Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
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
add comment

1 Answer

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!

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.