#### Answer

\[ = \sec {e^x}\tan {e^x} \cdot {e^x}\]

#### Work Step by Step

\[\begin{gathered}
y = \sec \,{e^x} \hfill \\
\hfill \\
y = f\,\left( u \right) = \sec u \hfill \\
\hfill \\
set\,\,u = g\,\left( x \right) = {e^x} \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
{\text{ }}\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{d}{{du}}\,\left( {\sec u} \right) \cdot \frac{d}{{dx}}\,\left( {{e^x}} \right) \hfill \\
\hfill \\
= \sec u\tan u \cdot {e^x} \hfill \\
\hfill \\
substitute\,\,\,back\,\,u = {e^x} \hfill \\
\hfill \\
= \sec {e^x}\tan {e^x} \cdot {e^x} \hfill \\
\hfill \\
\end{gathered} \]