Second derivative of $f(f(\cdots f(x)\cdots )?$ 
For convenience, let's write $f_n(x)=f(f(\cdots f(x)\cdots )$ where
  $f$ is iterated $n$ times. Suppose: $$f(0)=0,\quad f'(0)=\alpha,\quad f''(0)=\beta$$
  What is $f''_n(0)?$

I've found $f'_n(0),$ using the chain rule:
$$f'_n(x)=\prod_{k=1}^n f'(f_{k-1}(x))\Rightarrow f'_n(0)=\prod_{k=1}^n f'(0)=\alpha^n$$
But I'm stuck on $f_n''(0)$. I am sure there is a pattern, but I can't spot it.
 A: Let's take $f^{\prime\prime}_4(x)$ as an example.
We have $f^\prime_4(x)=f^\prime[f(f(f(x)))]\cdot f^\prime[f(f(x))]\cdot f^\prime[f(x)]\cdot f^\prime(x)$, by the chain rule.
The product rule gives $\frac{d}{dx}p(x)\cdot q(x)\cdot r(x)\cdot s(x)=$$p^\prime(x)\cdot q(x)r(x)s(x)+q^\prime(x)\cdot p(x)r(x)s(x)+r^\prime(x)\cdot p(x)q(x)s(x)+s^\prime(x)\cdot p(x)q(x)r(x)$.
So we just calculate the derivative of each term of the product, and evaluate the other terms. All terms evaluate down to $\alpha$, so our example reduces to $\alpha^3\cdot[p^\prime(x)+q^\prime(x)+r^\prime(x)+s^\prime(x)]$.
$p^\prime(x)=f^{\prime\prime}[f(f(f(x)))]\cdot f^\prime[f(f(x))]\cdot f^\prime[f(x)]\cdot f^\prime(x)=\beta\alpha^3$.
$q^\prime(x)=f^{\prime\prime}[f(f(x))]\cdot f^\prime[f(x)]\cdot f^\prime(x)=\beta\alpha^2$.
$r^\prime(x)=f^{\prime\prime}[f(x)]\cdot f^\prime(x)=\beta\alpha$.
$s^\prime(x)=f^{\prime\prime}(x)=\beta$.
The answer is then $\beta\alpha^3(\alpha^3+\alpha^2+\alpha+1)$. This last sum can be written as $\frac{\alpha^4-1}{\alpha-1}$, when $\alpha\neq1$.
Generalizing, we have $\beta\frac{(\alpha^{n-1})(\alpha^n-1)}{\alpha-1}$, or $n\beta$ when $\alpha=1$.
That's not a proof, but it's good enough to spot the pattern.
Now somebody generalize it to $f^{(m)}_n(0)$.  ;-)
A: Note that if $g(x)=ax+bx^2+O(x^3)$ and $h(x)=cx+dx^2+O(x^3)$ then 
$$\begin{align}
g(h(x)) &=a(cx+dx^2+O(x^3))+b(cx+dx^2+O(x^3))^2+O(x^3)\\
&=acx+(ad+bc^2)x^2+O(x^3)
\end{align}$$
so if we let $g=f$ and $h=f_n$ we get the recurrence 
$$f_{n+1}''(0)=\alpha f_n''(0)+\beta f_n'(0)^2=\alpha f_n''(0)+\beta\alpha^{2n}.$$
