Derivative of Inverse Hyperbolic Secant

Theorem
Let $S$ denote the open real interval:
 * $S := \openint 0 1$

Let $x \in S$.

Let $\sech^{-1} x$ denote the inverse hyperbolic secant of $x$.

Then:
 * $\map {\dfrac \d {\d x} } {\sech^{-1} x} = \dfrac {-1} {x \sqrt{1 - x^2} }$

Proof
$\sech^{-1} x$ is defined only on the half-open real interval $\hointl 0 1$.

Thus on $\hointl 0 1$:

When $x = 1$, however, $\sqrt{1 - x^2} = 0$ and so $\dfrac {-1} {x \sqrt {1 - x^2} }$ is undefined.

Hence $\sech^{-1} x$ can be defined only on $\openint 0 1$.