Derivative of Inverse Hyperbolic Secant

Theorem
Let $S$ denote the open real interval:
 * $S := \left({0 \,.\,.\, 1}\right)$

Let $x \in S$.

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

Then:
 * $\dfrac {\mathrm d}{\mathrm d x} \left({\operatorname{sech}^{-1} x}\right) = \dfrac {-1} {x \sqrt{1 - x^2} }$

Proof
$\operatorname{sech}^{-1} x$ is defined only on the half-open real interval $\left({0 \,.\,.\, 1}\right]$.

Thus on $\left({0 \,.\,.\, 1}\right]$:

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

Hence $\operatorname{sech}^{-1} x$ can be defined only on $\left({0 \,.\,.\, 1}\right)$.