Let \(y= \operatorname{sech} (x)\):
\[\begin{gathered} y = \operatorname{sech}(x) \\ \operatorname{arsech}(y) = x \end{gathered}\]
This means:
\[\begin{gathered} \frac{dy}{dx} = - \operatorname{sech}(x) \tanh(x) \\ \frac{d}{dy} \operatorname{arsech}(y) = \frac{dx}{dy} = \frac{1}{- \operatorname{sech}(x) \tanh(x) }\end{gathered}\]
Since \(\operatorname{sech}^2(x) = 1-\tanh^2(x)\):
\[\begin{align} \frac{dx}{dy} &= \frac{-1}{\operatorname{sech}(x) \sqrt{1- \operatorname{sech}^2(x)}} \\ &= \frac{-1}{y \sqrt{1-y^2}} \end{align}\]
Since \(y \in (0,1]\), then \(\sqrt{1-y^2}\) will always be positive. Also, the derivative is not defined at \(y=1\).