Answer
a ) $\nabla r=\dfrac{\bf{r}}{r}$
b) $\nabla \times r=0$
c) $\nabla (1/r)=\dfrac{-r}{r^3}$
d) $\nabla \ln r=\dfrac{r}{r^2}$
Work Step by Step
a) $\nabla r=\dfrac{\partial r }{\partial x}i+\dfrac{\partial r }{\partial y}j+\dfrac{\partial r}{\partial z}k$
$=\dfrac{x }{\sqrt{x^2+y^2+z^2}} i+\dfrac{y}{\sqrt{x^2+y^2+z^2}} j+\dfrac{z}{\sqrt{x^2+y^2+z^2}} k$
$\nabla r=\dfrac{(xi+yj+zk)}{\sqrt{x^2+y^2+z^2}} =\dfrac{\bf{r}}{r}$
b) $\nabla \times r=\begin{vmatrix}i&j&k\\\dfrac{\partial}{\partial x}&\dfrac{\partial }{\partial y}&\dfrac{\partial }{\partial z}\\x&y&z\end{vmatrix}=(\dfrac{\partial z}{\partial y}-\dfrac{\partial y}{\partial z})i-(\dfrac{\partial z}{\partial x}-\dfrac{\partial x}{\partial z})j+(\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{\partial y})k=0$
c) $\nabla (1/r)=\dfrac{\partial (1/r) }{\partial x}i+\dfrac{\partial (1/r) }{\partial y}j+\dfrac{\partial (1/r)}{\partial z}k=(\dfrac{-1}{r^2}) \nabla r$
Since, $\nabla r=\dfrac{\bf{r}}{r}$
Then, we have $\nabla (1/r)=\dfrac{-r}{r^3}$
d) $\nabla \ln r=\dfrac{\partial (\ln r) }{\partial x}i+\dfrac{\partial (\ln r) }{\partial y}j+\dfrac{\partial (\ln r)}{\partial z}k=(\dfrac{1}{r}) \nabla r$
Since, $\nabla r=\dfrac{\bf{r}}{r}$
Then, we have $\nabla \ln r=\dfrac{r}{r^2}$
