Does it make sense to speak in a total derivative of a functional? Part II I am trying to derive the Noether theorem from the following integral action:
\begin{equation}
S=\int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left(  \phi_{r},\partial_{\nu}%
\phi_{r},x\right) , \tag{II.1}\label{eq1}%
\end{equation}
where $\phi_{r}\equiv\phi_{r}\left(  x\right) $ represents the $r$-th field of
set $\left\{  \phi_{r}\right\}  $, while $\partial_{\nu}\phi_{r}\equiv
\partial\phi_{r}/\partial x^{\nu}$ represents its fisrt partial derivatives.
The functional $\mathcal{L}\left( \phi_{r},\partial_{\nu}\phi_{r},x\right) $
is the Lagrangian density of the theory and has, as usual, energy density
dimension, whereas $d^{D}x$ is the volume element of $D$-dimensional
spacetime. For simplicity, we have represented by the dependence of the
spacetime coordinates by $x$, such that $x \equiv x^{\mu}$.
I have assumed that the total variation occurs under the following coordinate
transformation
\begin{equation}
x^{\prime\mu}=f^{\mu}\left(  \varepsilon,x^{\nu}\right)  , \tag{II.2}\label{eq2}%
\end{equation}
where $\varepsilon$ is a paramenter responsible for variation. Naturaly, which
if $\varepsilon=0$, so
\begin{equation}
x^{\mu}=f^{\mu}\left(  0,x^{\nu}\right) . \tag{II.3}%
\end{equation}
If the parameter $\varepsilon$ is sufficiently small, it is possible to use a
Taylor expansion to rewrite the expression (\ref{eq2}) as
\begin{equation}
x^{\prime\mu}\approx x^{\mu}+\varepsilon\xi^{\mu}\left(  x\right)  ,
\tag{II.4}\label{eq4}%
\end{equation}
where only the terms of the first order in $\varepsilon$ are considered.
$\xi^{\mu}\left(  x\right)  $ It is a field vector, contravariant, which in
general can be defined by
\begin{equation}
\xi^{\mu}\left(  x\right)  =\dfrac{\partial x^{\prime\mu}}{\partial
\varepsilon}\rule[-0.35cm]{0.02cm}{0.9cm}_{\varepsilon=0}.\tag{II.5}%
\end{equation}
In the literature, it is common to denote $\varepsilon\xi^{\mu}$ by
$\delta{x^{\mu}}$, i.e., $\varepsilon\xi^{\mu}\equiv\delta{x^{\mu}}$.
At this point, I will denote the integral action (\ref{eq1}) in terms of the
prime coordinates, such that,
\begin{equation}
S^{\prime}=\int_{\mathbb{\Omega}^{\prime}}d^{D}x^{\prime}~\mathcal{L}\left(
\phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right)
\text{.} \tag{II.6}\label{eq6}%
\end{equation}
As it is known, the volume element in the coordinates prime is connected to
the volume element of the nonprime coordinates by means of the following
expression
\begin{equation}
d^{D}x^{\prime}=\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert
d^{D}x,\tag{II.7}\label{eq7}%
\end{equation}
where the Jacobian $\left\vert \dfrac{\partial x^{\prime}}{\partial
x}\right\vert $ can be calculated by the following expression
\begin{equation}
\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert =\dfrac{\left(
-1\right)  ^{s}}{D!}\epsilon_{\alpha_{1}\alpha_{2}\cdots\alpha_{D-1}\alpha
_{D}}\epsilon^{\beta_{1}\beta_{2}\cdots\beta_{D-1}\beta_{D}}\dfrac{\partial
x^{\prime\alpha_{1}}}{\partial x^{\beta_{1}}}\dfrac{\partial x^{\prime
\alpha_{2}}}{\partial x^{\beta_{2}}}\cdots\dfrac{\partial x^{\prime
\alpha_{D-1}}}{\partial x^{\beta_{D-1}}}\dfrac{\partial x^{\prime\alpha_{D}}%
}{\partial x^{\beta_{D}}}.\tag{II.8}\label{eq8}%
\end{equation}
