If `f: R->(0,\ 2)` defined by `f(x)=(e^x-e^(-x))/(e^x+e^(-x))+1` is invertible, find `f^(-1)dot`

Explanation

Let `y=f(x)=frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}+1`



`therefore y-1=frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=frac{e^{x}-frac{1}{e^{x}}}{e^{x}+frac{1}{e^{x}}} `


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


`therefore y e^{2 x}+y-e^{2 x}-1=e^{2 x}-1 `


`e^{2 x}(y-2)=-y `


`e^{2 x}=frac{y}{x-y}`



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