#### Answer

\[{f^,}\,\left( x \right) = \frac{{{{\cos }}\,\,\left[ {{{\tan }^{ - 1}}\ln \,\left( x \right)} \right]}}{{x\,\left( {1 + {{\ln }^2}x} \right)}}\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( x \right) = \sin \,\,\left[ {{{\tan }^{ - 1}}\,\left( {\ln x} \right)} \right] \hfill \\
\hfill \\
Use\,\,\,\,\frac{d}{{dx}}\,\,\left[ {{{\tan }^{ - 1}}u} \right] = \frac{{{u^,}}}{{1 + {u^2}}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \cos \,{\left( {{{\tan }^{ - 1}}\,\left( {\ln x} \right)\,\left( {{{\tan }^{ - 1}}\,\left( {\ln x} \right)} \right)} \right)^,} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \cos \,\left( {{{\tan }^{ - 1}}\,\left( {\ln x} \right)} \right)\,\left( {\frac{{\frac{1}{x}}}{{1 + {{\ln }^2}x}}} \right) \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{{{\cos }}\,\,\left[ {{{\tan }^{ - 1}}\ln \,\left( x \right)} \right]}}{{x\,\left( {1 + {{\ln }^2}x} \right)}} \hfill \\
\hfill \\
\end{gathered} \]