Let \(y= \sinh(x)\):
\[ y = \frac{e^x - e^{-x}}{2}\]
In order to find the formula for the inverse of \(\sinh\) we need to make \(x\) the subject of the above formula.
\[ 2y = e^x - e^{-x}\]
If we multiply both sides by \(e^x\):
\[\begin{gathered} 2ye^x = (e^x)^2 - 1 \\ - (e^x)^2 + 2ye^x + 1 = 0 \end{gathered}\]
Let \(v = e^x\):
\[- v^2 + 2yv + 1 = 0 \]
We can use the quadratic formula here:
\[v=\frac{-(2y) ± \sqrt{(2y)^2 - 4(-1)(1)}}{2(-1)} = \frac{-2y ± \sqrt{4y^2 + 4}}{-2} \]
Simplifying:
\[v = \frac{-2y ± 2\sqrt{y^2 + 1}}{-2} = y ∓ \sqrt{y^2 + 1} \]
Since \(v=e^x\):
\[x = \ln(y ∓ \sqrt{y^2 + 1}) \]
The domain of \(\ln(a)\) is \(a \in (0,\infty)\). This means \(y ∓ \sqrt{y^2 + 1}\) cannot be negative. Since \(y - |\sqrt{y^2 + 1}|\) is negative for all \(y>0\) and \(y + |\sqrt{y^2 + 1}|\) is positive for all \(y>0\), then the input for \(\ln\) can only be \(y+|\sqrt{y^2 + 1}|\):
\[x = \ln(y + | \sqrt{y^2 + 1}|) \]