Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 20

Show that $\displaystyle \frac{d}{dx} \left( \sec^{-1} x \right) = \frac{1}{x\sqrt{x^2 - 1}}$
If we let $y = \sec^{-1} x$, then
$\sec y = x$

By Implicit Differentiation,
$\displaystyle \frac{d}{dx} \sec y = \frac{d}{dx} (x)$

$\displaystyle \sec y \tan y \left( \frac{dy}{dx} \right) = 1$
$\displaystyle \frac{dy}{dx} = \frac{1}{\sec y \tan y}$

By applying Pythagorean Identity,

$
\begin{equation}
\begin{aligned}
1 + \tan^y &= \sec^2 y\\
\\
\tan y &= \sqrt{\sec^2 y - 1}
\end{aligned}
\end{equation}
$


Thus,
$\displaystyle \frac{dy}{dx} = \frac{1}{\sec y \sqrt{\sec^2 y - 1}}$

But $\sec y = x$, therefore,
$\displaystyle \frac{dy}{dx} = \frac{1}{x \sqrt{x^2 - 1}}$