Derivative of Arc secant or secant inverse

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, ${{dy} \over {dx}} = {1 \over {x\sqrt {x^2  - 1} }}$