Hyperbolic inverse Function

Question

how to find the inverse function of the arcsinh x also the domainof arcsinh

Answer 1

suppose $$ \frac{e^{x}-e^{-x}}{2}=y $$ Then $$ e^{x}-e^{-x}=2y \Leftrightarrow e^{2x}-2ye^x-1=0 $$ And $\Delta=4\left(y^2+1\right) \geq 0$ then with $X=e^x$ $$ X= \frac{2y \pm \sqrt{4\left(y^2+1\right)}}{2}=y\pm \sqrt{y^2+1} $$ Then the solution is the positive one ( I let you think about that ) so $$ e^x=y+ \sqrt{y^2+1} $$ Finally

$$ x=\ln\left(y+ \sqrt{y^2+1}\right) $$

But you could reverse the thing and evaluate $\text{arcsinh}\left(\text{sinh}\left(x\right)\right)$ to show it is identity.