# functional derivative of an integral of the function itself

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.

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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
@StefanoBorini I think that $\frac{d}{dn(x)}\int_{\mathbb{R}^3}n(x)dx=\frac{dN}{dn(x)}=0$. – Mohsen Shahriari Jun 12 '15 at 11:15