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## Main Question or Discussion Point

Ok, there are a couple of other threads about this, but they don't seem to answer my question.

If I take the double derivative of 1/r, I'll get 2/r^3, but if I take the laplacian, I get something different. Why?

Namely:

[itex]\frac{d}{dr}\frac{d}{dr}(\frac{1}{r}) = \frac{d}{dr} (\frac{-1}{r^{2}}) = \frac{(-1)(-2)}{r^{3}} = \frac{2}{r^{3}}[/itex]

However:

[itex]∇^{2}(\frac{1}{r}) = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{1}{r})] = 0 [/itex] or [itex]- \frac{δ(r)}{r^{2}}[/itex] or [itex]-4\piδ(r)[/itex]

Which of these answers is correct depends on who you ask - and the situation you are in. I am not interested in this debate. My question is not which one of these answers is correct; my question is more general.

Why they are different at all. Why is it easily differentiable in one case and not the other? What basic calculus definition did I forget or never learn?

If I take the double derivative of 1/r, I'll get 2/r^3, but if I take the laplacian, I get something different. Why?

Namely:

[itex]\frac{d}{dr}\frac{d}{dr}(\frac{1}{r}) = \frac{d}{dr} (\frac{-1}{r^{2}}) = \frac{(-1)(-2)}{r^{3}} = \frac{2}{r^{3}}[/itex]

However:

[itex]∇^{2}(\frac{1}{r}) = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{1}{r})] = 0 [/itex] or [itex]- \frac{δ(r)}{r^{2}}[/itex] or [itex]-4\piδ(r)[/itex]

Which of these answers is correct depends on who you ask - and the situation you are in. I am not interested in this debate. My question is not which one of these answers is correct; my question is more general.

Why they are different at all. Why is it easily differentiable in one case and not the other? What basic calculus definition did I forget or never learn?