can someone please help me with this rule:
$$ f(x) \cdot g(x)^{(n)} = \sum_{k=0}^{n}(-1)^k \cdot {n \choose k}\cdot (f(x)^{(k)}\cdot g(x))^{(n-k)} $$
or you can write:
$$ f(x) \cdot \frac{d^n g(x)}{dx^n} = \sum_{k=0}^{n}(-1)^k \cdot {n \choose k}\frac{d^{n-k}}{dx^{n-k}}(f(x)^{(k)}\cdot g(x)) $$
Where the powers with $()$ are derivations of the function.
Where does it come from and how can I proof it? I tried mathematical induction without success.
Thank You!
Your formula is wrong. See http://en.wikipedia.org/wiki/Product_rule and try induction with the true formula.
HINT: At first this question seems to be naive but it is not. Denote $u=f(x),v=g(x)$ then question can be simplified as follow $$uv^{(n)}=\sum_{k=0}^{n}(-1)^k\binom{n}{k}(u^{(k)}v)^{(n-k)}=uv^{(n)}+\sum_{k=1}^{n}(-1)^k\binom{n}{k}(u^{(k)}v)^{(n-k)}$$ or simply as $$\sum_{k=1}^{n}(-1)^k\binom{n}{k}(u^{(k)}v)^{(n-k)}=0$$ from Leibnitz rule we have that $$(u^{(k)}v)^{(n-k)}=\sum_{i=0}^{n-k}\binom{n-k}{i}u^{(k+i)}v^{(n-(k+i))}$$ the real question is to prove that $$\sum_{i=0}^{n-k}\sum_{k=1}^{n}(-1)^k\binom{n}{k}\binom{n-k}{i}u^{(k+i)}v^{(n-(k+i))}=0$$
$$fg''=(fg)''-(f'g)'+(f''g)=\left(f'g+fg'\right)'-f''g-f'g'+f''g=$$
$$=f''g+\color{red}{f'g'}+f'g'+{fg''}-\color{green}{f''g}-\color{red}{f'g'}+ \color{green}{f''g} =$$$$=f''g+f'g'+fg''$$
which is false in general, of course...
– DonAntonio Jan 14 '14 at 17:30