My attempt is to show that, for the function $h_a(x) = \operatorname{sech}^a(x)$, we can establish the veracity of
$$\lim_{\vert x \vert \rightarrow +\infty}(1 + \vert x \vert)^m\vert h_a^{(q)}(x)\vert = 0$$ $\forall q,m \in \mathbb{N}_0$.
If one creates the sequence of functions given recurrently by
\begin{equation} F_a^q = F_a^{q-1} + \frac{\frac{d}{dx}(F_a^{q-1})}{F_a^{q-1}} \end{equation}
Putting the first element $F_a^1 = -a\tanh(x)$ - which explain in part my choice of symbols, if someone prefer can change, this is irrelevant - knowing that $a>0$ is a real number and $F_a^q : \mathbb{R} \rightarrow \mathbb{R}$, i.e, $x \in \mathbb{R}$. We can show that (the number $q$ between bracket's means the $q$-derivative with respect to the $x$ variable)
\begin{equation}\label{recorrencia dos h's a's} h_a^{(q+1)}(x) = F_a^q(x)h_a^{(q)}(x) \end{equation}
With that in mind it is true that
\begin{equation} h_a^{(q+1)} = \prod_{k=1}^q F_a^k(x)h_a(x) \end{equation}
I am trying to prove that, for a given $q$ it is true that, given a $m \in \mathbb{N}$,
$$\lim_{\vert x\vert \rightarrow +\infty}(1 + \vert x \vert )^m \prod_{i=1}^q\vert F_a^i(x)\vert e^{-a\vert x \vert} = 0$$
This comes from the relation: $0<(\operatorname{sech}x)^a<2^a\,e^{-a|x|}.$ The $\prod_{i=1}^q\vert F_a^i(x)\vert$ part is a sum of product-terms of $\tanh$ and $\operatorname{sech}$ so how can I prove that this is bounded? I mean, rigorously show that for any $q$ this product give a bounded result, such that the limit be correct?