Answer
\[ = - \frac{1}{{{{\sin }^2}x}} = - {\csc ^2}x\]
Work Step by Step
\[\begin{gathered}
show\,\,that\,\,\frac{d}{{dx}}\,\left( {\cot x} \right) = - {\csc ^2}x \hfill \\
\hfill \\
use\,\,the\,\,identity\,\,\,\cot \,x = \frac{{\cos x}}{{\sin x}} \hfill \\
\hfill \\
Use\,\,the\,\,quotiente\,\,rule \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {\frac{{\cos x}}{{\sin x}}} \right] = \frac{{\,\left( {\sin x} \right)\,\left( {\cos x} \right)' - \,\left( {\cos x} \right)\,\left( {\sin x} \right)'}}{{{{\sin }^2}x}} \hfill \\
\hfill \\
then \hfill \\
\frac{d}{{dx}}\,\,\left[ {\frac{{\cos x}}{{\sin x}}} \right] = \frac{{\,\left( {\sin x} \right)\,\left( { - \sin x} \right) - \,\left( {\cos x} \right)\,\left( {\cos x} \right)}}{{{{\sin }^2}x}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {\cot x} \right] = \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}} \hfill \\
\hfill \\
= - \frac{1}{{{{\sin }^2}x}} = - {\csc ^2}x \hfill \\
\end{gathered} \]
