#### Answer

\[{f^,}\,\left( x \right) = \frac{1}{{x\sqrt {{x^2} - 1} }}\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( x \right) = {\cos ^{ - 1}}\,\left( {\frac{1}{x}} \right) \hfill \\
\hfill \\
use\,\,\frac{d}{{dx}}\,\,\left[ {{{\cos }^{ - 1}}u} \right] = \frac{{ - {u^,}}}{{\sqrt {1 - {u^2}} }} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{ - \,\left( { - \frac{1}{{{x^2}}}} \right)}}{{\sqrt {1 - \,{{\left( {\frac{1}{x}} \right)}^2}} }} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{\frac{1}{{{x^2}}}}}{{\frac{{\sqrt {{x^2} - 1} }}{x}}} \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{1}{{x\sqrt {{x^2} - 1} }} \hfill \\
\end{gathered} \]