$f(x){d^ng(x)\over dx^n}= \sum_{k=0}^{n}(-1)^{k}(_{k}^{n}){d^{n-k}\over dx^{n-k}}(f^{k}(x)g(x))$ 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?  
 A: 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.