Here, the parameter $s$ corresponds to the number of negative eigenvalues of
the metric.
To follow, we must then take the partial derivatives of Eq. (\ref{eq4}), which
leads us to:
\begin{equation}
\dfrac{\partial x^{\prime\alpha_{i}}}{\partial x^{\beta_{i}}}\approx
\delta_{\beta_{i}}^{\alpha_{i}}+\varepsilon\partial_{\beta_{i}}\xi^{\alpha
_{i}}.\tag{II.9}\label{eq9}%
\end{equation}
Substituting (\ref{eq9}) into (\ref{eq8}), we have found, after laborious calculations, that
\begin{multline}
\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert \approx\left(
-1\right)  ^{s}\left[  \dfrac{1}{D!}\epsilon_{\alpha_{1}\alpha_{2}\cdots
\alpha_{D-1}\alpha_{D}}\epsilon^{\alpha_{1}\alpha_{2}\cdots\alpha_{D-1}%
\alpha_{D}}\right.\\ \left.+\dfrac{1}{\left(  D-1\right)  !}\varepsilon\epsilon_{\alpha
_{1}\alpha_{2}\cdots\alpha_{D-1}\alpha_{D}}\epsilon^{\alpha_{1}\alpha
_{2}\cdots\alpha_{D-1}\beta_{D}}\partial_{\beta_{D}}\xi^{\alpha_{D}}\right]
\tag{II.10}\label{eq10}%
\end{multline}
Using relations
\begin{equation}
\epsilon_{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}%
\epsilon^{\beta_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}=\left(
-1\right)  ^{s}\left(  D-1\right)  !\delta_{\alpha_{1}}^{\beta_{1}}%
,\tag{II.11}\label{eq11}%
\end{equation}
and
\begin{equation}
\epsilon_{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}%
\epsilon^{\alpha_{1}\alpha_{2}\alpha_{3}\cdots\alpha_{D-1}\alpha_{D}}=\left(
-1\right)  ^{s}D!,\tag{II.12}\label{eq12}%
\end{equation}
we can show, without much difficulty, that:
\begin{equation}
\left\vert \dfrac{\partial x^{\prime}}{\partial x}\right\vert =\left(
-1\right)  ^{2s}\left(  1+\varepsilon\partial_{\alpha_{D}}\xi^{\alpha_{D}%
}\right) .\tag{II.13}\label{eq13}%
\end{equation}
Now, whatever the value of $s$, $\left( -1\right) ^{2s}=+1$, and so that, we
have that the volume elements are relationship by:
\begin{equation}
d^{D}x^{\prime}=\left(  1+\varepsilon\partial_{\alpha}\xi^{\alpha}\right)
d^{D}x.\tag{II.14}\label{eq14}%
\end{equation}
Now, returning Eq. (\ref{eq6}) and making use of Eq. (\ref{eq14}), we have:%
\begin{equation}
S^{\prime}=\int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left(  \phi_{r}^{\prime
},\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right)  +\varepsilon
\int_{\mathbb{\Omega}}d^{D}x~\mathcal{L}\left(  \phi_{r}^{\prime}%
,\partial_{\nu}^{\prime}\phi_{r}^{\prime},x^{\prime}\right)  \partial_{\alpha
}\xi^{\alpha}.\tag{II.15}\label{eq15}%
\end{equation}
To follow, we make use of Taylor's expansion to write
\begin{equation}
\phi_{r}^{\prime}\left(  x^{\prime}\right)  =\phi_{r}^{\prime}\left(
x+\varepsilon\xi\right)  \approx\phi_{r}^{\prime}\left(  x\right)
+\varepsilon\xi^{\mu}\left(  x\right)  \partial_{\mu}\phi_{r}^{\prime}\left(
x\right)  .\tag{II.16}\label{eq16}%
\end{equation}
We now denote the functional variation of the $\phi_{r}$ field at the same
point in space-time by
\begin{equation}
\phi_{r}^{\prime}\left(  x\right)  =\phi_{r}\left(  x\right)  +\varepsilon
\zeta_{r}\left(  x\right)  .\tag{II.17}\label{eq17}%
\end{equation}
Here, we point out that it is usual in the literature to identify
$\varepsilon\zeta_{r}\left(  x\right)  $ with $\delta{\phi}$, i.e.,
$\varepsilon\zeta_{r}\left(  x\right)  \equiv\delta{\phi\big(x\big)}$.
Substituting (\ref{eq17}) into (\ref{eq16}), we have:
\begin{equation}
