Derivative of Real Area Hyperbolic Cosine

Theorem
Let $x \in \R_{>1}$ be a real number.

Let $\cosh^{-1} x$ be the inverse hyperbolic cosine of $x$.

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

Proof
Note that when $y = 0$, $\cosh y$ is defined and equals $1$.

But from Derivative of Hyperbolic Cosine:
 * $\valueat {\dfrac \d {\d y} \cosh y} {y \mathop = 0} = \sinh 0 = 0$

Thus $\dfrac {\d y} {\d x} = \dfrac 1 {\sinh y}$ is not defined at $y = 0$.

Hence the limitation of the domain of $\map {\dfrac \d {\d x} } {\cosh^{-1} x}$ to exclude $x = 1$.

Now it is necessary to determine the sign of $\dfrac {\d y} {\d x}$.

From:
 * Real Inverse Hyperbolic Cosine is Strictly Increasing
 * Derivative of Strictly Increasing Real Function is Strictly Positive

it follows that $\map {\dfrac \d {\d x} } {\cosh^{-1} x} > 0$ on $\R_{>1}$.

Thus:
 * $\dfrac {\d y} {\d x} = \dfrac 1 {\sqrt {\cosh^2 y - 1} }$

where $\sqrt {\cosh^2 y - 1}$ denotes the positive square root of $\cosh^2 y - 1$.

Hence by definition of $x$ and $y$ above:
 * $\map {\dfrac \d {\d x} } {\cosh^{-1} x} = \dfrac 1 {\sqrt {x^2 - 1} }$