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