While studying general relativity I came across the following derivative:
$$ \frac{\partial g_{ef,gh}}{\partial g_{ab,cd}} \tag{1}$$
where, $g_{ab}$ is the metric tensor and the comma notation implies:
$$ \partial_{\mu}(g_{\alpha \beta})=g_{\alpha \beta , \mu}.\tag{2} $$
I'm aware of the following formulae:
$$ \frac{\partial g_{ef}}{\partial g_{ab}} = \frac{1}{2}(\delta^{e}_{a}\delta^{f}_{b} + \delta^{e}_{b}\delta^{f}_{a} ) , \tag{3}$$
cf. e.g. this Phys.SE post. But, how does it generalize when I have partial derivatives of the metric tensor involved such as $g_{ab,c}$ or $g_{ab,cd}$. Do we get something like this?
$$ \frac{\partial g_{ef,gh}}{\partial g_{ab,cd}} \stackrel{?}{=} \frac{1}{4}(\delta^{e}_{a}\delta^{f}_{b} + \delta^{e}_{b}\delta^{f}_{a} ) (\delta^{g}_{c}\delta^{h}_{d} + \delta^{g}_{d}\delta^{h}_{c}) .\tag{4}$$
If I had to guess, I would say no as an entity like $g_{ab,c}$ is not even a tensor (it is a partial derivative of a tensor) so there is probably a bit more subtlety involved here.
