Answer
$$\frac{{dy}}{{dx}} = \frac{{\left( {{2^{\cos x + \ln x}}\ln 2} \right)\left( {1 - x\sin x} \right)}}{x}$$
Work Step by Step
$$\eqalign{
& y = {2^{\cos x + \ln x}} \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{2^{\cos x + \ln x}}} \right] \cr
& {\text{use the formula }}\frac{d}{{dx}}\left[ {{b^u}} \right] = {b^u}\ln b \cdot \frac{{du}}{{dx}}\,\,\,\left( {See\,\,page\,\,\,429} \right) \cr
& {\text{consider }}b = 2{\text{ and }}u = \cos x + \ln x.\,{\text{Thus}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = {2^{\cos x + \ln x}}\ln 2 \cdot \frac{d}{{dx}}\left[ {\cos x + \ln x} \right] \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = {2^{\cos x + \ln x}}\ln 2 \cdot \left( { - \sin x + \frac{1}{x}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = {2^{\cos x + \ln x}}\ln 2 \cdot \left( {\frac{{1 - x\sin x}}{x}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{2^{\cos x + \ln x}}\ln 2} \right)\left( {1 - x\sin x} \right)}}{x} \cr} $$
