Gradient in Polar Coordinates ($\mathbb{R}^n$)

Question

I am told that $\textbf{x} = r\omega$ where $r = |x|$ and $\omega \in S^{n-1}$ , then $(\textbf{x} \cdot \nabla)f = r\partial_r f$. Why is this so?

Answer 1

If $n=2$, you can see this with the usual polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$. By the ordinary chain rule:

$$\frac{\partial}{\partial r}f(r\cos(\theta),r\sin(\theta)) = \cos(\theta) \frac{\partial f}{\partial x} + \sin(\theta) \frac{\partial f}{\partial y} $$

Multyply both sides by $r$, and you get $r \, \partial_r f = x f_x + y f_y = \mathbf{x} \cdot \nabla f$.

I'm sure the case for $n>2$ is similar, just using higher-dimensional spherical coordinates.