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I am trying to express the $k$-th derivative of $g(x) = exp(f(x))$ in a meaningful way. My logic got me so far. For the first derivative, the result would be $$ g' = f' g $$ Second derivative would be $$ g'' = (f'' + f'^2) g $$ Third derivative is $$ g''' = f''' + 3 f'' f' + f'^3 $$ and so on.

The correspondence that I made was that each derivative can be expressed as $$ f ''' \equiv (0,0,3) $$ $$ f'f'' = (0,1,2) $$ and $$ f'^3 \equiv (1, 1, 1). $$ This is identical to representing the partitions of the integer 3. This is similar to the partitions of a set; a set of size 3, $\left\{a,b,c\right\}$, can be partitioned into $$ \left\{a,b,c\right\} \equiv(0,0,3) $$ $$ \left\{a\right\} \left\{b,c\right\}, \left\{b\right\} \left\{a,c\right\}, \left\{c\right\} \left\{a,b\right\} \equiv(0,1,2) $$ $$ \left\{a\right\} \left\{b\right\} \left\{c\right\} \equiv(1,1,1). $$ The coefficient, for example, of $f'f''$, is the number of partitions corresponding to $(0,1,2)$. Then, I know that these coefficients can be calculated with the help of multinomial coefficients.

My question is: is there a proof of (or a reference about) this correspondence between the derivatives of $g$ and counting the number of partitions?

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    Have you tried induction? Seems like a reasonable idea here. – 2017-02-06
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    Yes it is: https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno's_formula – 2017-02-06
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    I'm not sure, but this might relate to the Butcher group: https://en.wikipedia.org/wiki/Butcher_group – 2017-02-07

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The paper Deriving Faa di Bruno’s formula for the derivative of a composite function via compositions of integers by Steffen Egger could be of interest to you.

It derives the $n$th derivative of composite functions by means of partitions resp. compositions, their ordered companions.

Note: If we consider an integer $n$ as (ordered) sum with $k$ parts $$n=\pi_1+2\pi_2+\cdots +k\pi_k$$ we can represent it as $k$-tuple \begin{align*} (\pi_1,\pi_2,\ldots,\pi_k) \end{align*} Similarly we can represent \begin{align*} f^{\prime\prime\prime} &\equiv (0,0,1)\\ f^{\prime}\cdot f^{\prime\prime}&\equiv (1,1,0)\\ \left(f^{\prime}\right)^3&\equiv (3,0,0) \end{align*}