The question is simple: How do I find the function derivative of $$(\delta/\delta \phi(x)) (\partial_\mu \phi(x))~?$$ As far as I can tell, I cannot use any of the standard computational rules for the functional derivative.

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    $\begingroup$ $\frac{\delta}{\delta\phi(x)}$ commutes with $\frac{\partial}{\partial y^\mu}$. $\endgroup$ – AccidentalFourierTransform Nov 9 '17 at 12:34
  • $\begingroup$ So the answer is the Heavyside step function? $\endgroup$ – Mikkel Rev Nov 9 '17 at 12:38
  • $\begingroup$ Nope. ${}{}{}{}$ $\endgroup$ – AccidentalFourierTransform Nov 9 '17 at 12:40
  • $\begingroup$ $\partial_\mu (\delta\phi(x) /\delta \phi(x)) = \partial_\mu \delta^{(4)}(0) = \partial_\mu \infty = \infty$? $\endgroup$ – Mikkel Rev Nov 9 '17 at 12:45
  • $\begingroup$ Hi @Marius Jonsson : Is this taken from some reference? Which page? $\endgroup$ – Qmechanic Nov 9 '17 at 12:45

The expression $\frac{\delta \partial_\mu\phi(y)}{\delta \phi(y)}$ is mathematically meaningless.

By definition, given a functional $F$ associating reals (or, more generally, complex numbers) $F[\phi]$ to smooth functions $\phi$, we say that the distribution $\frac{\delta F}{\delta \phi(x)}$ is the functional derivative of $F$, if $$\frac{d}{d\alpha}|_{\alpha=0} F[\phi + \alpha f] = \int \frac{\delta F}{\delta \phi(x)} f(x) dx$$ for every compactly-supported smooth function $f$.

In the considered case, one has to compute the functional derivative of the functional $F$ associating $\partial_\mu \phi(y)$ to $\phi$, i.e., $$\partial_\mu \phi(y) := \int \partial_\mu \phi(x) \delta(x-y) dx\:.$$ We have $$\frac{d}{d\alpha}|_{\alpha=0} F[\phi + \alpha f] = \frac{d}{d\alpha}|_{\alpha=0} \int \partial_\mu (\phi(x)+ \alpha f(x)) \delta(x-y) dx = \int \partial_\mu f(x) \delta(x-y) dx$$ $$= -\int f(x) \partial_\mu\delta(x-y) dx\:.$$ We conclude that $$\frac{\delta \partial_\mu\phi(y)}{\delta \phi(x)} = - \partial^{(x)}_\mu\delta(x-y) = \partial^{(y)}_\mu\delta(x-y)\:. $$ So $\frac{\delta }{\delta \phi(x)}$ and $\partial^{(y)}_\mu$ commute as said by @AccidentalFourierTransform.

In summary, $\frac{\delta \partial_\mu\phi(y)}{\delta \phi(y)}$ is not defined because the value at a fixed point of a non-regular distribution has no meaning.

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    $\begingroup$ +1 I would like to mention the following: if we were to insist that we want that expression evaluated at $x=y$ ("in a formal sense" e.g., with some regularisation scheme in mind), then one may argue that the expression vanishes, inasmuch as the derivative of the Dirac delta is odd. $\endgroup$ – AccidentalFourierTransform Nov 9 '17 at 14:06
  • $\begingroup$ Right, indeed: ex falso quodlibet $\endgroup$ – Valter Moretti Nov 9 '17 at 14:10
  1. As correctly pointed out in the answer by Valter Moretti, it is mathematically ill-defined to apply (the traditional definition of) the functional/variational derivative (FD) $$ \frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)} \tag{1}$$ to the same spacetime point.

  2. However, it is very common to introduce a 'same-spacetime' FD as $$ \frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots. \tag{2} $$ which obscures/betrays its variational origin, but is often used for notational convenience. (The ellipsis $\ldots$ in eq. (2) denotes possible contributions from higher-order spacetime derivatives.) See e.g. this, this, & this Phys.SE posts.

    If we interpret OP's expression via eq. (2), then OP's Lagrangian density ${\cal L}=\partial_{\mu} \phi$ is a total space-time derivative, so that OP's 'same-spacetime' FD vanishes, cf. e.g. this Phys.SE post.


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