Du löst also \(y= \frac{e^x-e^{-x}}{e^x+e^{-x}}\) nach \(y\) auf:
\(y= \frac{e^x-e^{-x}}{e^x+e^{-x}} \quad \left.\right|\cdot\frac{e^x}{e^x}\)
\(\Leftrightarrow\)
\(y = \frac{e^{2x}-1}{e^{2x}+1} \quad \left.\right|\cdot \left(e^{2x}+1\right)\)
\(\Leftrightarrow\)
\(y \left( e^{2x}+1 \right) =e^{2x}-1 \quad \left.\right| +1 - ye^{2x}\)
\(\Leftrightarrow\)
\(1+y =e^{2x}\left(1-y\right) \quad \left.\right| \div \left(1-y\right)\)
\(\Leftrightarrow\)
\(\frac{1+y}{1-y} = e^{2x} \)
\(\Leftrightarrow\)
\(\frac 12\ln \frac{1+y}{1-y} = x \)
\(\Rightarrow\)
\(\operatorname{artanh}x = \frac 12\ln \frac{1+x}{1-x} \)
