Rule
$f(x){d^ng(x)\over dx^n}= \sum_{k=0}^{n}(-1)^{k}(_{k}^{n}){d^{n-k}\over dx^{n-k}}(f^{k}(x)g(x))$
Above rule looks transformation of Leibniz's rule.
Tried to prove it but has a problem to transform the parameters of Combination.
Anyone can prove this rule?
Supposing that we seek to simplify
$$\sum_{k=0}^n {n\choose k} (-1)^k \left(\frac{d}{dx}\right)^{n-k} (f^{(k)}(x) g(x))$$
we obtain from Leibniz the form
$$\sum_{k=0}^n {n\choose k} (-1)^k \sum_{q=0}^{n-k} {n-k\choose q} f^{(n-q)}(x) g^{(q)}(x)$$
or
$$\sum_{q=0}^n f^{(n-q)}(x) g^{(q)}(x) \sum_{k=0}^{n-q} {n\choose k} (-1)^k {n-k\choose q} \\ = f(x) g^{(n)}(x) + \sum_{q=0}^{n-1} f^{(n-q)}(x) g^{(q)}(x) \sum_{k=0}^{n-q} {n\choose k} (-1)^k {n-k\choose q}.$$
With $q\lt n$ we introduce
$${n-k\choose q} = {n-k\choose n-k-q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-k-q+1}} (1+z)^{n-k} \; dz$$
The pole vanishes when $k\gt n-q$ so we may extend the inner sum to $k=n$ and get
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-q+1}} (1+z)^{n} \sum_{k=0}^n {n\choose k} (-1)^k \frac{z^k}{(1+z)^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-q+1}} (1+z)^{n} \left(1-\frac{z}{1+z}\right)^n \; dz = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-q+1}} \; dz.$$
Now with $q\lt n$ we get $2\le n-q+1\le n+1$ so this is zero and we have the claim.
Remark. Looking at the links in the sidebar we see that this question also appeared here at this MSE link. No proof was given however.
