Answer
\[{g^,}\,\left( z \right) = - \frac{1}{{{z^2} + 1}}\]
Work Step by Step
\[\begin{gathered}
g\,\left( z \right) = {\tan ^{ - 1}}\,\left( {\frac{1}{z}} \right) \hfill \\
\hfill \\
Use\,\,the\,\,formula\,\,\,\,\frac{d}{{dz}}\,\,\left[ {{{\tan }^{ - 1}}u} \right] = \frac{{{u^,}}}{{1 + {u^2}}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{g^,}\,\left( z \right) = \frac{{ - \frac{1}{{{z^2}}}}}{{1 + \,{{\left( {\frac{1}{z}} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{g^,}\,\left( z \right) = \frac{{ - \frac{1}{{{z^2}}}}}{{\frac{{{z^2} + 1}}{{{z^2}}}}} \hfill \\
\hfill \\
{g^,}\,\left( z \right) = - \frac{1}{{{z^2} + 1}} \hfill \\
\end{gathered} \]
