Let \(y= \tanh (x)\):
\[\begin{gathered} y = \tanh(x) \\ \operatorname{artanh}(y) = x \end{gathered}\]
This means:
\[\begin{gathered} \frac{dy}{dx} = \operatorname{sech}^2(x) \\ \frac{d}{dy} \operatorname{artanh}(y) = \frac{dx}{dy} = \frac{1}{\operatorname{sech}^2(x)} \end{gathered}\]
Since \(\operatorname{sech}^2(x) = 1 - \tanh^2(x)\):
\[\begin{align} \frac{dx}{dy} &= \frac{1}{1 - \tanh^2(x)} \\ &= \frac{1}{1-y^2} \end{align}\]
Since \(y = \tanh(x)\), then \(|y| \lt 1 \).