# Functional derivative of normal function

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1. Apr 22, 2015

### ChrisPhys

I can't convince myself whether the following functional derivative is trivial or not:

$\frac \delta {\delta \psi(x)} \big[ \partial_x \psi(x)\big],$

where $\partial_x$ is a standard derivative with respect to $x$.

One could argue that

$\partial_x \psi(x) = \int dx' [\partial_{x'} \psi(x')] \delta (x - x') = - \int dx' \psi(x') \partial_{x'} \delta (x - x'),$

assuming there is no boundary term in integration by parts.

In this case, the functional derivative would give

$\frac \delta {\delta \psi(x)}\Big[ - \int dx' \psi(x') \partial_{x'} \delta (x - x') \Big] = - \partial_{x'} \delta (x - x')\Big|_{x'=x} = 0.$

Any thoughts? Is this rigorous?

2. Apr 22, 2015

### Orodruin

Staff Emeritus
You really should use different variables for the function you are differentiating and the function you are taking the derivative with respect to ...

What makes you think that the last expression evaluates to zero? In the distribution sense it is the derivative of the delta distribution.

3. Apr 22, 2015

### ChrisPhys

I just viewed the $\delta$ function as the limit of a Gaussian, whose derivative at zero is zero. Is that an error?

4. Apr 22, 2015

### Orodruin

Staff Emeritus
This then falls back on your original problem with not using different variables for the function you are differentiating and the function you are differentiating with respect to. Your question should be: "What is $\delta (\partial_\mu \psi(x))/\delta \psi(y)$?"
The result should be a distribution which picks out the derivative of a function, i.e., the derivative (in the distributional sense) of the delta distribution.

5. Apr 23, 2015

### ChrisPhys

@Orodruin
Thank you, that makes sense. I appreciate your help.

6. Apr 30, 2015

### Gracie thomas

Hi
To informally guess the derivative, I use the 'little o' notation and then check the details. However, for the details. I think it is easiest to use composition rule, ie, if f,g are (Fréchet) differentiable with appropriate domains/ranges, then Df∘g(x)=Df(g(x))Dg(x).
Thanks.

7. Apr 30, 2015

### Orodruin

Staff Emeritus