What is the inverse of $f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$?

Question

please help me to find out the inverse this function,
$$f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ I know that, let
$$y=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$$ and if I find $x=\cdots$ then that is the inverse. But I can't calculate this. Is this the only way. is there any other way to figure it out?

Answer 1

Applying Componendo and dividendo on $$\frac y1=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}},$$

we get

$$\frac{y+1}{y-1}=\frac{e^x}{e^{-x}}=e^{2x}$$

$$2x=\ln\left(\frac{y+1}{y-1}\right)$$

Answer 2

You have $$ y = \frac{e^x+e^{-x}}{e^x-e^{-x}}. $$ Multiplying the top and bottom by $e^x$, you get $$ y = \frac{e^{2x}+1}{e^{2x}-1}. $$ The advantage of this form is that $x$ now appears ONLY within the expression $e^{2x}$, so we can treat $e^{2x}$ as the unknown that we're trying to solve for.

Now clearing fractions, we get $$ (e^{2x}-1)y = e^{2x}+1 $$ so $$ e^{2x}y - y = e^{2x}+1 $$ and then $$ e^{2x}y -e^{2x} = y+1. $$ From this we get $$ e^{2x}(y-1) = y+1. $$ Then $$ e^{2x} = \frac{y+1}{y-1}, $$ $$ 2x = \ln \frac{y+1}{y-1}, $$ $$ x = \frac 1 2 \ln \frac{y+1}{y-1}. $$