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