Lie bracket simplification Could someone help me simplify the following:

Let $$X= -x^1\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2} \qquad Y = x^2\frac{\partial}{\partial x^1}$$
  Calculate $[X,Y]$

This is what I've got:
Using the identity $[X,g\cdot Y] = X(g)\cdot Y +g\cdot [X,Y]$ and anticommutativity I get:
\begin{align}
[X,Y] &= \left( -x^1\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}\right) (x^2) \cdot \frac{\partial}{\partial x^1} + x^2\cdot \left[-x^1\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^1}\right]\\
&=   x^2 \cdot \frac{\partial}{\partial x^1} - x^2\cdot \left[\frac{\partial}{\partial x^1}, -x^1\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}\right]
\end{align}
How can I continue?
Solution should be $(x^1+x^2)\dfrac{\partial}{\partial x^1}$ but I don't see how I could get there...
 A: Firstly, the commutator is linear in that $[A,B+C]=[A,B]+[A,C]$.
Now note that $[\partial_1,x^2 \partial_2]=0$ since they are in different variables. Then either expand out the commutator directly, or use the identity
$$ [A,BC] = [A,B]C + B[A,C] $$
to get the others.
A: In
$$\begin{gathered}
  {L_{[X,Y]}}(f) = X(Y(f)) - Y(X(f)) \hfill \\
   = ( - x\frac{\partial }{{\partial x}} + y\frac{\partial }{{\partial y}})(y\frac{{\partial f}}{{\partial x}}) - y\frac{\partial }{{\partial x}}( - x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}}) \hfill \\ 
\end{gathered}$$
set $f=x$
$${L_{[X,Y]}}(x) = ( - x\frac{\partial }{{\partial x}} + y\frac{\partial }{{\partial y}})(y\frac{{\partial x}}{{\partial x}}) - y\frac{\partial }{{\partial x}}( - x\frac{{\partial x}}{{\partial x}} + y\frac{{\partial x}}{{\partial y}}) = 2y$$
now set $f=y$
$${L_{[X,Y]}}(y) = ( - x\frac{\partial }{{\partial x}} + y\frac{\partial }{{\partial y}})(y\frac{{\partial y}}{{\partial x}}) - y\frac{\partial }{{\partial x}}( - x\frac{{\partial y}}{{\partial x}} + y\frac{{\partial y}}{{\partial y}}) =  - y\frac{{\partial y}}{{\partial x}} = 0$$
This shows:
$${L_{[X,Y]}} = 2y\frac{\partial }{{\partial x}}$$
