Answer
$$\frac{{dy}}{{dx}} = \left[ {\left( {{x^3} + {x^{1/3}}} \right)\left( {\ln 5} \right) + 3{x^2} + \frac{1}{{3{x^{2/3}}}}} \right]{5^x}$$
Work Step by Step
$$\eqalign{
& y = \left( {{x^3} + \root 3 \of x } \right){5^x} \cr
& {\text{write }}\root 3 \of x {\text{ as }}{x^{1/3}} \cr
& y = \left( {{x^3} + {x^{1/3}}} \right){5^x} \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}\left[ {\left( {{x^3} + {x^{1/3}}} \right){5^x}} \right] \cr
& {\text{By the product rule}} \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} + {x^{1/3}}} \right)\frac{d}{{dx}}\left[ {{5^x}} \right] + {5^x}\frac{d}{{dx}}\left[ {{x^3} + {x^{1/3}}} \right] \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} + {x^{1/3}}} \right)\left( {{5^x}\ln 5} \right) + {5^x}\left( {3{x^2} + \frac{1}{3}{x^{ - 2/3}}} \right) \cr
& {\text{Factor out }}{5^x} \cr
& \frac{{dy}}{{dx}} = \left[ {\left( {{x^3} + {x^{1/3}}} \right)\left( {\ln 5} \right) + \left( {3{x^2} + \frac{1}{3}{x^{ - 2/3}}} \right)} \right]{5^x} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \left[ {\left( {{x^3} + {x^{1/3}}} \right)\left( {\ln 5} \right) + 3{x^2} + \frac{1}{{3{x^{2/3}}}}} \right]{5^x} \cr} $$
