Answer
\[{f^,}\,\left( t \right) = - \frac{{2{{\cos }^{ - 1}}t}}{{\sqrt {1 - {u^2}} }}\]
Work Step by Step
\[\begin{gathered}
f\,\left( t \right) = \,{\left( {{{\cos }^{ - 1}}t} \right)^2} \hfill \\
\hfill \\
Use\,\,\frac{d}{{dt}}\,\,\left[ {{{\cos }^{ - 1}}u} \right] = \frac{{ - {u^,}}}{{\sqrt {1 - {u^2}} }} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{f^,}\,\left( t \right) = 2\,\left( {{{\cos }^{ - 1}}\,t} \right)\,{\left( {{{\cos }^{ - 1}}t} \right)^,} \hfill \\
\hfill \\
{f^,}\,\left( t \right) = 2\,\left( {{{\cos }^{ - 1}}\,t} \right)\,\left( {\frac{{ - 1}}{{\sqrt {1 - {u^2}} }}} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( t \right) = - \frac{{2{{\cos }^{ - 1}}t}}{{\sqrt {1 - {u^2}} }} \hfill \\
\end{gathered} \]
