Let $\mathfrak{S}\_n$ be the permutation group on $[n]$. Denote the cardinality of $\{\pi\in\mathfrak{S}\_n: \pi^2=id\}$, the set of [involutions](http://mathworld.wolfram.com/PermutationInvolution.html), by $I(n)$.
It is well-known that these numbers have the exponential generating function
$$\sum\_{n\geq0}I(n)\frac{x^n}{n!}=e^{x+\frac12x^2}.$$


After these preparations, we wish to tackle the problem of computing arbitrary order derivatives of the function $f(x)=e^{x+\frac12x^2}$, which sure get complicated very quickly. 



> 
> **Question.** Instead, is it true we could bypass this with 
>  $$\frac{D^mf(x)}{f(x)}=(x+I)^m?$$
> 
> 
> 


In [*umbral notation*](http://www.mathe2.uni-bayreuth.de/axel/papers/gessel:applications_of_the_classical_umbral_calculus.pdf), one writes $I^k=I(k)$.


I wish to see combinatorial arguments; if not, other cute proofs will do.



permutations rescuing chain/product rules?
Here is what I would consider a "cute proof." Write
 \begin{eqnarray\*} \sum\_{m\geq 0} D^mf(x) \frac{t^m}{m!} & = &
 f(x+t)\\ & = & f(x)e^{t+\frac 12t^2 +tx}. \end{eqnarray\*}
We get your formula by taking the coefficient of $t^m/m!$ in the
product
 $$ e^{t+\frac 12t^2}e^{tx}. $$
Note also that directly from the Exponential Formula we get the equivalent formulation
 $$ (x+I)^m = \sum\_w (1+x)^{c\_1(w)}, $$
where the sum is over all involutions $w\in\mathfrak{S}\_m$, and
 $c\_1(w)$ is the number of fixed points of $w$.


