# Why does the following equality hold about $r$ component of laplace operator?

I am wondering why the following equality holds ? $$\frac{1}{r^2} \frac {\partial}{\partial r}(r^2 \frac {\partial}{\partial r})= \frac {1}{r} \partial_r^2 r$$ . I tried making some substitutions but it doesn't seem to work . :(

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For displayed equations, you need to put the punctuation inside the dollar signs (ideally spaced apart by \;), since, as you can see, otherwise it ends up on the following line. –  joriki Nov 6 '12 at 12:30

This sort of identity is proved simply by moving all derivatives through to the right by applying the product rule and then comparing:

\begin{align} \frac1{r^2}\def\fr{\frac{\partial}{\partial r}}\fr\left(r^2\fr\right) &= \frac1{r^2}\left(\fr r^2\right)\fr+\frac1{r^2}r^2\fr\fr \\ &= \frac2r\fr+\frac{\partial^2}{\partial r^2} \end{align}

and

\begin{align} \frac1r\fr\fr r &= \frac1r\fr\left(\fr r\right)+\frac1r\fr r\fr \\ &= \frac1r\fr+\frac1r\left(\fr r\right)\fr+\frac1rr\fr\fr \\ &= \frac2r\fr+\frac{\partial^2}{\partial r^2}\;. \end{align}

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