#### Answer

\[ = - \csc x\cot x\]

#### Work Step by Step

\[\begin{gathered}
\hfill \\
\frac{d}{{dx}}\,\left( {\csc x} \right) = - \csc x\cot x \hfill \\
\hfill \\
use\,\,trigonometric\,\,identity\,\,\csc x = \frac{1}{{\sin x}} \hfill \\
then \hfill \\
\hfill \\
\frac{d}{{dx}}\,\left( {\frac{1}{{\sin x}}} \right) = \frac{{0 - \,\left( {\cos x} \right)}}{{{{\sin }^2}x}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \,\left( { - \frac{{\cos x}}{{\sin x}}} \right)\,\left( {\frac{1}{{\sin x}}} \right) \hfill \\
\hfill \\
= - \csc x\cot x \hfill \\
\hfill \\
\end{gathered} \]