Answer
\[{g^,}\,\left( y \right) = {e^{y + 1}}{y^{e - 1}} + {e^y}{y^e}\]
Work Step by Step
\[\begin{gathered}
g\,\left( y \right) = {e^y} \cdot {y^e} \hfill \\
\hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
{g^,}\,\left( y \right) = {e^y}\,{\left( {{y^e}} \right)^,} + \,\left( {{y^e}} \right)\,{\left( {{e^y}} \right)^,} \hfill \\
\hfill \\
differentiate \hfill \\
\hfill \\
{g^,}\,\left( y \right) = {e^y}\,\,\left( {e{y^{e - 1}}} \right) + {y^e}\,\left( {{e^y}} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
{g^,}\,\left( y \right) = {e^{y + 1}}{y^{e - 1}} + {e^y}{y^e} \hfill \\
\end{gathered} \]
