The inverse of the function $f: R→ {x∈R:x<1}$ given by $f (x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ is 

$\frac{1}{2}\log\frac{1+x}{1-x}$

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

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

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

$⇒ye^{2x}+y=e^{2x}-1$

so $(y-1)e^{2x}=-y-1$

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

so $2x=\log\left|\frac{1+y}{1-y}\right|$

$x=\frac{1}{2}\log\left|\frac{1+y}{1-y}\right|$

so $f^{-1}(x)=\frac{1}{2}\log\left|\frac{1+x}{1-x}\right|$