Are Lie Brackets and Their Derivatives Equivalent in Vector Field Calculations?

[pleasehelpmeno]

Hi i have two questions:
1) When asked to prove \mathcal{L}_{u}\mathcal{L}_{v}W - \mathcal{L}_{v}\mathcal{L}_{u}W = \mathcal{L}_{[u,v]}.

I achieved [u,v]w = \mathcal{L}_{[u,v]}. This was found by appliying a scalar field <b> to the LHS and simplifying and expanding using + and scalar linearitys to get [u,v]w but I am not sure if these are equivalent.

2) When asked to calculate the Lie bracket [X,Y] where X=5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x} and Y= y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y} is this equivalent to:
\left((5x^{2}\frac{\partial}{\partial t} -4t\frac{\partial}{\partial x})\frac{\partial (y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})}{\partial x^{a}}-(y\frac{\partial}{\partial t} + t\frac{\partial}{\partial y})\frac{\partial (5x^{2}\frac{\partial}{\partial t}-4t\frac{\partial}{\partial x})}{\partial x}\right)\frac{\partial}{\partial x^{b}}

and if so can it be expanded any further I am not so sure but i don't fully understand \frac{\partial}{\partial x^{a}} derivatives.

 

[pleasehelpmeno]

never mind figured it out

 

[dextercioby]

I'm sure you're missing out the vector field W in the RHS of 1).

 

[pleasehelpmeno]

yeah i figured them out and yeah missed it off, thx