Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 18

Show that $\displaystyle \frac{d}{dx} (\sec x) = \sec x \tan x$

Get the reciprocal of $\sec x$

$\displaystyle \sec x = \frac{1}{ \cos x}$

Use Quotient Rule


$
\begin{equation}
\begin{aligned}


\frac{d}{dx} (\sec x) =& \frac{\displaystyle \cos x \frac{d}{dx} (1) - \left[ 1 \frac{d}{dx} (\cos x) \right]}{(\cos x)^2}
&& \text{}
\\
\\
\frac{d}{dx} (\sec x) =& \frac{\cos x (0) - (- \sin x)}{\cos ^2 x}
&& \text{Simplify the equation}
\\
\\
\frac{d}{dx} (\sec x) =& \frac{\sin x}{\cos ^2 x}
&& \text{Factor out $\large \frac{1}{\cos x}$ and $\large \frac{\sin x}{\cos x}$}
\\
\\
\frac{d}{dx} (\sec x) =& \left( \frac{1}{\cos x} \right) \left( \frac{\sin x}{\cos x} \right)
&& \text{Get the identities}
\\
\\
\frac{d}{dx} (\sec x) =& \sec x \tan x
&& \text{}
\\
\\

\end{aligned}
\end{equation}
$