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I have two functionals, $F(h, \nabla h)$ and $G(h, \nabla h)$. I'd like to calculate

$\frac{\delta F}{\delta G}$

Since functional derivatives also follow the chain rule, would I be correct in thinking:

$\frac{\delta F}{\delta G} = \frac{\delta F}{\delta h} \cdot \left(\frac{\delta G}{\delta h}\right)^{-1}$

like this link?

Thanks

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Mike
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    $F$ isn't necessarily a function of $G$ so this doesn't really make sense - you'd expect $\delta F/\delta G [G(h_s)] = \frac d{ds}|_{s=0} F(h_s)/G(h_s)$ for every family $h_s$, but the RHS depends on much more than just $G(h_s)$. – Anthony Carapetis Jun 23 '16 at 00:58
  • Thanks, but I'm not sure I understand the problem. In the link I mentioned, they say $\frac{dg(x)}{df(x)}=\frac{dg(x)}{x} \cdot \left( \frac{df(x)}{dx} \right)^{-1}$. Although in the link, g is an explicit function of f(x), the formula does not require this. It is still meaningful to as how one function of x varies with respect to another. – Mike Jun 23 '16 at 14:30
  • Formula above should be: $\frac{dg(x)}{df(x)}=\frac{dg(x)}{dx} \cdot \left( \frac{df(x)}{dx} \right)^{-1}$
    • – Mike Jun 23 '16 at 14:36
  • This works only in the single variable case, where $g$ is always locally a function of $f$ wherever $f'(x) \ne 0$. In higher (or infinite) dimensions $\delta G/\delta h$ cannot be invertible. – Anthony Carapetis Jun 23 '16 at 15:20