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

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

But from Derivative of Hyperbolic Cosine Function:
 * $\left.{\dfrac {\mathrm d} {\mathrm d y} \cosh y}\right\vert_{y \mathop = 0} = \sinh 0 = 0$

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

Hence the limitation of the domain of $\dfrac {\mathrm d}{\mathrm d x} \left({\cosh^{-1} x}\right)$ to exclude $x = 1$.

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

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

it follows that $\dfrac {\mathrm d}{\mathrm d x} \left({\cosh^{-1} x}\right) > 0$ on $\R_{>1}$.

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

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

Hence by definition of $x$ and $y$ above:
 * $\dfrac {\mathrm d} {\mathrm d x} \left({\cosh^{-1} x}\right) = \dfrac 1 {\sqrt {x^2 - 1} }$