# Higher mixed partial derivatives of $e^{f(x)}$

I have a function $g(x) = e^{-f(x)}$, $x = (x_1,x_2,...,x_n)$. Is there some compact and beautiful formula for the derivative $\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}...\partial x_n^{\alpha_n}}g(x)?$ Maybe in the case of $n=2$?

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I doubt it. Once you take any first partial derivative of $g$, you get a product of functions and I can only see things getting uglier. – Ravi Jan 27 '12 at 21:35
For $n=1$ and $f(x)=x^2$ you have to introduce Hermite polynomials. So we are not optimistic for a nice general formula. – Davide Giraudo Jan 27 '12 at 21:38
But maybe it can be written in nice form via symbols $\sum$ and $\prod$ (something like Faa di Bruno formula)? – Nimza Jan 27 '12 at 21:48
Here I've found some generalisation of Faa di Bruno formula. I'll try with it. combinatorics.org/Volume_16/PDF/v16i1n21.pdf – Nimza Jan 27 '12 at 22:05
The short answer is "yes". I'll post a longer answer below. – Michael Hardy Jan 28 '12 at 0:05

You suggest $n=2$. Here it is if $n=3$: \begin{align} \frac{\partial^3}{\partial x_1\;\partial x_2\;\partial x_3} e^{f(x_1,x_2,x_3)} & = e^{f(x_1,x_2,x_3)} \left( \frac{\partial^3 f}{\partial x_1\;\partial x_2\;\partial x_3} \right. \\ \\ \\ \\ & {} + \frac{\partial f}{\partial x_1} \cdot \frac{\partial^2 f}{\partial x_2\;\partial x_3} + \frac{\partial f}{\partial x_2} \cdot \frac{\partial^2 f}{\partial x_1\;\partial x_3} + \frac{\partial f}{\partial x_3} \cdot \frac{\partial^2 f}{\partial x_1\;\partial x_2} \\ \\ \\ \\ & \left.{} + \frac{\partial f}{\partial x_1} \cdot \frac{\partial f}{\partial x_2} \cdot \frac{\partial f}{\partial x_3} \right). \end{align} The idea is to run through the list of all (unordered) partitions of the set of variables. If two or more of the variables happen to be the same variable, e.g. if one wants $\displaystyle\frac{\partial^3 }{\partial x_1\;\partial x_2^2}e^{f(x_1,x_2)}$, then just identify the two indices $2$ and $3$, so that some of the terms in the expansion become the same as each other.

Here's an excercise: Proof this by induction on the number of variables.

And look up the term "exponential formula" in Richard Stanley's Enumerative Combinatorics, volume 2. He gives the formal-power-series version of the formula, where the partial derivatives are coefficients in the power series.

Later note: OK, I've momentarily stopped being lazy and located the paper I wrote about this and closely related topics: http://www.combinatorics.org/Volume_13/v13i1toc.html

It's the very first paper in the volume for the year 2006.

In particular, see the bottom of page 3.

Quoting: $$\frac{\partial^n}{\partial x_1\cdots\partial x_n} e^y = e^y \sum_\pi \prod_{B\in\pi} \frac{\partial^{|B|}y}{\prod_{j\in B}\partial x_j}$$ where the sum is over all partitions $\pi$ of the set $\{1,\ldots, n\}$ and the product is over all the parts $B$ of the partition $\pi$, and $|B|$ is the number of members of $B$.

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One can generalize this to arbitrary mixed partials with the same exact formula by considering multisets and subsequently partitions of multisets. (And as it turns out, I see that's what you actually did in your paper. Nice.) – anon Jan 28 '12 at 0:47
Great thanks, your article combinatorics.org/Volume_13/PDF/v13i1r1.pdf perfectly clarifies situation! – Nimza Jan 28 '12 at 10:00