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How do I prove these identities for vector calculus? It has to do with divergence, but I can't even figure out where to start, let alone complete it:

Let $\vec{r}(x, y, z) = (x, y, z)$ and $r = \sqrt{x^2 + y^2 + z^2} = \|\vec{r}\|$.

  • (a) $\vec{\nabla} \frac{1}{r} = −\frac{\vec{r}}{r^3}$, $r \not= 0$; and, in general, $\vec{\nabla}(r^n) = n(r^n−2)\vec{r}$ and $\vec{\nabla}\log (r) = \frac{\vec{r}}{r^2}$.

  • (b) $\nabla^2 \frac{1}{r} = 0$, $r \not= 0$; and, in general, $\nabla^2 r^n = n(n + 1)(r^n−2)$.

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    Start off by reading on what the nabla operator on a scalar field does. It seems like its difficult, but just start from definitions and you'll be good to go as long as you understand what a vector is, and know how to differentiate – 2017-02-09
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    In general $\vec{\nabla}f(r) = f'(r)\vec{\nabla} r$ where $\vec{\nabla} r = (\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z})$. Since $r$ is symmetric in $x,y,z$ it suffices to compute one of them, say $\frac{\partial r}{\partial x}$, to know them all (just replace $x$ and $y$ in the result to get $\frac{\partial r}{\partial y}$). Then proving all the identities in (a) above simply reduces to computing a normal derivative of $f(r) = \log(r)$ and $f(r) = r^n$. – 2017-02-09

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