How to proof that bracket of two vector field can be computed by second derivation Can some one give a hint how can I proof that 

where $\phi$ indicated the flow of vector fields.
 A: For the vector field $X$ and the associated dynamical system $\dot{x}=X(x)$ we have 
$$x(t+h)=x(t)+\int_t^{t+h}{X(x(s))ds}\\ =x(t)+\int_t^{t+h}{[X(t)+\frac{dX}{dx}(x(t))X(t)s]ds}+0(h^2)\\ =x(t)+X(t)h+\frac{1}{2}\frac{dX}{dx}(x(t))X(x(t))h^2+o(h^2)$$
We calculate the evolution of a initial point $x_0$ under the various flows. Initially under $\Phi_t^X$ 
$$x(t)=x_0+X(x_0)t+\frac{1}{2}\frac{dX}{dx}(x_0)X(x_0)t^2+o(t^2)$$
Then  under $\Phi_t^Y$ 
$$x(2t)=x(t)+Y(x(t))t+\frac{1}{2}\frac{dY}{dx}(x(t))Y(x(t))t^2+o(t^2)\\=x_0+[X(x_0)+Y(x_0)]t+\left[\frac{1}{2}\frac{dX}{dx}(x_0)X(x_0)+\frac{dY}{dx}(x_0)X(x_0)+\frac{1}{2}\frac{dY}{dx}(x_0)Y(x_0)\right]t^2+o(t^2)$$
Then  under $\Phi_{-t}^X$ 
$$x(3t)=x(2t)-X(x(2t))t+\frac{1}{2}\frac{dX}{dx}(x(2t))X(x(2t))t^2+o(t^2)\\ =x_0+Y(x_0)t+\left[-\frac{dX}{dx}(x_0)Y(x_0)+\frac{dY}{dx}(x_0)X(x_0)+\frac{1}{2}\frac{dY}{dx}(x_0)Y(x_0)\right]t^2+o(t^2)$$
And finally  under $\Phi_{-t}^Y$ 
$$x(4t)=x(3t)-Y(x(3t))t+\frac{1}{2}\frac{dY}{dx}(x(3t))Y(x(3t))t^2+o(t^2)\\ =x_0+\left[-\frac{dX}{dx}(x_0)Y(x_0)+\frac{dY}{dx}(x_0)X(x_0)\right]t^2+o(t^2)\\ =x_0+\mathcal{L}_XY(x_0)t^2+o(t^2)$$
