Calculus: Early Transcendentals (2nd Edition)

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