Binomial Theorem of Differentiation? I noticed that $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ and it's had me scratching for a little bit. It's easy to see how the cross terms add up but can anyone draw a direct comparison between the behavior being exhibited here and the binomial expansion? It seems to work for the case equivalent to the multinomial theorem as well ($\frac{d^{m}}{dx^{m}} \prod_{k=1}^{n} f_k(x)$). Can someone produce a proof for this more general case (preferably by induction)? Any and all insights are welcome!
 A: Taylor series expansion of $f(x+h)$
$$
f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....
$$
Taylor series expansion of $g(x+h)$
$$
g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....
$$
Taylor series expansion of $f(x+h)g(x+h)$
$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....
$$

$$
f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....)
$$
If you order $h^n$, it will give you binom expansion
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ 
Because it has exactly the same coefficient of
$$
=(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)}
$$
$$
e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+....
$$
If you order $h^n$ of $e^{hx}e^{hy}$  and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.
$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n  \frac{x^i y^{n-i}}{i! (n-i)!} $$ 
$$(x+y)^n=\sum_{i=0}^n  \frac{n! x^i y^{n-i}}{i! (n-i)!} $$ 
$$(x+y)^n=\sum_{i=0}^n  {n \choose i} x^i y^{n-i} $$ 
Thus it is also true
$$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ 
A: This is something you typically do by induction (I have fixed a misprint in your notation), starng with
$$
\frac{d}{dx} \left( f(x) g(x) \right) = \frac{d}{dx}\left( f(x) \right)g(x) + f(x) \frac{d}{dx} \left( g(x) \right).
$$
Here is as it goes:
\begin{align}\frac{d^{n+1}}{dx^{n+1}} f(x)g(x)&= \frac{d}{dx} \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x) \\&= \sum_{i=0}^n {n \choose i} f^{(i+1)}(x)g^{(n-i)}(x) + \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i+1)}(x) \\&= \sum_{i=1}^{n+1} {n \choose i-1} f^{(i)}(x)g^{()n+1)-i)}(x) + \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n+1)-i)}(x)\\&= \sum_{i=0}^{n+1} \left( {n \choose i-1} + {n \choose i}\right) f^{(i)}(x)g^{(n+1)-i)}(x)\\&=\sum_{i=0}^{n+1} \left( {n+1 \choose 1}\right) f^{(i)}(x)g^{(n+1)-i)}(x).\end{align}
