Given:

$\overline{r}=xi+yj+zk$

To prove:

$\nabla \log r=\dfrac{\overline{r}}{r^2}$

According to the generalised formula,

$\nabla \phi=\dfrac{\partial\phi}{\partial x}i+\dfrac{\partial\phi}{\partial y}j+\dfrac{\partial\phi}{\partial z}k$

Here,$\phi=\log r....(1)$,

$r^2=x^2+y^2+z^2...(2)$

Differentiate The equation (1) with respect to r we get:

$\dfrac{d\phi}{dr}=\dfrac{1}{r}$

Diffrentiating the equation (2) withe respect to x,yand z we get:

$\dfrac{\partial r}{\partial x}=\dfrac{x}{r},\dfrac{\partial r}{\partial y}=\dfrac{y}{r},\dfrac{\partial r}{\partial z}=\dfrac{z}{r}.$

$\nabla \log \phi=\dfrac{d\phi}{d r}\times\dfrac{\partial r}{\partial x} i+\dfrac{d\phi}{d r}\times\dfrac{\partial r}{\partial y}j+\dfrac{d\phi}{d r}\times\dfrac{\partial r}{\partial z}k $

$\nabla\log \phi=\dfrac{1}{r}\times\dfrac{x}{r}i+\dfrac{1}{r}\times\dfrac{y}{r}j+\dfrac{1}{r}\times\dfrac{z}{r}k$

taking common we get

$\nabla\log \phi=\dfrac{1}{r^2}\big(xi+yj+zk\big)$

$\nabla\log \phi=\dfrac{1}{r^2}\big(\overline{r}\big)$