Why ${\rm arcosh}(\cosh x) =x $?

Question

I'd like to know why ${\rm arcosh}(\cosh x) =x$. Also I have read that the derivative of ${\rm arcosh}(\cosh x) = \sinh x/|\sinh x|$. Why?

Thanks all

Answer 1

It's not true that $\def\arcosh{\operatorname{arcosh}}\arcosh(\cosh x)=x$, because this holds only for $x\ge0$.

The function $x\mapsto \cosh x$ is invertible in the interval $[0,\infty)$ (where it assumes all the values in $[1,\infty)$), because it is increasing.

Therefore one can define a function $$ \arcosh\colon [1,\infty)\to[0,\infty) $$ such that $\arcosh(\cosh x)=x$ and $\cosh(\arcosh y)=y$, for all $x\ge0$ and all $y\ge1$.

Since $\cosh(-x)=\cosh x$, we easily have $$ \arcosh(\cosh x)=|x|. $$

The $\arcosh$ function can be described explicitly by solving for $x$ the equation $$ y=\cosh x=\frac{e^x+e^{-x}}{2} $$ that becomes $$ e^{2x}-2ye^x+1=0 $$ that is, $$ e^x=y+\sqrt{y^2-1} $$ or, finally, $$ \arcosh y=\log\bigl(y+\sqrt{y^2-1}\bigr) $$ which is valid only for $y\ge 1$.