Derivative of Real Area Hyperbolic Cosine

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Theorem

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

Let $\arcosh x$ be the real area hyperbolic cosine of $x$.


Then:

$\map {\dfrac \d {\d x} } {\arcosh x} = \dfrac 1 {\sqrt {x^2 - 1} }$


Proof

\(\ds y\) \(=\) \(\ds \arcosh x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \cosh y\) Definition of Real Area Hyperbolic Cosine
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds \sinh y\) Derivative of Hyperbolic Cosine
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {\sinh y}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \pm \frac 1 {\sqrt {\cosh^2 y - 1} }\) Difference of Squares of Hyperbolic Cosine and Sine


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 $\map {\dfrac \d {\d x} } {\arcosh x}$ to exclude $x = 1$.


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

From:

Real Area Hyperbolic Cosine is Strictly Increasing
Derivative of Strictly Increasing Real Function is Strictly Positive

it follows that $\map {\dfrac \d {\d x} } {\arcosh x} > 0$ 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:

$\map {\dfrac \d {\d x} } {\arcosh x} = \dfrac 1 {\sqrt {x^2 - 1} }$

$\blacksquare$


Sources