Let \(y= \sinh (x)\):
\[\begin{gathered} y = \sinh(x) \\ \operatorname{arsinh}(y) = x \end{gathered}\]
This means:
\[\begin{gathered} \frac{dy}{dx} = \cosh(x) \\ \frac{d}{dy} \operatorname{arsinh}(y) = \frac{dx}{dy} = \frac{1}{\cosh(x)} \end{gathered}\]
Since \(\cosh^2(x) - \sinh^2(x) = 1\):
\[\begin{align} \frac{dx}{dy} &= \frac{1}{\sqrt{1+\sinh^2(x)}} \\ &= \frac{1}{\sqrt{1+y^2}} \end{align}\]
Since \(\cosh(x)\) is postive for all values of \(x\), then \(\sqrt{1+y^2}\) will always be positive as well.