Answer
$$\frac{{dy}}{{dx}} = \frac{{{x^2}}}{{\ln 10}} + 3{x^2}{\log _{10}}x $$
Work Step by Step
$$\eqalign{
& y = {x^3}{\log _{10}}x \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^3}{{\log }_{10}}x} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = {x^3}\frac{d}{{dx}}\left[ {{{\log }_{10}}x} \right] + {\log _{10}}x\frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& {\text{use the rules }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}{\text{ and }}\cr
&\frac{d}{{dx}}\left[ {{{\log }_a}x} \right] = \frac{1}{{\left( {\ln a} \right)x}}.{\text{ Then}} \cr
& \frac{{dy}}{{dx}} = {x^3}\left( {\frac{1}{{\left( {\ln 10} \right)x}}} \right) + {\log _{10}}x\left( {3{x^2}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2}}}{{\ln 10}} + 3{x^2}{\log _{10}}x \cr} $$
