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 .

Consider two vector fields $X$ and $Y$. Note that starting from the point $p$, the vector field $X$ generates the curve $\phi(p,t)$ via the following equation
\begin{align*}
\frac{d}{dt}(f(\phi(p,t))|_t = Xf|_t
\end{align*} with $\phi(p,0)=p$.
Also, note that for small $t$, $f(\phi(p,t))=f(p)+Xf|_pt$.
We can define the Lie derivative of $Y$ with respect to $X$ as
\begin{align*}
(\mathcal{L}_XY)_pf = &\lim_{t \to 0} \frac{(Yf)(\phi(p,t))-(\phi_{*t}Y_p)f}{t} \\
= &\lim_{t \to 0} \frac{(Yf)(p)+X(Yf)|_pt-Y_p(f+X_pft)}{t}\\
=&\lim_{t \to 0} \frac{(Yf)(p)+(XY)_pft-(Yf)_p-(YX)_pft}{t}\\
=&(XY-YX)_pf.
\end{align*}
In the first line of the above equation $\phi_{*t}Y_p$ denotes the push forward of $Y_p$ along the curve $\phi(p,t)$.
