Let $y = f(x) = \sec ^{ - 1} x$
Then $x = \sec y$.
Differentiating both sides with respect to $x$, we get
$1 = (\sec y\tan y){{dy} \over {dx}}$
Note that we used the chain rule to differentiate $\sec y$ with respect to $x$ because $\sec y$ is a function of $y$, and $y$ in turn is a function of $x$.
or, ${{dy} \over {dx}} = {1 \over {\sec y\tan y}}$
or, ${{dy} \over {dx}} = {1 \over {\sec y\sqrt {\sec ^2 y - 1} }}$
or, ${{dy} \over {dx}} = {1 \over {x\sqrt {x^2 - 1} }}$
