Lagrangian for Klein-Gordon equation is given by $$L=\frac{1}{2}\partial_\mu \phi \partial^\mu\phi - m^2\phi^2/2.$$
To derive Klein-Gordon Equation I have to Compute derivative in Euler-Lagrange equation. \begin{equation} \frac{\partial L}{\partial \phi}-\partial_\mu \left(\frac{\partial L }{ \partial \partial_\mu \phi}\right)=0\tag{2.3} \end{equation} see Peskin equation 2.3. or $$ \frac{1}{2} \frac{\partial}{\partial \partial_\mu\phi} (\partial_\nu \phi\partial^\nu\phi)$$
= $$\frac{1}{2}\left(\frac{\partial \partial_\nu \phi}{\partial \partial_\mu \phi}\partial^\nu\phi+\frac{\partial \partial^\nu\phi}{\partial \partial_\mu \phi}\partial_\nu\phi\right)$$ $$\frac{1}{2}\left(\frac{\partial \partial_\nu \phi}{\partial \partial_\mu \phi} \partial^\nu \phi +\frac{\partial \partial^\nu\phi}{\partial \partial_\mu \phi}\partial^\lambda\phi\eta_{\lambda\nu}\right).$$ now I am taking the $$\frac{\partial \partial_\nu \phi}{\partial \partial_\mu \phi} = \frac{\partial_\nu \partial \phi}{\partial_\mu \partial \phi} = \frac{\partial x^\mu}{\partial x^\nu} =\delta^\mu_\nu.$$
similarly second term becomes
similarly second term becomes $$\eta^{\nu \mu} \eta_{\lambda\nu} = \delta^\mu_\lambda$$
and final derivative becomes $$\frac{1}{2}(\delta^\mu_\nu \partial^\mu\phi + \delta^\mu_\lambda \partial^\lambda\phi)=\partial^\mu\phi.$$
I have following doubt can I take $$\partial (\partial_\nu\phi) = \partial_\nu (\partial \phi)?$$ many books gives $$\delta(\partial_\nu \phi) = \partial_\nu (\delta\phi).$$ is it true for $\partial$ also? is $\partial$ and variation $\delta$ are same?
Is it correct to take $\partial$ (the derivative which appears in Euler Lagrange equation) inside?
or in other way can i assume that $\partial_\mu \phi$ and $\partial_\nu \phi$ is independent if $\mu \ne \nu$? Here derivation of Euler Lagrange equation is given with $\delta$ variation notation, which seems to be correct as we vary the action.
Since Lagrangian density L is $L(\phi, \partial_\mu\phi)$ in phase space $\phi$, $\partial_\mu \phi$ are field variable and derivative and can be treated independently in phase space, or $\partial_\mu \phi$ and $\partial_\nu\phi$ are two independent variable if $\mu \ne \nu$. Hence partial derivative of $\partial_\mu\phi$ with respect to $\partial_\nu\phi$ will give a $\delta^\nu_\mu$. This explains the derivation without exchanging the $\delta$ or $\partial$.
