{Calculate }}{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right) \hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&t&{{t^2}} \\
{{t^2}}&{ - 2t}&1
\end{array}} \right| \hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
t&{{t^2}} \\
{ - 2t}&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&{{t^2}} \\
{{t^2}}&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&t \\
{{t^2}}&{ - 2t}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right) = \left( {t + 2{t^3}} \right){\mathbf{i}} - \left( {1 - {t^4}} \right){\mathbf{j}} + \left( { - 2t - {t^3}} \right){\mathbf{k}} \hfill \\
\hfill \\
{\text{Calculate the derivative}} \hfill \\
\frac{d}{{dt}}\left[ {{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right)} \right] = \left( {1 + 6{t^2}} \right){\mathbf{i}} - \left( { - 4{t^3}} \right){\mathbf{j}} + \left( {} \right){\mathbf{k}} \hfill \\
\frac{d}{{dt}}\left[ {{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right)} \right] = \left( {1 + 6{t^2}} \right){\mathbf{i}} + 4{t^3}{\mathbf{j}} - \left( {2 + 3{t^2}} \right){\mathbf{k}} \hfill \\
or \hfill \\
\frac{d}{{dt}}\left[ {{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right)} \right] = \left\langle {1 + 6{t^2},4{t^3}, - 2 - 3{t^2}} \right\rangle \hfill \\
\end{gathered} \]
