Derivative of Inverse Hyperbolic Cotangent

Theorem
Let $S$ denote the union of the unbounded open real intervals:
 * $S := \openint \gets {-1} \cup \openint 1 \to$

Let $x \in S$.

Let $\coth^{-1} x$ be the inverse hyperbolic cotangent of $x$.

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

Also presented as
This result can also be (and usually is) reported as:
 * $\map {\dfrac \d {\d x} } {\coth^{-1} x}\ = \dfrac 1 {1 - x^2}$

but this obscures the fact that $x^2 > 1$ in order for it to be defined.