#### Answer

\[{f^,}\,\left( x \right) = {e^{x\sqrt {3x + 2} }}\,\left( {\frac{{3x}}{{2\sqrt {3x + 2} }} + \sqrt {3x + 2} } \right)\]

#### Work Step by Step

\[\begin{gathered}
f\,\left( x \right) = {e^{x\sqrt {3x + 2} }} \hfill \\
Find\,\,the\,\,derivative \hfill \\
{f^,}\,\left( x \right) = \,\,\left[ {{e^{x\sqrt {3x + 2} }}} \right] \hfill \\
Use\,\,the\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{a^{g\,\left( x \right)}}} \right] = \,\left( {\ln a} \right){a^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = {e^{x\sqrt {3x + 2} }}\,\left( {x\sqrt {3x + 2} } \right) \hfill \\
Use\,\,product\,\,rule \hfill \\
{f^,}\,\left( x \right) = {e^{x\sqrt {3x + 2} }}\,\left( {x\,{{\left( {\sqrt {3x + 2} } \right)}^,} + \sqrt {3x + 2} \,{{\left( x \right)}^,}} \right) \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = {e^{x\sqrt {3x + 2} }}\,\left( {\frac{{3x}}{{2\sqrt {3x + 2} }} + \sqrt {3x + 2} } \right) \hfill \\
\end{gathered} \]