Answer
\[{f^,}\,\left( y \right) = - \frac{{2y}}{{{y^4} + 2{y^2} + 2}}\]
Work Step by Step
\[\begin{gathered}
f\,\left( y \right) = {\cot ^{ - 1}}\,\left( {\frac{1}{{{y^2} + 1}}} \right) \hfill \\
\hfill \\
Use\,\,\frac{d}{{dy}}\,\,\,\left[ {{{\cot }^{ - 1}}\,u} \right] = \frac{{ - {u^,}}}{{1 + {u^2}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{f^.}\,\left( y \right) = \frac{{ - \frac{{2y}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}}{{1 + \,{{\left( {\frac{1}{{{y^2} + 1}}} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( y \right) = \frac{{\frac{{ - 2y}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}}{{\frac{{\,{{\left( {{y^2} + 1} \right)}^2} + 1}}{{\,{{\left( {{y^2} + 1} \right)}^2}}}}} \hfill \\
\hfill \\
{f^,}\,\left( y \right) = - \frac{{2y}}{{{y^4} + 2{y^2} + 2}} \hfill \\
\hfill \\
\end{gathered} \]
