- #1

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we know it F(x)= f(x).g(x) → F'(x)= f'(x).g(x)+g'(x).f(x)

and i want to calculate nth order.

**n>1**

f

^{n}(x).g(x)+ [itex]\sum^{n}_{k=1}[/itex] binomial coefficient

[itex]\left(n k\right)[/itex]. f

^{(n-k)}(x).g

^{k}(x)+ g

^{n}.f(x)

is it true? and i found myself but i'm sure this formula is exist.:uhh: