Prove that the Lie derivative of a vector field equals the Lie bracket: $\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y]$ Let $X$ and $Y$ be vector fields on a smooth manifold $M$, and let $\phi_t$ be the flow of $X$, i.e. $\frac{d}{dt} \phi_t(p) = X_p$. I am trying to prove the following formula:
$\frac{d}{dt} ((\phi_{-t})_* Y)|_{t=0} = [X,Y],$
where $[X,Y]$ is the commutator, defined by $[X,Y] = X\circ Y - Y\circ X$. 
This is a question from these online notes: http://www.math.ist.utl.pt/~jnatar/geometria_sem_exercicios.pdf .

Here is a simple proof which I found in the book "Differentiable Manifolds: A Theoretical Phisics Approach" of G. F. T. del Castillo. Precisely it is proposition 2.20.
We denote $(\mathcal{L}_XY)_x=\frac{d}{dt}(\phi_t^*Y)_x|_{t=0},$ where $(\phi^*_tY)_x=(\phi_{t}^{-1})_{*\phi_t(x)}Y_{\phi_t(x)}.$
Recall also that $(Xf)_x=\frac{d}{dt}(\phi_t^*f)_x|_{t=0}$ where $\phi_t^*f=f\circ\phi_t$ and use that $(\phi^*_tYf)_x=(\phi^*_tY)_x(\phi^*_tf).$
We claim that $(\mathcal{L}_XY)_x=[X,Y]_x.$ 
Proof:$$(X(Yf))_x=\lim_{t\to 0}\frac{(\phi_t^*Yf)_x-(Yf)_x}{t}=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(Yf)_x}{t}=\star$$
Now we add and subtract $(\phi^*_tY)_xf.$ Hence 
$$\star=\lim_{t\to 0}\frac{(\phi^*_tY)_x(\phi^*_tf)-(\phi^*_tY)_xf+(\phi^*_tY)_xf-Y_xf}{t}=$$
$$=\lim_{t\to 0}(\phi^*_tY)_x\frac{(\phi^*_tf)-f}{t}+\lim_{t\to 0}\frac{(\phi^*_tY)_x-Y_x}{t}f=Y_xXf+(\mathcal{L}_XY)_xf.$$
So we get that $XY=YX+\mathcal{L}_XY.$
