Answer
$$\frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4\left( {3\cos x - {e^x}} \right)$$
Work Step by Step
$$\eqalign{
& y = {4^{3\sin x - {e^x}}} \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{4^{3\sin x - {e^x}}}} \right] \cr
& {\text{use the formula }}\frac{d}{{dx}}\left[ {{b^u}} \right] = {b^u}\ln b \cdot \frac{{du}}{{dx}}\,\,\,\left( {See\,\,page\,\,\,429} \right) \cr
& {\text{consider }}b = 4{\text{ and }}u = 3\sin x - {e^x}.\,{\text{Thus}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4 \cdot \frac{d}{{dx}}\left[ {3\sin x - {e^x}} \right] \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4\left( {3\cos x - {e^x}} \right) \cr} $$