Confusion regarding the $\partial_{\mu}$ operator I've been confused about the $\partial_{\mu}$ operator.
Peskin and Schroeder defines it as $\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$
For example, the Euler Lagrange equation of motion is 
$$ \frac{\partial}{\partial x^{\mu}}\Big( \frac{\partial \mathcal{L}}{\partial (\partial \phi/\partial x^{\mu})}\Big) - \frac{\partial\mathcal{L}}{\partial \phi} = 0$$
If we apply this to the Lagrangian 
$$ \mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi)^2  - \frac{1}{2}m^2 \phi^2$$
Then 
$$  \frac{\partial \mathcal{L}}{\partial (\partial \phi/\partial x^{\mu})} = \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} =  \partial_{\mu}\phi$$
And thus the first term should be $$ \partial_{\mu} \partial_{\mu} \phi$$
But the first term in the Klein Gordon equation is 
$$ \partial^{\mu} \partial_{\mu} \phi$$
Where am I going wrong here? How do contravariant and covariant tensors and the Minkowski metric tensor apply?
[Edit]
Follow up question:
$$\partial_{\mu}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\Bigg) = \partial_{t}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{t} \phi)}\Bigg) + \partial_{x}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{x} \phi)}\Bigg)$$
        But if
        $$ \frac{\partial \mathcal{L}}{\partial (\partial_{t} \phi)} = \partial_{t} \phi = \partial^{t} \phi$$
        And
        $$ \frac{\partial \mathcal{L}}{\partial (\partial_{x} \phi)} = -\partial_{x} \phi = \partial^{x} \phi$$
        Then 
        $$\partial_{\mu}\Bigg(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}\Bigg) = \partial_{t} \partial^{t} \phi + \partial_{x} \partial^{x} \phi$$
        We seem to missing a minus sign here. Where's the mistake?
 A: Perhaps it is useful to just expand these terms without the fancy notation.
$$
\frac{1}{2} (\partial_\mu \phi)^2 = \frac{1}{2} ( \partial_\mu \phi) (\partial^\mu \phi) = \frac{1}{2} \left( (\partial_t \phi)^2 - (\partial_x \phi)^2 \right)
$$
Therefore
$$
\frac{\partial \frac{1}{2} (\partial_\mu \phi)^2}{\partial (\partial_t \phi)} = \frac{\partial \frac{1}{2} (\partial_t \phi)^2}{\partial (\partial_t \phi)} = \partial_t \phi
$$
$$
\frac{\partial \frac{1}{2} (\partial_\mu \phi)^2}{\partial (\partial_x \phi)} = \frac{\partial -\frac{1}{2} (\partial_x \phi)^2}{\partial (\partial_x \phi)} = -\partial_x \phi
$$
These two things can be summarized as
$$
\frac{\partial \frac{1}{2} (\partial_\nu \phi)^2}{\partial (\partial_\mu \phi)} = \partial^\mu \phi
$$
The operator $\partial^\mu$ is different from the operator $\partial_\mu$. Differential operators exist naturally with the lowered indices. Raising them involved multiplying them by the metric. So in the mostly minus convention $(+---)$ we have
$$
\partial_\mu = (\partial_t, \partial_x, \partial_y, \partial_z)
$$
$$
\partial^\mu = (\partial_t, -\partial_x, -\partial_y, -\partial_z)
$$
