Answer
\[\frac{{dy}}{{dx}} = {x^3}\ln 3 \cdot {3^x} + {3^{x + 1}}{x^2}\]
Work Step by Step
\[\begin{gathered}
y = {x^3} \cdot {3^x} \hfill \\
\hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {{x^3}} \right)\,{\left( {{3^x}} \right)^,} + \,\left( {{3^x}} \right)\,{\left( {{x^3}} \right)^,} \hfill \\
\hfill \\
Use\,\,the\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{a^u}} \right] = {a^u}\ln a \cdot {u^,} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {{x^3}} \right)\,\left( {{3^x}\ln x} \right) + {3^x}\,\left( {3{x^2}} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {x^3}\ln 3 \cdot {3^x} + {3^{x + 1}}{x^2} \hfill \\
\end{gathered} \]
