Assume $x(t) \in C^{(n)}[0,\infty)$. I am trying to find a formula for the $n^{th}$ derivative of
$$Q_j(t) = e^{-x(t)}\,\frac{x(t)^j}{j!}, $$
in terms of $Q_0(t), Q_1(t), ..., Q_j(t)$. (Or any other useful recursive formula in terms of previous derivatives, etc.)
So far I have tried this:
$$ Q^{'}_0(t) = -x^{'}(t)\,Q_0(t),$$ and for $j\geq1$, $$ Q_j^{'}(t) = x^{'}(t)\,\left(Q_{j-1}(t)-Q_j(t)\right).$$ Or we can assume $0=Q_{-1}(t)=Q_{-2}(t)=...$ and write it in a general form for $j\geq0$, $$ Q_j^{'}(t) = x^{'}(t)\,\left(Q_{j-1}(t)-Q_j(t)\right).$$
Therefore, for $j\geq0$, the second derivative is \begin{equation} \begin{split} Q_j^{"}(t) &= x^{"}(t)\,\left(Q_{j-1}(t)-Q_j(t)\right) + (x^{'}(t))^2\,\left(Q_{j-2}(t)-2\,Q_{j-1}(t)+Q_{j}(t)\right). \end{split} \end{equation}
And for the 3rd derivative I got this,
\begin{equation} \begin{split} Q^{(3)}_j(t) &= x^{(3)}(t)\,\left(Q_{j-1}(t)-Q_j(t)\right)+ 3\,x^{'}(t)\,x^{"}(t)\,\left(Q_{j-2}(t)-2\,Q_{j-1}(t)+Q_{j}(t)\right) \\ & + (x^{'}(t))^3\,\left(Q_{j-3}(t)-3\,Q_{j-2}(t)+3\,Q_{j-1}(t)+ Q_{j}(t)\right). \end{split} \end{equation}
Can we see a pattern here?