Lie bracket and flows on manifold Suppose that $X$ and $Y$ are smooth vector fields with flows $\phi^X$ and $\phi^Y$ starting at some $p \in M$ ($M$ is a smooth manifold). Suppose we flow with $X$ for some time $\sqrt{t}$ and then flow with $Y$ for this same time. Then we flow backwards along $X$ for the same time, and then flow backwards along $Y$. All in all, we can define a curve dependent on $t$ as follows $$\alpha(t):= \phi_{-\sqrt(t)}^Y \circ \phi_{-\sqrt(t)}^X \circ \phi_{\sqrt(t)}^Y \circ \phi_{\sqrt (t)}^X$$
It is an exercise to show that $\frac{d}{dt}|_{t=0} \alpha(t) = [X,Y](p)$. In theory, this should be workable with just the chain rule, (assuming I know how to do these derivatives properly, which is something I'd like some clarification on) but this process is going to be excruciatingly long and painful, and it's really just something I want to avoid if I can help it. Is there another way to do this computation that will be less painful and more illustrative of why exactly this works out?
 A: I'm not sure if this is the answer you're looking for, since this is by computation. But it does not involve the chain rule, at least.
So... both sides of the equation are elements of the tangent space $T_pM$. To see that they are equal, we compute their action on a function $f:M \to \mathbb R$. 
Now, by definition
$$
[X,Y](f)(p) = \lim_{h \to 0} \frac 1h \left[ (Yf) \circ \phi_h^X(p)-Y(f)\right] - \lim_{h \to 0} \frac 1h \left[ (Xf) \circ \phi_h^Y(p)-X(f)\right]
$$
The first term is equal to:
$$
\lim_{h \to 0} \frac 1h \left[ (\lim_{k \to 0}\frac 1k [f \circ \phi_k^Y(p)-f(p)]) \circ \phi_h^X(p)-[\lim_{k \to 0}\frac 1k f \circ \phi_k^Y(p)-f(p) ]\right] 
$$
Setting $k=h$ (the functions are differentiable, so this shouldn't change the answer), we get
$$
= \lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^Y \circ \phi_h^X (p) - f \circ \phi_h^X(p) -f \circ \phi_h^Y(p) + f(p) \right]
$$
Doing the same for the other term, we get:
$$
\lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^X \circ \phi_h^Y (p) - f \circ \phi_h^Y(p) -f \circ \phi_h^X(p) + f(p) \right]
$$
Subtracting them, most terms cancel and we get
$$
\lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) - f \circ \phi_h^X \circ \phi_h^Y(p) \right] 
$$
Now we use that $(\phi_h^X)^{-1}=\phi_{-h}^X$ to get
$$
\lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right] 
$$
But this is just the derivative of $\alpha$ (we traverse in the opposite direction, but that's okay)! Putting $t=h^2$ we get the result.
So all in all, this computation was not painfree, but it is clear why we need the square root signs. 
A: I've been working on this problem for a while. Even though this question is from quite a long time ago, I haven't seen an answer which is similar to this one, so I'm posting it mainly because this question appears rather commonly (I first saw it as an exercise in Warner's book).
I'm not going to prove that $\alpha'(0)$ exists, but rather that $\alpha$ defines a derivation at $t=0$ that is equal to the Lie bracket. More precisely, for any $f\in \mathfrak{F}(M)$:
$$
\lim_{t\to 0^{+}}\frac{f(\alpha(t))-f(p)}{t}=[X,Y]_{p}(f).
$$
If the derivative did exist, then this equation would imply that $\alpha'(0)=[X,Y]_{p}$, but paraphrasing Warner's book, the $\sqrt{t}$ makes it so that the derivative may not exist.
First, define $\beta(t)=\Phi_{-t}^{Y}\circ \Phi_{-t}^{X} \circ \Phi_{t}^{Y} \circ \Phi_{t}^{X}(p)$, so that $\alpha(t)=\beta(\sqrt{t})$ for every $t>0$ sufficiently small. Since the square root is a homeomorphism from $[0,\infty)$ to itself, it suffices to prove that
$$
\lim_{s\to 0^{+}}\frac{f(\alpha(s^{2}))-f(p)}{s^{2}}=\lim_{s\to 0^{+}}\frac{f(\beta(s))-f(p)}{s^{2}}=[X,Y]_{p}(f).
$$
By using L'Hôpital's rule twice, this limit can be rewritten as
$$
\lim_{s\to 0^{+}}\frac{(f\circ \beta)''(s)}{2}=\frac{1}{2}(f\circ \beta)''(0).
$$
So, in the end we have reduced the problem to computing the following second derivative:
$$
\frac{d^{2}}{dt^{2}}|_{t=0}f(\beta(t))=\frac{d^{2}}{dt^{2}}|_{t=0}f(\Phi_{-t}^{Y}\circ \Phi_{-t}^{X} \circ \Phi_{t}^{Y} \circ \Phi_{t}^{X}(p)).
$$
A common trick for evaluating this kind of derivatives is to "separate the $t$'s". If we define
$$
H(a,b,c,d)=f(\Phi_{a}^{Y}\circ \Phi_{b}^{X} \circ \Phi_{c}^{Y} \circ \Phi_{d}^{X}(p)), g(t)=H(-t,-t,t,t),
$$
then $(f\circ \beta)''(0)=g''(0)$. By using the chain rule twice, this equals
$$
g''(0)=H_{aa}(0)+H_{bb}(0)+H_{cc}(0)+H_{dd}(0)+2H_{ab}(0)-2H_{ac}(0)-2H_{ad}(0)-2H_{bc}(0)-2H_{bd}(0)+2H_{cd}(0).
$$
There are a lot of second derivatives to compute, but they are relatively easy (I can add an example if needed), and by the end you'll get
$$
g''(0)=YYf(p)+XXf(p)+YYf(p)+XXf(p)+2XYf(p)-2YYf(p)-2XYf(p)-2YXf(p)-2XXf(p)+2XYf(p) \\
=2XYf(p)-2YXf(p)=2[X,Y]_{p}(f).
$$
Now, plug this result into the original limit, and we get
$$
\lim_{t\to 0^{+}}\frac{f(\alpha(t))-f(p)}{t}=[X,Y]_{p}(f).
$$
Hope this helps! If anyone believes there's some error in the proof, I'd like to know, since I haven't seen this argument anywhere else.
EDIT: I'm aware that this solution uses explicitly the chain rule (which the author wanted to avoid), but since it is used in the simplest way I could think of (that is, using it with a function from $\mathbb{R}^{4}$ to $\mathbb{R}$), I hope it's forgivable.
