permutations rescuing chain/product rules? 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, 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, one writes $I^k=I(k)$.
I wish to see combinatorial arguments; if not, other cute proofs will do.
 A: 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 $w$.
A: Taylor's theorem is often an effective way of finding formulas for higher derivatives.
With $f(x) = e^{x+x^2/2}$, we have
$$
\begin{aligned}
\sum_{m=0}^\infty D^m f(x)\frac{y^m}{m!} &=f(x+y)\\
  &= e^{x+y +(x+y)^2/2}\\
  &=f(x) e^{xy}f(y).
\end{aligned}
$$
Thus $D^m f(x)/f(x)$ is the coefficient of $y^m/m$ in $e^{xy}f(y)$, which is 
$$\sum_{i=0}^m \binom{m}{i} x^i I(m-i)=(x+I)^m.$$
It's not hard to give a corresponding combinatorial proof. 
We can also give an umbral proof, though the proof isn't really any shorter and the computation is similar. We first show that
for any polynomial $p$ we have (umbrally) 
$$p(I)e^{xI}=p(x+I) e^{x+x^2/2}.\tag{$*$}$$
To do this we check that $(*)$ holds for $p(z) = e^{uz}$, where both sides are equal to $f(u+x)$. Then by equating coefficients of $u^m/m!$ we see that $(*)$ holds for $p(z) = z^m$, so by linearity it holds for any polynomial. 
Then 
 $$
D^m f(x) = I^m e^{xI}=(x+I)^m e^{x+x^2/2} = (x+I)^m f(x).
$$
