General formula needed for this product rule expression (differential operator) Let $D_i^t$, $D_i^0$ for $i=1,\dots,n$ be differential operators. (For example $D_1^t = D_x^t$, $D_2^t = D_y^t,\dots$, where $x$, $y$ are the coordinates).
Suppose I am given the identity
$${D}_a^t (F_t u) = \sum_{j=1}^n F_t({D}_j^0 u){D}_a^t\varphi_j$$
where $\varphi_j$ are smooth functions and $F_t$ is some nice map. So
$$
D^t_bD^t_a(F_t u) = \sum_j D^t_b\left(F_t({D}_j^0 u)\right){D}_a^t\varphi_j+\sum_j F_t({D}_j^0 u)D^t_b{D}_a^t\varphi_j
$$
and because
$${D}_b^t (F_t (D_j^0u)) = \sum_{k=1}^n F_t({D}_k^0 D_j^0u){D}_b^t\varphi_k,$$ we have
$$D^t_bD^t_a(F_t u) =  \sum_{j,k=1}^n F_t({D}_k^0 D_j^0u){D}_b^t\varphi_k+\sum_j F_t({D}_j^0 u)D^t_b{D}_a^t\varphi_j
.$$
My question is how do I generalise this and obtain a rule for
$$D^t_{\alpha} (F_t u)$$ where
$\alpha$ is a multiindex of order $n$ (or order $m$)? My intention is to put the derivatives on $u$ and put the $F_t$ outside, like I demonstrated above.
Can anyone help me with getting the formula for this? It's really tedious to write out multiple derivatives so it's hard to tell for me.

This is probably better served as a comment, but I can't add one because of lack of reputation. If I understand correctly you are probably looking for the multivariate version of Faà di Bruno's formula, beware that the Wikipedia entry uses slightly different notation.
