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

$$f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ 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?

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This is the hyperbolic cotangent function, and its inverse is the inverse hyperbolic cotangent function. – Ayesha Feb 9 '14 at 3:21
Replace $e^x$ with some other letter, say, $u$; notice that $e^{-x}$ is just $1/u$; now you have a formula for $y$ in terms of $u$; can you solve that formula for $u$? – Gerry Myerson Feb 9 '14 at 5:20

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

we get

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

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

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Where did that absolute value come from? If $e^{2x}=\dfrac{y+1}{y-1}$ then $2x = \ln\dfrac{y+1}{y-1}$. ${}\qquad{}$ – Michael Hardy Feb 9 '14 at 5:27
@MichaelHardy,sorry for the wrong notation. – lab bhattacharjee Feb 9 '14 at 5:45

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

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

therefore,

$$f^{-1}(x)= \operatorname{arccoth}(x)$$

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