Dear Feng Qi,

a simple induction shows that the formula you are looking for is

$$

\frac{d^m}{d x^m} (f_1 (x) \cdots f_n (x) )= \sum_{i_1 + \cdots + i_n = m} C_{m; i_1, \ldots, i_n} \frac{d^{i_1} f_1}{dx^{i_1}} (x) \cdots \frac{d^{i_n} f_n}{dx^{i_n}} (x)

$$

where $i_1, \ldots, i_n$ are non-negative integers, and $C_{m; i_1, \ldots, i_n} := \frac{n!}{i_1! \cdots i_n!}$ is the multinomial coefficient (which I forgot in a previous version of my answer).

Best

Luca

Dear Feng Qi, you can find it on:

https://en.wikipedia.org/wiki/General_Leibniz_rule (More than two factors)

with the coefficients: https://en.wikipedia.org/wiki/Multinomial_theorem

or http://www.jstor.org/stable/2312259?seq=1#page_scan_tab_contents p.739 etc.

Best regards, Viera

Great! Thanks all.

For higher derivatives of a function of a single variable, variations on the Leibniz formula suffice.  If the function has several variables, use generalizations of the binomial theorem (trinomial, etc.).  For products of many functions, it may be more compact to work with the log, which is a sum.

Dear Feng Qi,. You can use the polynomial formula for (f_1 + f_2 + ... f_m)^n and replace f_j^k by derivative f_j^{(k)}.

Thank all a lot for your ideas and recommended references. Thanks!