Derivatives of the commutator of flows (or, what are those higher derivatives doing in my tangent space?!) Let $p \in M$ be a a point in a manifold and let $\varphi^X_t$ and $\varphi^Y_t$ be the local flows of the vector fields $X$ and $Y$ respectively. Define the commutator of flows: $\alpha(t)= \varphi^Y_{-t} \varphi^X_{-t}\varphi^Y_t\varphi^X_t$. I'm trying to prove:
$$\left .\frac{d}{dt} \right|_{t=0}\alpha(\sqrt{t})=[X.Y]_p$$
I managed to prove that $\left .\frac{d}{dt}\right |_{t=0}\alpha(t) =0$ and that it implies that: 
$$\left .\frac{d}{dt}\right |_{t=0}\alpha(\sqrt{t})=2\left .\frac{d^2}{dt^2}\right |_{t=0}\alpha(t)$$ 
But trying to compute the second derivations I got stuck with expressions like these:
$$\left .\frac{d}{dt}\right |_{t=0}\ (X_{\varphi^X_t} \cdot \varphi^Y_{-t}\varphi^X_{-t}\varphi^Y_t)$$
And realized I'm not so sure as to how the second derivations work. Can I use the product rule on the the above expression? More alarmingly the commutator is itself a second derivation! so how come it's a vector field? (obviously it is).
If all commutators are vector fields (and all commutators of them are too and etc..) this means that all these higher derivatives are actually first order derivatives? What's going on? 
Are there more higher derivatives like these that aren't commutators but are still first order derivations?
 A: Here is one way to make calculations like this a bit easier:
In coordinates, write $ X = X^i\;\partial_i $. Then:
$$
\varphi^X_t(p) \approx p^i + t \; X^i(p) + \frac{t^2}{2} X^j(p) \; \partial_j X^i(p) + O(t^3) 
$$
You can check then that
$$
\frac{d}{dt} \varphi^X_t(p) = X^i(p) + t X^j(p) \partial_j X^i(p) + O(t^2) \\
= X^i(p^i + t \; X^i(p)) + O(t^2) \\
= X^i(\varphi^X_t(p)) + O(t^2) 
$$
where I taylor expanded $ X^i $.
Composing the next flow, (and taylor expanding $ Y^i( \varphi^X_t(p)) $  and dropping terms of $ O(t^3) $), I get something like:
$$
 \varphi^Y_t \varphi^X_t(p) = p^i + t X^i + \frac{t^2}{2} X^j \partial_j X^i + t Y^i + \frac{t^2}{2} Y^j \partial_j Y^i + t^2 X^j \partial_j Y^i 
$$
Since I am tired of typing up the long formulas, you can finish off by composing the last two flows. The lucky thing is that the terms $ \frac{t^2}{2} X^j \partial_j X^i $ will cancel away. The only problem is that I am left with $ 2 [X, Y] t^2 $...so I am off by a factor of 2.
PS. I am dying to know what typo suggested that flows are polite?