\phi_{r}^{\prime}\left(  x^{\prime}\right)  \approx\phi_{r}\left(  x\right)
+\varepsilon\left[  \zeta_{r}\left(  x\right)  +\xi^{\mu}\left(  x\right)
\partial_{\mu}\phi_{r}\left(  x\right)  \right]  ,\tag{II.18}\label{eq18}%
\end{equation}
where we can identity the total variation of the $\phi$ by
\begin{equation}
\zeta_{r}\left(  x\right)  +\xi^{\mu}\left(  x\right)  \partial_{\mu}\phi
_{r}\left(  x\right)  =\frac{\tilde{\delta}{\phi}}{\varepsilon}.\tag{II.19}%
\label{eq19}%
\end{equation}
Similarly, knowing that
\begin{equation}
\partial_{\nu}^{\prime}=\left[  \delta_{\nu}^{\rho}-\varepsilon\partial_{\nu
}\xi^{\rho}\left(  x\right)  \right]  \partial_{\rho},\tag{II.20}\label{eq20}%
\end{equation}
we can show that
\begin{equation}
\partial_{\nu}^{\prime}\phi_{r}^{\prime}\left(  x^{\prime}\right)
\approx\partial_{\nu}\phi_{r}\left(  x\right)  +\varepsilon\partial_{\nu}%
\zeta_{r}\left(  x\right)  +\varepsilon\xi^{\mu}\left(  x\right)
\partial_{\nu}\partial_{\mu}\phi_{r}\left(  x\right)  .\tag{II.21}\label{eq21}%
\end{equation}
Now, from of the Eq.(\ref{eq4}), (\ref{eq18}) and (\ref{eq21}), we can, by
means of Taylor expansion, to write
\begin{equation}
\mathcal{L}\left(  \phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime
},x^{\prime}\right)  =\mathcal{L}\left(  \phi_{r}+\varepsilon\left(  \zeta
_{r}+\xi^{\mu}\partial_{\mu}\phi_{r}\right)  ,\partial_{\nu}\phi
_{r}+\varepsilon\left(  \partial_{\nu}\zeta_{r}+\xi^{\mu}\partial_{\nu
}\partial_{\mu}\phi_{r}\right)  ,x+\varepsilon\xi\right)  ,\tag{II.22}%
\label{eq22}%
\end{equation}
\begin{multline}
\mathcal{L}\left(\phi_{r}^{\prime},\partial_{\nu}^{\prime}\phi_{r}^{\prime
},x^{\prime}\right)  \approx \mathcal{L}\left(\phi_{r},\partial_{\nu}\phi
_{r},x\right)  +  \varepsilon\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\left(  \zeta_{r} + \xi^{\mu}\partial_{\mu}\phi_{r}\right) \\  + \varepsilon
\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\left(
\partial_{\nu}\zeta_{r}+\xi^{\mu}\partial_{\mu}\partial_{\nu}\phi_{r}\right)
+\varepsilon\partial_{\mu}\mathcal{L}\xi^{\mu}.\tag{II.23}\label{eq23}%
\end{multline}
We now use (\ref{eq23}) in (\ref{eq15}) and after some development, we get
\begin{multline}
\dfrac{S^{\prime}-S}{\varepsilon} \approx \int_{\mathbb{\Omega}}d^{D}x~\left\{
\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\zeta_{r} + \dfrac{\partial
\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\nu}\zeta_{r}\right. \\ \left. + \xi^{\mu
}\left(  \dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\partial_{\mu}\phi
_{r}+\dfrac{\partial\mathcal{L}}{\partial\partial_{\nu}\phi_{r}}\partial_{\mu
}\partial_{\nu}\phi_{r}+\partial_{\mu}\mathcal{L}\right)  +\partial_{\mu}%
\xi^{\mu}\mathcal{L}\right\}.\tag{II.24}\label{eq24}%
\end{multline}
    This is where my doubt lies! The first two terms lead to the Euler-Lagrange equation plus a term of total divergence. The other terms must be written in the form of a total divergence that will also "absolve" the divergence term that comes from the Euler-Lagrange equation. 
    The term in parentheses suggests that we may write that term as a total derivative in relation to L. However, I am not sure that this is correct. Such doubt motivated the exposition and inquiries exposed in the post: Does it make sense to speak in a total derivative of a functional? Part I.
 A: *

