Answer
\[{f^,}\,\left( x \right) = \frac{1}{{2x\sqrt {x - 1} }}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {\sec ^{ - 1}}\sqrt x \hfill \\
\hfill \\
Use\,\,the\,\,formula\,\,\,\frac{d}{{dx}}\,\,\left[ {{{\sec }^{ - 1}}u} \right] = \frac{{{u^,}}}{{\left| u \right|\sqrt {{u^2} - 1} }} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{\,{{\left( {\sqrt x } \right)}^,}}}{{\left| {\sqrt x } \right|\sqrt {\,{{\left( {\sqrt x } \right)}^2} - 1} }} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{\frac{1}{{2\sqrt x }}}}{{\sqrt x \sqrt {x - 1} }} \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{1}{{2x\sqrt {x - 1} }} \hfill \\
\end{gathered} \]
