Answer
$$\frac{{dy}}{{dx}} = 2{e^{\cot x}}{\csc ^2}x$$
Work Step by Step
$$\eqalign{
& y = - 2{e^{\cot x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ { - 2{e^{\cot x}}} \right] \cr
& {\text{constant multiple rule}} \cr
& \frac{{dy}}{{dx}} = - 2{D_x}\left[ {{e^{\cot x}}} \right] \cr
& {\text{use the chain rule }}{D_x}\left( {{e^u}} \right) = {e^u} \cdot {D_x}\left( u \right).{\text{ consider }}u = \cot x \cr
& \frac{{dy}}{{dx}} = - 2{e^{\cot x}} \cdot {D_x}\left( {\cot x} \right) \cr
& {\text{recall that }}{D_x}\left( {\cot x} \right) = - {\csc ^2}x \cr
& \frac{{dy}}{{dx}} = - 2{e^{\cot x}}\left( { - {{\csc }^2}x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 2{e^{\cot x}}{\csc ^2}x \cr} $$
