If $$\delta S = \int \sqrt g F[\phi] \delta \phi\tag{1}$$
Then is it natural to define the functional derivative as follows,
$$\frac{\delta S}{\delta \phi} = F[\phi].\tag{2}$$
In particular does this definition satisfy the commutativity of the functional derivatives.
I understand that this is not the standard definition of a functional derivative, but if I define this way this makes a certain calculation I am doing much easier to control. So to Clarify what I want to know is that if I define the 'functional derivative' this way then if
$$S = \int \sqrt{g} L[\phi, g_{\mu \nu}]\tag{3}$$ then
is the following true?
$$\frac{\delta}{\delta\phi} \frac{\delta}{\delta g_{\mu \nu}} S = \frac{\delta}{\delta g_{\mu \nu}} \frac{\delta}{\delta\phi} S \tag{4} $$
This is my procedure for computing the second functional derivative suppose
$$\frac{\delta}{\delta \phi}S = E[\phi, g_{\mu \nu}]\tag{5}$$
and $$\frac{\delta}{\delta g_{\mu \nu}}S = E^{\mu \nu}[\phi, g_{\mu \nu}]\tag{6}$$
(equations of motion) Then to compute second functional derivative we write, eg.
$$E[\phi, g_{\mu \nu}](x) = \int \sqrt{g} d^4 y E(y) \hat \delta(x-y)\tag{7}$$
where $$\hat \delta (x-y) = \frac{\delta(x-y)}{\sqrt{g}}\tag{8} $$ is the generalised delta function. After this we can use the same definition to compute
$$\frac{\delta} {\delta g_{\mu \nu}} \int \sqrt{g} d^4 y E(y) \hat \delta(x-y)\tag{9} .$$ Of course now the result would involve delta functions.
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