Computation in QFT I'm always a mess with the upstairs and downstairs notation. To be specific, say I want to calculate the Euler-Lagrange equations of 
\begin{equation}
\mathcal{L} = \frac{1}{2}\partial^\mu\phi \partial_\mu \phi = \frac{1}{2}(\partial_\mu\phi)^2.\tag{1}
\end{equation}
So the partial derivative with respect to $\phi$ is zero and
$$
\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi.
\tag{2}$$
Why the index $\mu$ is upstairs and no downstairs? If I use the metric $(+,-,-,-)$ and $\partial_\mu = (\frac{\partial}{\partial t}, \nabla)$, then $\partial^\mu$ should have a $-\nabla$, right? Why in this case I can make $\partial_\mu \phi = \partial^\mu \phi$?
 A: You can always write the objects with upper indices as the objects with lower indices contracted with the metric tensor
$$\partial^\mu\phi=\eta^{\mu\nu}\partial_\nu\phi$$
Then you can rewrite the Lagrangian as
$$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi)$$
Taking the derivative we get
$$\frac{\partial\mathcal{L}}{\partial (\partial_\alpha\phi)}=\frac{1}{2}\eta^{\alpha\nu}(\partial_\nu\phi)+\frac{1}{2}\eta^{\mu\alpha}(\partial_\mu\phi)=\partial^\alpha\phi$$
You can expect that the result should have the upper index from the definition of this type of derivative,
$$\frac{\partial v^\nu}{\partial v^\mu}=\delta_\mu^\nu,\quad \frac{\partial v_\nu}{\partial v_\mu}=\delta^\mu_\nu$$
The indices should be located in such a way so that $\delta$-symbols were tensors.
A: At a very basic level, taking a derivative with a down (up) index gives an up (down) index. For instance, let
$$
f(v,w) = v^\mu w_{\mu} = v_{\mu} w^\mu.
$$
Then
$$
\frac{\partial f}{\partial v_\mu} = w^\mu,
\quad
\frac{\partial f}{\partial v^\mu} = w_\mu.
$$
If you understand this example, the QFT computation should not be a problem.
