Question about operation with derviative of product in Noether proof I have been studying Noether's theorem proof, I have a problem understanding one of the last steps, because I don't understand the calculation.
Noether theorem assures us that the action under a transformation has $\delta S=0$
$$
\delta S= \mathcal{L}^{'}(x^{'})- \mathcal{L}(x)  = \mathcal{L}^{'}(x^{'})- \mathcal{L}^{'}(x) + \mathcal{L}^{'}(x)- \mathcal{L}(x).
$$
The first part $\mathcal{L}^{'}(x^{'})- \mathcal{L}^{'}(x)$ is the variation due to the change in coordinates and the second, $\mathcal{L}^{'}(x)- \mathcal{L}(x)$, is the functional variation due to the transformation. My question is about the analysis of the second part.
$$
\mathcal{L}^{'}(x)- \mathcal{L}(x) = \frac{\partial\mathcal{L}}{\partial\phi}\delta_F \phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta_F (\partial_\mu\phi )
$$
With $\delta_F \phi = \delta \phi - \partial_\mu\phi\delta x^\mu$
After some calculation we get to my problem:
$$
\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi
- \frac{\partial\mathcal{L}}{\partial\phi}(\partial_\mu\phi)\delta x^\mu 
+ \partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi \right] 
- \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] \delta\phi 
- \partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}(\partial_\nu\phi)\delta x^\nu \right] 
- \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] (\partial_\nu\phi)\delta x^\nu . 
$$
My book says that all of the above is equal to:
$$
\partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi - \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} (\partial_\nu\phi)\delta x^\nu    \right] .
$$
But i don't see how I get the terms $\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi
$ and $- \frac{\partial\mathcal{L}}{\partial\phi}(\partial_\mu\phi)\delta x^\mu $ from it. I suppose it must be in the form of $\partial_\mu(ABC)$ because we have 3 terms with $\delta\phi$ and 3 with $\delta x^\mu$, but I can't seem to find my way through this calculation.
 A: First of all, the last sign of the long expression should be a $+$, see: Since $\delta_F\phi$ accounts for the change of the field at the same point, $[\delta_F,\partial_\mu]=0$, then
$$\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta_F(\partial_\mu\phi)=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta_F\phi)=\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta_F\phi\right]-\left[\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right]\delta_F\phi=\partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi \right] 
- \partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}(\partial_\nu\phi)\delta x^\nu \right]- \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] \delta\phi 
+ \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] (\partial_\nu\phi)\delta x^\nu.$$
Then, the functional derivative of the Lagrangian (not the density) is
$$\frac{\delta L}{\delta\phi}=\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)},$$
so putting together your first term with the fourth, and the second with the last one, your expression can be converted into
$$
\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi
- \frac{\partial\mathcal{L}}{\partial\phi}(\partial_\mu\phi)\delta x^\mu 
+ \partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi \right] 
- \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] \delta\phi 
- \partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}(\partial_\nu\phi)\delta x^\nu \right] 
+ \left[ \partial_\mu  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right] (\partial_\nu\phi)\delta x^\nu=\\=\frac{\delta L}{\delta\phi}\Big(\delta\phi-(\partial_\nu\phi)\delta x^\nu\Big)+\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi-\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}(\partial_\nu\phi)\delta x^\nu\right].$$
Now, if you consider that equations of motion are satisfied, the first term vanishes, because
$$\frac{\delta L}{\delta\phi}=0.$$
