Massless Scalar Field: Taking $ \partial / ( \partial( \partial_\mu \phi ) ) $ The Lagrangian density for a massless scalar field is $ \mathcal{L} = \frac{1}{2} ( \partial_\mu \phi ) ( \partial^\mu \phi ) $. In order to derive the equations of motion for such a Lagrangian, we plug it into the Euler-Lagrange equation. In particular we need to compute the quantity $ \partial_\mu \frac{ \partial \mathcal{L} }{  \partial( \partial_\mu \phi ) } $. I have two questions: 


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*Is the following derivation correct (ignoring the $1/2$) :


$$ \begin{align*}
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\partial_\mu \frac{ \partial}{ \partial( \partial_\mu \phi ) } [ ( \partial_\nu \phi ) ( \partial^\nu \phi ) ] 
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&= \partial_\mu \frac{ \partial}{ \partial( \partial_\mu \phi ) } [ ( \partial_\nu \phi ) g^{\nu\alpha} ( \partial_\alpha \phi ) ]
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\\ &= \partial_\mu g^{ \nu\alpha } [ ( \partial_\alpha \phi )\delta_\nu^\mu + ( \partial_\nu \phi ) \delta_\alpha^\mu ]  \quad \text{(Using product rule)}
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\\ &= \partial_\mu  [ (\partial^\nu \phi ) \delta_\nu^\mu  + ( \partial^\alpha \phi ) \delta_\alpha^\mu ] \quad \text{(Evaluating metric on both terms)} 
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\\ &= 2 \partial_\mu \partial^\mu \phi \quad \text{(Evaluating delta sum)}
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\end{align*} $$


*If it was correct, was such a derivation necessary, or is there a shorter way? Using the notation $ \mathcal{L} = ( \partial_\mu \phi )^2 $ makes it look very tempting to just apply the "power rule" to get $2 \partial_\mu \phi $ when we take $ \partial / \partial ( \partial_\mu \phi) $, but it just doesn't feel right... is that legitimate math?

 A: I'll present a very explicit approach, which makes it obvious it's the correct answer. Notice we can express the kinetic term as,
$$T \equiv\partial_\mu \phi \partial^\mu \phi = (\partial_t\phi)^2 - \sum_{i=1}^{d-1}(\partial_i \phi)^2$$
since the metric $\eta = \mathrm{diag}(1,-1,\dots,-1)$. We see that,
$$\frac{\partial T}{\partial (\partial_t \phi)} = 2\partial_t\phi, \quad \frac{\partial T}{\partial(\partial_i \phi)} = -2\partial_i \phi$$
and noting that $\partial^\mu = (\partial_t,-\nabla)$ it immediately follows that,
$$\frac{\partial}{\partial(\partial_\mu \phi)} \partial_\nu \phi \partial^\nu \phi = 2 \partial^\mu \phi.$$
This should elucidate why we can quickly jot down the answer using an analogous 'power rule' though one should be careful when dealing with new terms, and double check explicitly.
A: One way could be to consider that this is the Lagrangian density of a massless real scalar field and therefore the equation will be Dalambertian phi equal to zero; Comparing terms with the Euler Lagrange equation you get the same without having to derive
