Are there functions $f,g:\mathbb{R} \to \mathbb{R}$ such that they differentiate each other? i.e.
$$f'(x) = (g \circ f)(x)$$
$$g'(x) = (f\circ g)(x)$$
I came up with this question a few years ago. A friend found the only example I know:
for $c\in\mathbb{R}$
$$f(x) = c$$
$$g(x) = cx - c^2 $$
After trying with some particular cases (with no success), I used the formula for the derivative of the inverse function and got that if $f$ and $g$ are bijective then,
$$f^{-1} = \int{}{dt \over g(t)}  $$
$$g^{-1} = \int {dt \over f(t)}$$
Assuming all conditions neccesary for this to be possible. I tried using this fact to construct the functions, again with no luck.
I would really appreciate any insight on how to tackle this problem.
 A: Some partial results, but too long for a comment.
The condition implies that $f,g$ are $C^\infty$.
In particular,
$$ f''=(g\circ f)'=f'\cdot g'\circ f=g\circ f \cdot f\circ g\circ f=(\operatorname{id}\cdot f)\circ g\circ f$$
so that every zero of $f'$ is also a zero of $f''$.
Let $h_0=f$ and recursively $$h_{n+1}=\begin{cases}g\circ h_n&n\text{ even}\\f\circ h_n&n\text{ odd}\end{cases}$$ So far, we have $f=h_0$, $f'=h_1$, $f''=h_1h_2$.
Note that
$$h_{n+1}'=\begin{cases}h_n'\cdot g'\circ h_n=h_n'\cdot f\circ g\circ h_n&n\text{ even}\\
h_n'\cdot f'\circ h_n=h_n'\cdot g\circ f\circ h_n&n\text { odd}\end{cases} $$
So at any rate,
$$ h_{n+1}'=h_n'h_{n+2}.$$
Thus by induction,
$$\tag1 h_n'=\prod_{k=1}^{n+1} h_k.$$
Let $$ H=\Bigl\{\,\prod_{k=1}^m h_k^{a_k}\Bigm| m\ge 1, a_1\ge a_2\ge a_m\ge1 \,\Bigr\}$$
If $\phi\in H$, then by the product rule and $(1)$, $\phi'$ is a finite sum of elements of $H$.
As $f'=h_1\in H$, we conclude that all $f^{(n)}$, $n\ge1$, are finite sums of elements of $ H$.
In particular, for all $x$ with $f'(x)=0$, we have $f^ {(n)}(x)=0$ for all $n\ge1$.
The same conclusion of course holds for $g$.
As a consequence of the identity theorem:

If $f$ is analytic, then $f$ is either constant (and so $g$ linear) or $f'$ has no zeroes (so in particular, $f$ is strictly monotonic). Same for $g$.

So if $f,g$ are analytic and neither is constant, then both are monotonic, hence all $h_n$ are strictly monotonic. If both $f,g$ are increasing, then all derivatives are increasing.
