Answer
\[{f^,}\,\left( x \right) = \frac{{3{e^{\sqrt {3x} }}{{\sec }^2}\,\left( {{e^{\sqrt {3x} }}} \right)}}{{2\sqrt {3x} }}\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = \tan \,\left( {{e^{\sqrt {3x} }}} \right) \hfill \\
\hfill \\
Chain\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\
\hfill \\
f'\,\left( x \right) = {\sec ^2}\,\left( {{e^{\sqrt {3x} }}} \right)\frac{d}{{dx}}\,\,\left[ {{e^{\sqrt {3x} }}} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{f^,}\,\left( x \right) = {\sec ^2}\,\left( {{e^{\sqrt {3x} }}} \right)\,\left( {\frac{3}{{2\sqrt {3x} }}} \right){e^{\sqrt {3x} }} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{3{e^{\sqrt {3x} }}{{\sec }^2}\,\left( {{e^{\sqrt {3x} }}} \right)}}{{2\sqrt {3x} }} \hfill \\
\hfill \\
\end{gathered} \]
