Answer
\[\frac{{dy}}{{dx}} = \frac{{ - {{\csc }^2}x - {{\csc }^3}x + {{\cot }^2}x\csc x}}{{\,{{\left( {1 + {{\csc }^2}x} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{\cot x}}{{1 + \csc x}} \hfill \\
\hfill \\
{\text{using}}\,\,the\,\,quotient\,\,rule. \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\left( {1 + \csc x} \right)\left[ {\cot x} \right]' - \left( {\cot x} \right)\left[ {1 + \csc x} \right]'}}{{{{\left( {1 + \csc x} \right)}^2}}} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{\,\left( {1 + \csc x} \right)\,\left( { - {{\csc }^2}x} \right) - \cot x\,\left( { - \csc x\cot x} \right)}}{{\,{{\left( {1 + \csc x} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{ - {{\csc }^2}x - {{\csc }^3}x + {{\cot }^2}x\csc x}}{{\,{{\left( {1 + {{\csc }^2}x} \right)}^2}}} \hfill \\
\end{gathered} \]
