I have a continuous transformation on the field $\phi$ of the form
$$\phi(x)\rightarrow \phi'(x)=\phi(x)+\alpha\Delta\phi(x),\tag{1}$$
where $\alpha$ is a constant infinitesimal parameter and $\Delta\phi$ is a deformation of the field. Note that in this notation (Peskin and Schroeder one) $\delta\phi=\alpha\Delta\phi$
To have a symmetry my action should be invariant up to a surface term, so my lagrangian has to be invariant up to a 4-divergence:
\begin{equation} \mathcal{L}\rightarrow\mathcal{L}+\alpha\partial_\mu J^\mu.\tag{2} \end{equation}
Now we proceed varying the Lagrangian:
\begin{equation} \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta\phi)=\alpha\frac{\partial\mathcal{L}}{\partial\phi}\Delta\phi+\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)-\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Big)\Delta\phi.\tag{3} \end{equation}
Now, the first and the third term cancel out because of Euler-Lagrange equations.
If I want to satisfy my symmetry, the variation I just calculated has to be equal to the 4-divergence:
\begin{equation} \alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)=\alpha\partial_\mu J^\mu \rightarrow \alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-J^\mu\Big)=0 .\tag{4} \end{equation}
Thus, the quantity
$$j^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-J^\mu\tag{5}$$ is conserved.
And this is pretty clear to me. But what if $\alpha=\alpha(x)$?
$\textbf{My attempt}$:
Since $\alpha$ is a function of $x$, the quantity $\partial_\mu(\delta\phi)=\partial_\mu(\alpha\Delta\phi)$ becomes $\Delta\phi\partial_\mu \alpha+\alpha\partial_\mu \Delta\phi$
\begin{equation} \delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu(\delta\phi)=\frac{\partial\mathcal{L}}{\partial\phi}\alpha\Delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\partial_\mu \alpha+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\alpha\partial_\mu \Delta\phi.\tag{6} \end{equation}
The first and the third term give rise to a term $\alpha\Delta\mathcal{L}$, exactly like the one we get with constant $\alpha$.
And from here my ideas start to get fuzzy.
So, I got my action varying like:
$$ \delta S=\int d^4x\Big(\alpha\Delta\mathcal{L}+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\partial_\mu \alpha\Big)=\int d^4x\Big(\alpha\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\Big)+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi(\partial_\mu \alpha)\Big)$$ $$=\int d^4x \partial_\mu \Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\alpha\Big).\tag{7} $$
Now, if I have a symmetry, my action is invariant up to a boundary term, hence an integral of $\alpha\partial_\mu J^\mu$. So I get the same result as conserved current $j^\mu$.
$\textbf{My question}$:
In Tong notes (http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf) at page 19 there is a total different approach. He says that the Lagrangian varies like \begin{equation} \delta\mathcal{L}=(\partial_\mu \alpha)h^\mu\tag{8} \end{equation} where $h^\mu$ is the conserved current. Using my notation, I guess that this true only for the cases where $\delta\mathcal{L}=0$ for $\alpha$=const, hence for $J^\mu=0$.
But My professor, and also some other notes, says that since
\begin{equation} \delta S=\int d^4x(\partial_\mu \alpha) h^\mu\tag{9} \end{equation}
if I want to find the conserved current all I have to do is vary the Lagrangian I'm given assuming $\alpha=\alpha(x)$ and then simply search for the quantity "next to" $\partial_\mu \alpha$. Isn't this true only for the cases where
$$\delta\mathcal{L}=0\quad\text{for}\quad \alpha=\text{const}? \tag{10}$$
He didn't say anything about it, suggesting that this is the most general case.
