Answer
$$\frac{{dy}}{{dx}} = - \frac{{\left( {x\cot x + 1} \right)\csc x}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\csc x}}{x} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{\csc x}}{x}} \right] \cr
& {\text{use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{x \cdot {D_x}\left( {\csc x} \right) - \csc x \cdot {D_x}\left( x \right)}}{{{x^2}}} \cr
& {\text{use }}{D_x}\left( {\csc x} \right) = - \csc x\cot x \cr
& \frac{{dy}}{{dx}} = \frac{{x\left( { - \csc x\cot x} \right) - \csc x\left( 1 \right)}}{{{x^2}}} \cr
& {\text{multiply}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - x\csc x\cot x - \csc x}}{{{x^2}}} \cr
& {\text{factor}} \cr
& \frac{{dy}}{{dx}} = - \frac{{\left( {x\cot x + 1} \right)\csc x}}{{{x^2}}} \cr} $$
