Answer
$$\frac{d}{{dx}}\left[ {\cot x} \right] = - {\csc ^2}x$$
Work Step by Step
$$\eqalign{
& \cot x = \frac{{\cos x}}{{\sin x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{d}{{dx}}\left[ {\frac{{\cos x}}{{\sin x}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{\left( {\sin x} \right)\frac{d}{{dx}}\left[ {\cos x} \right] - \cos x\frac{d}{{dx}}\left[ {\sin x} \right]}}{{{{\left( {\sin x} \right)}^2}}} \cr
& {\text{then}} \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{\left( {\sin x} \right)\left( { - \sin x} \right) - \cos x\left( {\cos x} \right)}}{{{{\left( {\sin x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = \frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\left( {\sin x} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = - \frac{1}{{{{\left( {\sin x} \right)}^2}}} \cr
& {\text{use }}\csc x = \frac{1}{{\sin x}} \cr
& \frac{d}{{dx}}\left[ {\cot x} \right] = - {\csc ^2}x \cr} $$
