Answer
\[\begin{gathered}
\left( { - 2t{e^{2t}} - 2{t^2}{e^{2t}} + 2{e^{2t}} - 16{e^{ - t}}} \right){\mathbf{i}}\, + \left( {6{t^2}{e^{2t}} + 4{t^3}{e^{2t}} - 8{e^t}} \right){\mathbf{j}} \hfill \\
\, + \left( {12{t^2}{e^{ - t}} - 2t{e^t} - 4{t^3}{e^{ - t}} - {t^2}{e^t} + {e^t}} \right){\mathbf{k}} \hfill \\
\end{gathered} \]
Work Step by Step
\[\begin{gathered}
{\text{Let }}{\mathbf{u}}\left( t \right) = 2{t^3}{\mathbf{i}} + \left( {{t^2} - 1} \right){\mathbf{j}} - 8{\mathbf{k}}{\text{ and }}{\mathbf{v}}\left( t \right) = {e^t}{\mathbf{i}} + 2{e^{ - t}}{\mathbf{j}} - {e^{2t}}{\mathbf{k}} \hfill \\
{\text{Calculate the derivative }}{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right){\text{using the cross product rule}} \hfill \\
\frac{d}{{dt}}\left( {{\mathbf{u}}\left( t \right) \times {\mathbf{v}}\left( t \right)} \right) = {\mathbf{u}}'\left( t \right) \times {\mathbf{v}}\left( t \right) + {\mathbf{u}}\left( t \right) \times {\mathbf{v}}'\left( t \right) \hfill \\
{\mathbf{u}}'\left( t \right) = 6{t^2}{\mathbf{i}} + 2t{\mathbf{j}} \hfill \\
{\mathbf{v}}'\left( t \right) = {e^t}{\mathbf{i}} - 2{e^{ - t}}{\mathbf{j}} - 2{e^{2t}}{\mathbf{k}} \hfill \\
\hfill \\
{\mathbf{u}}'\left( t \right) \times {\mathbf{v}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{6{t^2}}&{2t}&0 \\
{{e^t}}&{2{e^{ - t}}}&{ - {e^{2t}}}
\end{array}} \right| \hfill \\
{\mathbf{u}}'\left( t \right) \times {\mathbf{v}}\left( t \right) = \left| {\begin{array}{*{20}{c}}
{2t}&0 \\
{2{e^{ - t}}}&{ - {e^{2t}}}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{6{t^2}}&0 \\
{{e^t}}&{ - {e^{2t}}}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
{6{t^2}}&{2t} \\
{{e^t}}&{2{e^{ - t}}}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}}'\left( t \right) \times {\mathbf{v}}\left( t \right) = - 2t{e^{2t}}{\mathbf{i}} + 6{t^2}{e^{2t}}{\mathbf{j}} + \left( {12{t^2}{e^{ - t}} - 2t{e^t}} \right){\mathbf{k}} \hfill \\
\hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}'\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{2{t^3}}&{{t^2} - 1}&{ - 8} \\
{{e^t}}&{ - 2{e^{ - t}}}&{ - 2{e^{2t}}}
\end{array}} \right| \hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}'\left( t \right) = \left| {\begin{array}{*{20}{c}}
{{t^2} - 1}&{ - 8} \\
{ - 2{e^{ - t}}}&{ - 2{e^{2t}}}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
{2{t^3}}&{ - 8} \\
{{e^t}}&{ - 2{e^{2t}}}
\end{array}} \right|{\mathbf{j}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left| {\begin{array}{*{20}{c}}
{2{t^3}}&{{t^2} - 1} \\
{{e^t}}&{ - 2{e^{ - t}}}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{u}}\left( t \right) \times {\mathbf{v}}'\left( t \right) = \left( { - 2{t^2}{e^{2t}} + 2{e^{2t}} - 16{e^{ - t}}} \right){\mathbf{i}} - \left( { - 4{t^3}{e^{2t}} + 8{e^t}} \right){\mathbf{j}} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( { - 4{t^3}{e^{ - t}} - {t^2}{e^t} + {e^t}} \right){\mathbf{k}} \hfill \\
\hfill \\
{\mathbf{u}}'\left( t \right) \times {\mathbf{v}}\left( t \right) + {\mathbf{u}}\left( t \right) \times {\mathbf{v}}'\left( t \right) \hfill \\
= \left( { - 2t{e^{2t}} - 2{t^2}{e^{2t}} + 2{e^{2t}} - 16{e^{ - t}}} \right){\mathbf{i}}\, + \left( {6{t^2}{e^{2t}} + 4{t^3}{e^{2t}} - 8{e^t}} \right){\mathbf{j}} \hfill \\
\, + \left( {12{t^2}{e^{ - t}} - 2t{e^t} - 4{t^3}{e^{ - t}} - {t^2}{e^t} + {e^t}} \right){\mathbf{k}} \hfill \\
\end{gathered} \]
