I am reading An Modern Introduction to Quantum Field Theory by Maggiore. I have difficulty following the calculation of $\delta ( d^4 x)$ and $\delta (\partial_\mu \phi_i)$. Also, wonder whether the infinitesimal parameter $\epsilon^a$ depends on $x$.
On section 3.2 Nother's Theorem, given the transformation $$ \begin{aligned} x^\mu &\longrightarrow x^\mu + \epsilon^aA^\mu_a (x) \\ \phi_i &\longrightarrow \phi_i + \epsilon^aF_{i,a}(\phi, \partial\phi), \end{aligned} $$ where $x^\mu, \;\phi_i$ the coordinate and the field respectively, and $\epsilon^a$ the infinitesimal parameters.
We would like to derive the conserved current $j^\mu_a$ from the variation of the action $$ \delta S = \int \delta(d^4x)L + d^4x \Big(\frac{\partial L}{\partial \phi_i} \delta \phi_i + \frac{\partial L}{\partial \partial_\mu\phi_i} \delta (\partial_\mu\phi_i)\Big) $$ induced by the above transformation. It claims that: $$ \begin{aligned} d^4 x \longrightarrow d^4x(1+ A^\mu_a\partial_\mu \epsilon^a) \end{aligned} $$ and $$ \begin{aligned} \delta (\partial_\mu\phi_i) &= \frac{\partial \phi'^i}{\partial x'^\mu} - \frac{\partial \phi^i}{\partial x^\mu} \\ &= \frac{\partial x^\nu}{\partial x'^\mu} \frac{\partial }{\partial x^\nu}\big(\phi_i + \epsilon^a F_{i,a} \big) - \frac{\partial \phi^i}{\partial x^\mu} \\ &= -(\partial_\mu \epsilon^a) (A^\nu_a\partial_\nu \phi_i - F_{i,a}). \end{aligned} $$ Here we collect terms proportional to $ \partial_\mu \epsilon^a$.
My attempts are: $$ \begin{aligned} \delta (\partial_\mu\phi_i) &= \frac{\partial }{\partial x'^\mu}\big(x'^\nu - \epsilon^a A^\nu_a(x) \big)\big(\partial_\nu \phi^i - \partial_\nu \epsilon^a F_{i,a} \big) - \partial_\mu \phi_i \\ &= \big(\delta^\nu_\mu - \frac{\partial}{\partial x'^\mu}(\epsilon^a A^\nu_a(x))\big) \big(\partial_\nu \phi^i - \partial_\nu \epsilon^a F_{i,a} \big) - \partial_\mu \phi_i. \end{aligned} $$ What do we do with the term $\frac{\partial}{\partial x'^\mu}(\epsilon^a A^\nu_a(x))$? I have no idea what to do for the $\delta (d^4 x)$.
Also, I assume it was global symmetry, i.e., the infinitesimal parameters are constant and doesn't depend on $x$. But wouldn't the term $\partial_\mu \epsilon^a$ just be zero then?
