Answer
$$\frac{d}{{dx}}\left[ {\csc x} \right] = - \csc x\cot x$$
Work Step by Step
$$\eqalign{
& \csc x = \frac{1}{{\sin x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = \frac{d}{{dx}}\left[ {\frac{1}{{\sin x}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = \frac{{\left( {\sin x} \right)\frac{d}{{dx}}\left[ 1 \right] - \frac{d}{{dx}}\left[ {\sin x} \right]}}{{{{\left( {\sin x} \right)}^2}}} \cr
& {\text{then}} \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = \frac{{\left( {\sin x} \right)\left( 0 \right) - \left( {\cos x} \right)}}{{{{\left( {\sin x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = - \frac{{\cos x}}{{{{\left( {\sin x} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = - \left( {\frac{1}{{\sin x}}} \right)\left( {\frac{{\cos x}}{{\sin x}}} \right) \cr
& \frac{d}{{dx}}\left[ {\csc x} \right] = - \csc x\cot x \cr} $$