Chain Rule for more than two functions, general formula. For a composition of $k$ functions, $F(x) = f_1 \circ f_2 \circ \dots \circ f_k(x)$, (I'm not sure if that is the correct notation for function composition of more than two functions), is there a general formula for finding $F'(x)$?
 A: Set  $f_{k+1}(x)=x$
$$F'(x)=\prod\limits_{i=1}^{k}f'_i(f_{i+1} \circ f_{i+2} \circ \dots \circ f_k(x))$$
A: I always find it easier to think about the chain rule using Leibniz-style notation.  In this style, the usual two step version of the chain rule is this:
$$
\frac{\textrm{d}y}{\textrm{d}x}=\frac{\textrm{d}y}{\textrm{d}u}\cdot\frac{\textrm{d}u}{\textrm{d}x}
$$
Now, if $v$ is an intermediate step between $y$ and $u$, then we also have this:
$$
\frac{\textrm{d}y}{\textrm{d}u}=\frac{\textrm{d}y}{\textrm{d}v}\cdot\frac{\textrm{d}v}{\textrm{d}u}
$$
Putting the two together gives:
$$
\frac{\textrm{d}y}{\textrm{d}x}=\frac{\textrm{d}y}{\textrm{d}v}\cdot\frac{\textrm{d}v}{\textrm{d}u}\cdot\frac{\textrm{d}u}{\textrm{d}x}
$$
As an example, suppose we want to find $\frac{\textrm{d}y}{\textrm{d}x}$ where 
$$y=\sqrt{\sin(2x+1)}$$
Then we can put $u=2x+1$ and $y=\sqrt{\sin(u)}$. It's easy to find $\frac{\textrm{d}u}{\textrm{d}x}$, but $\frac{\textrm{d}y}{\textrm{d}u}$ requires another application of the chain rule.  Put $v=\sin(u)$ so that $y=\sqrt{v}$.  Now everything is easy:
\begin{eqnarray}
\frac{\textrm{d}y}{\textrm{d}x} & = & \frac{\textrm{d}y}{\textrm{d}v}\cdot\frac{\textrm{d}v}{\textrm{d}u}\cdot\frac{\textrm{d}u}{\textrm{d}x} \\
& = & \frac{1}{2\sqrt{v}}\cdot \cos(u) \cdot 2 \\ 
& = & \frac{\cos(2x+1)}{\sqrt{\sin(u)}} \\
& = & \frac{\cos(2x+1)}{\sqrt{\sin(2x+1)}} 
\end{eqnarray}
A: You can compute it recursively: chain rule (Wikipedia)

To take the derivative of a composite of more than two functions,
  notice that the composite of f, g, and h (in that order) is the
  composite of f with g ∘ h.
E.g.: $(f \circ g \circ h)'(a) = f'((g \circ h)(a))\cdot (g \circ h)'(a) = f'((g \circ h)(a))\cdot g'(h(a))\cdot h'(a).$

A: This isn't quite an answer to the question asked, but maybe it's worth something.
Suppose $f(0) = 0$ and $f$ is twice differentiable in
a neighborhood of 0.  Then
$$
(\ \underbrace{f \circ \cdots \circ f }_n\ )''(0)
= f''(0)\left( f'(0)^{n-1} + f'(0)^n + \cdots + f'(0)^{2n-2}  \right)
\text{ for }n \ge 1.
$$
I found this $\ldots\ldots$ somewhere $\ldots\ldots$ and I expect it can be proved by induction on $n$.
A: You can group the last $k - 1$ functions and use chain rule on that to get
$$
(f_1 \circ (f_2 \circ \dotsb \circ f_k))'(x) = {f_1}'((f_2 \circ \dotsb \circ f_k)(x)) \cdot (f_2 \circ \dotsb \circ f_k)'(x)
$$
Now you can repeat the process for $(f_2 \circ \dotsb \circ f_k)'(x)$ to get
$$
(f_2 \circ \dotsb \circ f_k)'(x) = {f_2}'((f_3 \circ \dotsb \circ f_k)(x)) \cdot (f_3 \circ \dotsb \circ f_k)'(x)
$$
and hence
$$
(f_1 \circ (f_2 \circ \dotsb \circ f_k))'(x) = {f_1}'((f_2 \circ \dotsb \circ f_k)(x)) \cdot {f_2}'((f_3 \circ \dotsb \circ f_k)(x)) \cdot (f_3 \circ \dotsb \circ f_k)'(x).
$$
We can keep repeating this process until we reach the end when we get the formula given by Yangzhe Lau. You can make this more rigorous by proving it by induction.
