Inverse function of tanh(x)

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I have a problem while calculating inverse function of tanh(x).

I know it is y = sinh(x)/cosh(x) and then I should express x, but I am stuck with that.

Will you help me with this? thx a lot

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1 Answer

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Use that $$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}=y$$ so $$e^{2x}=\frac{y+1}{1-y}$$It is $$\frac{e^x+e^{-x}}{e^x-e^{-x}}\times \frac {e^x}{e^x}$$

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