#### Answer

\[{k^,}\,\left( t \right) = 4{t^3} - 4t\]

#### Work Step by Step

\[\begin{gathered}
k\,\left( t \right) = \,{\left( {{t^2} - 1} \right)^2} \hfill \\
Write\,\,as\,a\,\,product\,\, \hfill \\
k\,\left( t \right) = \,\left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right) \hfill \\
Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{k^,}\,\left( t \right) \hfill \\
{k^,}\,\left( t \right) = \left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} + \left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} \hfill \\
{k^,}\,\left( t \right) = 2\left( {{t^2} - 1} \right)\,\left( {{t^2} - 1} \right){\,^,} \hfill \\
Then \hfill \\
{k^,}\,\left( t \right) = 2\,\left( {{t^2} - 1} \right)\,\left( {2t} \right) \hfill \\
Multiplying \hfill \\
{k^,}\,\left( t \right) = 4t\,\left( {{t^2} - 1} \right) \hfill \\
{k^,}\,\left( t \right) = 4{t^3} - 4t \hfill \\
\hfill \\
\end{gathered} \]