Answer
$$\frac{{dy}}{{dx}} = \left[ {\left( {{x^2} + \sqrt x } \right)\left( {\ln 3} \right) + 2x + \frac{1}{{2\sqrt x }}} \right]{3^x}$$
Work Step by Step
$$\eqalign{
& y = \left( {{x^2} + \sqrt x } \right){3^x} \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}\left[ {\left( {{x^2} + \sqrt x } \right){3^x}} \right] \cr
& {\text{By the product rule}} \cr
& \frac{{dy}}{{dx}} = \left( {{x^2} + \sqrt x } \right)\frac{d}{{dx}}\left[ {{3^x}} \right] + {3^x}\frac{d}{{dx}}\left[ {{x^2} + \sqrt x } \right] \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = \left( {{x^2} + \sqrt x } \right)\left( {{3^x}\ln 3} \right) + {3^x}\left( {2x + \frac{1}{{2\sqrt x }}} \right) \cr
& {\text{Factor out }}{3^x} \cr
& \frac{{dy}}{{dx}} = \left[ {\left( {{x^2} + \sqrt x } \right)\left( {\ln 3} \right) + 2x + \frac{1}{{2\sqrt x }}} \right]{3^x} \cr} $$
