#### Answer

\[ = \sec \,x\,\left( {{{\sec }^2}x + {{\tan }^2}x} \right)\]

#### Work Step by Step

\[\begin{gathered}
y = \sec x\,\,\tan x \hfill \\
\hfill \\
Using\,\,product\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{d\sec x}}{{dx}}\tan x + \frac{{d\tan x}}{{dx}} \cdot \sec x \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \sec x \cdot \tan x \cdot \tan x + {\sec ^2}x \cdot \sec x \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \sec x \cdot {\tan ^2}x + {\sec ^3}x \hfill \\
\hfill \\
factor \hfill \\
\hfill \\
= \sec \,x\,\left( {{{\sec }^2}x + {{\tan }^2}x} \right) \hfill \\
\end{gathered} \]