Inverting Sinh x
To find\[sinh^{-1}x\]
let \[y=sinhx=\frac{e^x-e^{-x}}{2}\]
the multiplying by \[2e^x\]
gives\[2e^xy=e^{2x}-1 \rightarrow 2y(e^x)-e^{2x}+1=0 \rightarrow e^{2x}-2ye^x-1=0\]
This is a quadratic equation in
\[e^x\]
.\[e^x= \frac{2y \pm \sqrt{(-2y)^2- 4 \times 1 \times -1}}{2} = y \pm \sqrt{y^2+1}\]
.Hence
\[x= ln(y \pm \sqrt{y^2+1})= ln(y \pm \sqrt{y^2+1})\]
.Only the positive square root will return a valid value since the negative root will result in having to take the natural log of a negative number, which is impossible, so
\[x= ln(y + \sqrt{y^2+1})\]
.Then
\[sinh^{-1}x= ln(x + \sqrt{x^2+1})\]
.