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Imagine we have a product of functions $f_1\cdots f_m$. We know a rule to compute the derivative. On the other hand, we also have a rule or formula to compute the $n$-th derivative of $fg$ but my question is: Does anyone have a smart notation of way to write:

$$\frac{d^n}{dx^n}(f_1\cdots d_m) = \sum_{\text{indices}} \text{something}.$$

Because the derivatives become mixed up.

Thank you guys!

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Iterating Leibniz rule I found: $\sum_{k_1=0}^n \sum_{k_2=0}^{k_1}\cdots \sum_{k_{m-1}}^{k_{m-2}} C(n,k_1)C(k_1,k_2) \cdots C(k_{m-2},k_{m-1}) f_1^{(k_{m-1})} \cdots f_{m-1}^{(k_1-k_2)}f_m^{(n-k_1)}$. Can the sum or the product of combinatorial numbers be written in a shorter form? Is there a better way? – Derivator Jan 24 '13 at 18:25

Well, you can think of the expression as you choose $n$ of the functions to take a derivative at, i.e. perhaps you take the derivative of $f_1$ once and $f_2$ $n-1$ times and there are $n-1$ ways to do this. To prove this is equivalent simply consult the product rule on $n$ functions when only taking $1$ derivative. It immediately follows that: $$\frac{d^n}{dx^n} (f_1f_2...f_m) = \sum_{a_1 + a_2 + ... + a_m = n} \dbinom{n}{a_1,a_2,...,a_m} f_1^{(a_1)}...f_m^{(a_m)}$$ where $\displaystyle \dbinom{n}{a_1,a_2,...,a_m} $ denotes a multinomial coefficient. So we get a very familiar expression that looks quite similar to the multinomial theorem actually!

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Very nice expression! that's exactly what I needed! Thanks a lot :) Would you have a quick answer for a similar expression, when each $f_i = g_i\circ h$? a composition of a function with another function which is the same for all $g_i$. :) – Derivator Jan 24 '13 at 20:15
Ah! I think the answer is Faà di Bruno's formula :) thanks again! :D – Derivator Jan 24 '13 at 20:29

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