Answer
$$\frac{{dy}}{{dx}} = 4{e^{4x}}\cos {e^{4x}}$$
Work Step by Step
$$\eqalign{
& y = \sin {e^{4x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\sin {e^{4x}}} \right] \cr
& {\text{use the chain rule }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right).{\text{ consider }}u = {e^{4x}} \cr
& \frac{{dy}}{{dx}} = \cos {e^{4x}} \cdot {D_x}\left( {{e^{4x}}} \right) \cr
& {\text{use the chain rule }}{D_x}\left( {{e^u}} \right) = {e^u} \cdot {D_x}\left( u \right) \cr
& \frac{{dy}}{{dx}} = \cos {e^{4x}}\left( {{e^{4x}}} \right){D_x}\left( {4x} \right) \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = \cos {e^{4x}}\left( {{e^{4x}}} \right)\left( 4 \right) \cr
& \frac{{dy}}{{dx}} = 4{e^{4x}}\cos {e^{4x}} \cr} $$