*The parameter $s$ below eq. (\ref{eq8}) is non-standard. Noether's theorem and its Lagrangian formalism do in general not rely on a metric. Nevertheless, we only need eq. (\ref{eq14}), which is indeed correct.

*Note that the so-called vertical generator $\zeta_{r}\left( \phi_{r}(x),\partial\phi_{r}(x),x\right)$ in eq. (\ref{eq17}) depends on the field and derivatives thereof in important applications, not just $x$. (For a simple example from point mechanics, see e.g. this Phys.SE post.)

*The partial derivative $\partial_{\nu}\zeta_{r}$ in the main eq. (\ref{eq24}) should actually be a total derivative $d_{\nu}\zeta_{r}$. Then the main eq. (\ref{eq24}) leads to Noether's theorem by standard arguments. In particular, the parenthesis $(\ldots)$ in eq. (\ref{eq24}) is indeed the total spacetime derivative $d_{\mu}{\cal L}\equiv \frac{d {\cal L}}{dx^{\mu}}$, cf. OP's question.
A: Following the hints that have been given by @Qmechanic in his reply, and after consulting some references, such that as [1], [2] and [3], I have gone the following conclusions:


*

*The general form to equation \eqref{eq2} should be written as \begin{equation}
x^{\prime\mu}=f^{\mu}\left(  \varepsilon,x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\right)  , \tag{A}\label{A}%
\end{equation} which, automatically, generalized the equation \eqref{eq4} to\begin{equation}
x^{\prime\mu}\approx x^{\mu}+\varepsilon\xi^{\mu}\left(  x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\right).
\tag{B}\label{B}%
\end{equation}

*In \eqref{eq9}, $\dfrac{\partial {x^\prime}^\mu}{\partial {x}^\nu}$ is actually a total partial derivative because the field $\phi_r$ is dependent of $x^\mu$ and it cannot be held fixed such as it would be if we have considered function instead of functionals. The same is valid to $\xi\big(x,\phi_r\big(x\big),\partial_\nu\phi_r\big(x\big)\big)$ and also to  Jacobian in \eqref{eq8} [ See [1] in page 172, footnote 18].

*Once that the conclusion 2 to be valid is easy to conclude that the latter two terms in Eq. \eqref{eq24} leads to total (partial) derivative term, and thus, we can to conclude that situation 1 in Part III is the correct situation.
[1] I. M. Gelfand and S. V. Fomin, Calculus of Variation, Prentice-Hall, Inc, chapter 7;
[2] D. E. Neuenschwander, Emmy Noether's Wonderful Theorem, Johns Hopkins University Press, chapters 4 and 6;
[3] Nivaldo Lemos, Analytical Mechanics, Cambridge University Press, chapter 11.
