Very basic partial derivatives problem

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SUMMARY

The discussion centers on the confusion surrounding the expression for partial derivatives, specifically questioning the validity of stating that \(\frac{\partial f}{\partial f} = 1\). Participants clarify that while this expression lacks meaningful context, the correct approach involves functional derivatives, represented as \(\frac{\delta f(x_1, \cdots, x_n)}{\delta f(x_1', \cdots, x_n')} = \delta(x_1 - x_1') \cdots \delta(x_n - x_n')\), utilizing Dirac delta distributions. The original poster acknowledges a misunderstanding and refers to a related thread for a more accurate explanation.

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mnb96
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Hello,
I should feel ashamed to ask this, but it's giving me (and others) some troubles.

given [tex]f(x_1,\ldots,x_n)[/tex], is it wrong to say that:

[tex]\frac{\partial f}{\partial f}=1[/tex]

...?
 
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It doesn't make a lot of sense... the idea is that in the expression [itex]\frac{\partial f}{\partial x}[/itex], f is a function and x is a variable on which f does (or does not, or does indirectly) depend.

You can write
[tex]\frac{\delta f(x_1, \cdots, x_n)}{\delta f(x_1', \cdots, x_n')} = \delta(x_1 - x_1') \cdots \delta(x_n - x_n')[/tex]
where the delta's on the right hand side are Dirac delta distributions, but you are taking functional derivatives then.
 
uhm...it makes some sense in the following context:
https://www.physicsforums.com/showthread.php?t=365940
however, there must be a mistake but I cannot see it.

[EDIT]: in the thread mentioned above a solution to the problem is described in its correct context. Sorry for this kind of "double posting"; in the beginning I thought I was facing a different problem than the original.
 
Last edited:

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