Let \(y= \cosh (x)\):
\[\begin{gathered} y = \cosh(x) \\ \operatorname{arcosh}(y) = x \end{gathered}\]
This means:
\[\begin{gathered} \frac{dy}{dx} = \sinh(x) \\ \frac{d}{dy} \operatorname{arcosh}(y) = \frac{dx}{dy} = \frac{1}{\sinh(x)} \end{gathered}\]
Since \(\cosh^2(x) - \sinh^2(x) = 1\):
\[\begin{align} \frac{dx}{dy} &= \frac{1}{\sqrt{\cosh^2(x) -1}} \\ &= \frac{1}{\sqrt{y^2-1}} \end{align}\]
Since \(y \in (1,\infty) \), then \(\sqrt{y^2-1}\) will always be positive.