Answer
\[\frac{{dy}}{{dx}} = {e^x}\,\left( {\frac{3}{2}\sqrt x + \sqrt {{x^3}} } \right)\]
Work Step by Step
\[\begin{gathered}
y = {e^x}\sqrt {{x^3}} \hfill \\
\hfill \\
differentiate \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {{e^x}} \right)\,\left( {\frac{{3{x^2}}}{{2\sqrt {{x^3}} }}} \right) + \sqrt {{x^3}} {e^x} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {e^x}\,\left( {\frac{3}{2}{x^{\frac{1}{2}}}} \right) + \sqrt {{x^3}} {e^x} \hfill \\
\hfill \\
factor \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {e^x}\,\left( {\frac{3}{2}\sqrt x + \sqrt {{x^3}} } \right) \hfill \\
\hfill \\
\end{gathered} \]
