Derivative of Inverse Hyperbolic Tangent

Theorem
Let $S$ denote the open real interval:
 * $S := \openint {-1} 1$

Let $x \in S$.

Let $\tanh^{-1} x$ be the inverse hyperbolic tangent of $x$.

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