Taking derivative of a partial bell polynomial? I am trying to prove a statement that involves me taking the derivative of a bell polynomial.
Is there an elementary way to express:
$$
\frac{d}{dx}[ B_{n,k}(x_1,x_2,....,x_{n-k+1})]
$$
Where you could treat the $x$ terms within the bell polynomial as functions of $x$
I mean I have some idea of what you would do, like perhaps relate it to Faà di Bruno's formula? But that doesn't seem very plausible.
 A: Let us replace: $x_{i}=a_{i}(x)$.
$$B_{n,k}(a_{1}(x),a_{2}(x),...,a_{n-k+1}(x))=\sum_{^{\sum_{i=1}^{n-k+1}j_{i}=k}_{\sum_{i=1}^{n-k+1}i!j_{i}=n}}\frac{n!}{\prod_{i=1}^{n-k+1}i!^{j_{i}}j_{i}!}\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}$$
For abbreviation, we write
$$B_{n,k}(a_{1}(x),a_{2}(x),...,a_{n-k+1}(x))=\sum_{\pi(n,k)}\frac{n!}{\prod_{i=1}^{n-k+1}i!^{j_{i}}j_{i}!}\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}},$$
where $\pi(n,k)=\pi_{n,k}(j_{1},j_{2},...,j_{n-k+1})$ and $\pi(n,k)$ is the partition of integer $n$ with exactly $k$ parts.
$$\frac{d}{dx}B_{n,k}(a_1(x),a_2(x),...,a_{n-k+1}(x))=\sum_{\pi(n,k)}\frac{n!}{\prod_{i=1}^{n-k+1}i!^{j_{i}}j_{i}!}\left(\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}\right)'$$
$$\left(\prod_{i=1}^{n-k+1}f_{i}(x)\right)'=\sum_{l=1}^{n-k+1}\frac{f'_{l}(x)}{f_{l}(x)}\prod_{i=1}^{n-k+1}f_{i}(x)$$
$$\left(\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}\right)'=\sum_{l=1}^{n-k+1}\frac{\left(a_{l}(x)^{j_{l}}\right)'}{a_{l}(x)^{j_{l}}}\prod_{i=1}^{n-k+1}a_{l}(x)^{j_{l}}$$
$$\left(a_{l}(x)^{j_{l}}\right)'=j_{l}a_{l}(x)^{j_{l}-1}a'_{l}(x)=j_{l}a_{l}(x)^{j_{l}-1}a_{l+1}(x)$$
$$\left(\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}\right)'=\sum_{l=1}^{n-k+1}j_{l}\frac{a_{l+1}(x)}{a_{l}(x)}\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}$$
$$\frac{d}{dx}B_{n,k}(a_{1}(x),a_{2}(x),...,a_{n-k+1}(x))=\sum_{\pi(n,k)}\frac{n!}{\prod_{i=1}^{n-k+1}i!^{j_{i}}j_{i}!}\sum_{l=1}^{n-k+1}j_{l}\frac{a_{l+1}(x)}{a_{l}(x)}\prod_{i=1}^{n-k+1}a_{i}(x)^{j_{i}}$$
Another formula is written in Manipulation of Bell polynomials equation (1).
A: This is a very old question, but it still appears as a top hit when searching for derivatives of Bell polynomials, so might be good to answer it anyways.
I recently added the formula for partial derivatives of Bell polynomials to the Wikipedia page, which implies (using the chain rule) a nice formula for what you are asking:
$${\frac {d}{dx}}\left(B_{n,k}(a_{1}(x),\cdots ,a_{n-k+1}(x))\right)=\sum _{i=1}^{n-k+1}{\binom {n}{i}}a_{i}'(x)B_{n-i,k-1}(a_{1}(x),\cdots ,a_{n-i-k+2}(x)).$$
