Do those iterated increment rates always yield monotonic functions? For any $a\in{\mathbb R}$ and any non-empty open interval
$I$ containing $a$, we have an operator $T_a$ on $C^{\infty}(I,{\mathbb R})$ defined by
$$
T_a(f)(x)=\left\lbrace
\begin{array}{lcl}
\frac{f(x)-f(a)}{x-a} & \text{if} & x \neq a, \\
f'(a) & \text{if} & x = a
\end{array}\right.
$$
A simple computation shows that $T_a$ and $T_b$ commute for any 
two $a,b$. Let $f$ be the natural logarithm function.
Is it true that for any (not necessarily distinct) positive real numbers
$a_1,a_2,\ldots,a_n$, then $g=T_{a_1}T_{a_2}\ldots T_{a_n}f$ is increasing
if $n$ is even and decreasing if $n$ is odd ?
(If true, it would be an analogy with the fact that the odd-order higher derivatives of $f$ decreasing while the even-order ones are increasing).
If all the $a_i$ are equal, then $g$ can be interpreted 
as a Taylor remainder, written as an integral and the monotonicity property follows immediately. The general case is still unclear to me however.
 A: First use induction to prove that for any $f$ we have
$$
(T_a(f))^{(k)}(x)=\frac{f^{(k)}-k(T_a(f))^{(k-1)}(x)}{x-a}.
$$
Using this formula and again induction, arrive at
$$
(T_a(f))^{(k)}(x)=\frac{k!}{(a-x)^{k+1}}\left(f(a)-\sum_{j=0}^k\frac{f^{(j)}(x)}{j!}(a-x)^j \right).
$$
But then, by Taylor's theorem (Lagrange form of the remainder), we have
$$
(T_a(f))^{(k)}(x)=\frac{k!}{(a-x)^{k+1}}\frac{f^{(k+1)}(c)}{(k+1)!}(c-x)^{k+1},
$$
for some $c$ with either $a<c<x$ or $a>c>x$. We obtain that 
$$
sgn((T_a(f))^{(k)}(x))=sgn(f^{(k+1)}(c)).
$$
In particular, for $sgn(f^{(k)})=(-1)^{k+1}$ we have
$$
sgn(T_{a_1}(f))^{(k)}=(-1)^k,\quad sgn(T_{a_1}(T_{a_2}(f)))^{(k)}=(-1)^{k+1},\quad\dots\quad,\quad
sgn(T_{a_1}\dots T_{a_n}(f))^{(k)}=(-1)^{k+n+1}.
$$
For $k=1$, this shows that outside the points $a_1,\dots,a_n$, 
the assertion is true, at least, when all are different (use the commutativity to cover the whole interval $(0,+\infty)$). But then obviously the increasing/decreasing condition holds for all points, since 
$T_{a_1}\dots T_{a_n}(f)$ is continuous.
In the case that some are equal,  a density argument 
(given by Ewan Delanoy in a comment) concludes the proof.
Argument:
Let $(a_1,a_2,…,a_n)$ be an uple where some $a_i$ may be equal and let $x\in \Bbb R$. Then for every $\varepsilon>0$, there is a $(b_1,b_2,…,b_n)$ with $|b_i−a_i|<\varepsilon$, all the $b_i$ pairwise distinct and distinct from $x$, and 
$T_{b_1}…T_{b_n}f(x)→T_{a_1}…T_{a_n}f(x)$. This finishes the proof.
