simple question about $\nabla r$ In my physics notes, it says
$\nabla r = \underline{e_r} = \frac{\underline{r}}{r}$
and
$\nabla \frac{1}{r} = - \frac{\underline{r}}{r^3} = - \frac{1}{r^2} \underline{e_r}$

I don't quite understand why this is?
I know that $\nabla \varphi = (\frac{\delta\varphi}{\delta x},\frac{\delta\varphi}{\delta y},\frac{\delta\varphi}{\delta z})$ but I do not understand the above.
Thanks.
 A: In rectangular coordinates, 
$$(\nabla r)_k = \frac{\partial r}{\partial x_k}.$$
Therefore, you must calculate 
$$\frac{\partial r}{\partial x_k} = \frac{\partial}{\partial x_k}\left(\sum_k (x_k)^2\right)^\frac{1}{2}.$$
You will see that the result is 
$$\frac{\partial r}{\partial x_k} = \frac{x_k}{r}.$$
Consequently 
$$\nabla r = \left(\frac{\partial r}{\partial x}, \, \frac{\partial r}{\partial y}, \, \frac{\partial r}{\partial z} \right) = \left(\frac{x}{r}, \, \frac{y}{r}, \, \frac{z}{r}\right) = \frac{1}{r}(x, \, y, \, z) = \frac{\mathbf{r}}{r}.$$
Similarly, you can prove the formula
$$\nabla \left(\frac{1}{r}\right) = -\frac{\mathbf{r}}{r^3}$$
by showing that 
$$\left[\nabla\left(\frac{1}{r}\right)\right]_k = \frac{\partial}{\partial x_k}\left(\frac{1}{r}\right) = -\frac{x_k}{r^3}.$$
A: Note that $r^{2} = x^{2} + y^{2} + z^{2}$. Taking gradients and using the chain rule,
$$
2r\, \nabla r = 2(x, y, z) = 2\vec{r},\quad\text{or}\quad
\nabla r = \frac{\vec{r}}{r}.
$$
For the second, apply the chain rule and use the preceding identity.
