How to find the inverse function of $f(x)=\frac{\sinh(\ln(\cosh x))}{\sinh x}$

Question

How to find the inverse function of $f(x)=\frac{\sinh(\ln(\cosh x))}{\sinh x}$?

I' ve tried the following:

$y=\frac{\sinh(\ln(\cosh x))}{\sinh x}$ . Now I should express $x$ in terms of $y$.

Then: $$\sinh^-1 (\sinh x) =x=\sinh ^-1( \frac{\sinh(\ln(\cosh x))}{y})$$ Now I am stuck. How shall I continue and how do I get to the solution?

Answer 1

If $e^t = \cosh (x) $ then $$y = \dfrac{\sinh(t)}{\sinh(x)} = \dfrac{e^t - e^{-t}}{2 \sinh(x)} = \dfrac{\cosh^2 (x) - 1}{2 \sinh(x) \cosh(x)} = \dfrac{\tanh(x)}{2}$$

Answer 2

$f(x)=y=\dfrac{\sinh (\ln \cosh x)}{\sinh x}$

we want to solve for $x$

$\sinh x=\dfrac{\sinh (\ln \cosh x)}{y}$

$\implies x=\sinh ^{-1}\dfrac{\sinh (\ln \cosh x)}{y}=\sinh^{-1}\dfrac{y\sinh x}{y}=\sinh^{-1}(\sinh x)=x$

no inverse?

Answer 3

HINT

I would say

$f(x)=\frac{\sinh(\ln(\cosh x))}{\sinh x}=\frac{\cosh x-\frac{1}{\cosh x})}{2\sinh x}=\frac{\cosh^2x-1}{2\cosh x\sinh x}=\frac{\sinh^2x}{2\cosh x\sinh x}=\frac{1}{2}\tanh x$

Now calculate the inverse function.