Compact formula for the $n^\text{th}$ derivative of $\operatorname{sech}^2(x)$? In short, is there a clean formula for $\frac{d^n}{dx^n}\operatorname{sech}^2(x)?$

For convenience, let
\begin{align}
f(x)&=\operatorname{sech}^2(x),\\\\
g(x)&=\tanh(x).
\end{align}
Then
\begin{align}
f'(x)&=-2\operatorname{sech}^2(x)\tanh(x)& &=af(x)g(x),\\\\
g'(x)&=\operatorname{sech}^2(x)& &=f(x),
\end{align}
where $a=-2.$ We also have $g^2(x)=1-f(x).$
Hence $f^{(n)}(x)$ can always be written as a sum where each term is of the form $c_{k,p} [f(x)]^k[g(x)]^p$ where $p=0$ or $1$. So the question becomes: is there a formula for the $c_{k,p}$'s?
I would expect some pattern to emerge from repeated uses of the power rule combined with the recursion of the above formulas, but I'm not sure how to proceed. Also, I know sometimes coefficients of special polynomials can encode information about $n^\text{th}$ derivatives (e.g., Hermite polynomials and $e^{-x^2}$), and I would find it acceptable to express the answer in terms of such coefficients if possible.
 A: The $n$-th derivative of $\tanh(x)$ are obtained
in this article by Boyadzhiev.
Denoting $z=\tanh x$ and $C_{n}\left( \tanh x \right)=\frac{d^{n}}{dx^{n}}\left( \tanh x \right) $, we have
\begin{equation}
C_{n}\left(z \right)=2^{n}\left( 1+ z\right)\sum_{k=0}^{n}\frac{k!}{2^k}S\left( n,k \right)\left( z-1 \right)^k
\end{equation}
where $S(n,k)$ are the Stirling numbers of the second kind.
Then,
\begin{align}
 \frac{d^n}{dx^n}\operatorname{sech}^2(x)&=\frac{d^{n+1}}{dx^{n+1}}\tanh(x)\\
 &=C_{n+1}\left( \tanh(x) \right)\\
 &=2^{n+1}\left( 1+ \tanh(x)\right)\sum_{k=0}^{n+1}\frac{k!}{2^k}S\left( n+1,k \right)\left( \tanh(x)-1 \right)^k
\end{align}
Remarking that
$S(n+1,0)=0$ , the polynomials can also be expressed as
\begin{equation}
C_{n+1}\left(z \right)=\left( 1-z^2 \right)P_{n}(z) 
\end{equation}
where $P_{n}(z) $ is a polynomial of degree $n$:
\begin{align}
 \frac{d^n}{dx^n}\operatorname{sech}^2(x)&=\operatorname{sech}^2(x)P_n(\tanh x)\\
 P_n(\tanh x)&=-2^{n+1}\sum_{k=1}^{n+1}\frac{k!}{2^k}S\left( n+1,k \right)\left( \tanh(x)-1 \right)^{k-1}
\end{align}
These polynomials are identical to the $P_n$ of @RenéGy 's answer.
A: $\textbf{Fa'a di Bruno's formula}:$
if $g$ and $f$ are functions with a sufficient number of derivatives, then
\begin{multline}
 \frac{d^m}{dt^m}g(f(t))  
 =\sum \frac{m!}{b_1!b_2!\cdots b_m!}g^{(k)}(f(t))\left(\frac{f'(t)}{1!}\right)^{b_1}\left(\frac{f''(t)}{2!}\right)^{b_2}\cdots \left(\frac{f^{(m)}(t)}{m!}\right)^{b_m}
\end{multline}
where the sum is over all different solutions in nonnegative integers $b_1, b_2,\cdots, b_m$ of $b_1+2b_2+\cdots +mb_m=m$, and $k:=b_1+\cdots+b_m$. Let $g(x)=x^2$ and $f(x)=\text{sech}(x)$ then $h(x)= g\circ f=\text{sech}^2(x)$. Now, use the above formula to obtain the $nth$ derivative.
