# How is Leibniz's rule for the derivative of a product related to the binomial formula? [duplicate]

Possible Duplicate:
“Binomial theorem”-like identities

The binomial formula describes the expansion of the $n$th power of the sum $(a+b)$:

$$(a+b)^n = \sum_{k = 0}^n {n\choose k}a^kb^{n-k}$$

In calculus, there is a generalization of the product rule called Leibniz's rule, which describes the expansion of the $n$th derivative of the product of two functions $f$ and $g$:

$$(fg)^{(n)} = \sum_{k = 0}^n {n\choose k}f^{(k)}g^{(n-k)}$$

It's not difficult to prove using the product rule and induction, what I'm really trying understand here is why the formula is true. Is there a way to prove it using the binomial formula? How are these two formulas related? Is there a combinatorial proof for Leibniz's rule like there is for the binomial formula?

I understand this question is a bit vague, but I hope you'll agree that these two formulas look so similar it just feels like there should be some relation between the two. Of course this is not necessarily true, but I'm hoping it's not just a crazy coincidence.

## marked as duplicate by Qiaochu YuanApr 22 '12 at 21:30

• For the longest time I've been looking for a connection too. I was trying to interpret the pointwise mult. and differentiation in Leibniz's formula as ring operations of some sort (which if successful would make Leibniz pop out of the binomial expansion, for free) and mostly tried exponentiation (to pass from $+$ in binomial to $\times$ in Leibniz; maybe that was a dead end), but I was stymied trying to come up with a meaningful interpretation of differentiation as an $n$-ary operation (since mult. is supposed to act on sequences of terms) =\ I'm still hoping for something in that direction. – Vandermonde Mar 2 '16 at 4:54

Hint $\$ Let $\rm\: f(x) = {\it e}^{ax},\ g(x) = {\it e}^{bx}$

To understand the innate duality from a more advanced standpoint look up the Umbral calculus, e.g. see Steven Roman's book by that name, or the Rota-Odlyzko exposition.

There's a formal correspondence between the two rules. Both can be understood as special cases of:

Abstract binomial theorem. Assume that we have:

1. A vector space (or even a module) $V$.
2. A family of "general terms" $(T_{i,j}\in V)_{i,j\in\mathbb N}$.
3. A linear operatior $F: V\to V$, such that $F(T_{i,j}) = T_{(i+1),j} + T_{i,(j+1)}$.

Then $F^n(T_{0,0}) = \sum_{i=0}^n \binom{n}{i} T_{i,(n-i)}$.

The ordinary binomial theorem results if we take $F$ to be multiplication by $(a+b)$ and $T_{i,j}=a^ib^j$ -- in particular $T_{0,0}=1$.

The Leibniz rule results if we take $F$ to be $\frac{d}{dx}$ and $T_{i,j}=f^{(i)}g^{(j)}$ -- in particular $T_{0,0}=fg$.

• Great answer. This is a good general result to know. – Pedro Tamaroff Apr 23 '12 at 1:00
• Great generalization. Can you give me a reference of this? – Bumblebee Jan 3 '17 at 1:48