Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following

$$ \frac{d}{dn(x)} \int_{x \in \cal{R}^3} {n(x) dx} $$

I know that this additional relationship holds

$$ \int_{x \in \cal{R}^3}{n(x) dx} = N $$

where N is a constant.

My question is, what is the value of the above derivative, and what is the procedure for this case? I am not really experienced in functionals and functional derivatives.

share|improve this question
As written the "additional relationship" is nonsense. An indefinite integral is a set of functions often written in the form "$\ N(x)+C\ $" and cannot be equal to "a constant" $N$. –  Christian Blatter Jan 20 '12 at 21:34
@ChristianBlatter : ok sorry. The integration is over the 3D space. –  Stefano Borini Jan 20 '12 at 21:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.