Answer
\[ = \sec x\tan x\]
Work Step by Step
\[\begin{gathered}
\frac{d}{{dx}}\,\left( {\sec x} \right) = \sec x\tan x \hfill \\
\hfill \\
use\,\,trigonometric\,\,identity{\text{ }}\sec x = \frac{1}{{\cos x}} \hfill \\
\hfill \\
\frac{d}{{dx}}\,\left( {\frac{1}{{\cos x}}} \right) \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\frac{d}{{dx}}\,\left( {\frac{1}{{\cos x}}} \right) = \frac{{ - \left( { - \sin x} \right)}}{{{{\cos }^2}x}} \hfill \\
\hfill \\
or \hfill \\
\hfill \\
= \,\left( {\frac{{\sin x}}{{\cos x}}} \right)\,\left( {\frac{1}{{\cos x}}} \right) \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \sec x\tan x \hfill \\
\hfill \\
\end{gathered} \]
