## Calculus, 10th Edition (Anton)

$$\frac{{dy}}{{dx}} = \left[ {\left( {{x^3} + {x^{1/3}}} \right)\left( {\ln 5} \right) + 3{x^2} + \frac{1}{{3{x^{2/3}}}}} \right]{5^x}$$
\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}