Gradient, divegence and curl of functions of the position vector

Question

For scalar functions $f$ of the position vector $\vec{r}$, it seems as if the following relations apply:

• $\nabla f(\vec{a}\cdot\vec{r})=\vec{a}f'(\vec{a}\cdot\vec{r})$
• $\nabla \cdot \vec{b}f(\vec{a}\cdot\vec{r})=\vec{a}\cdot\vec{b}f'(\vec{a}\cdot\vec{r})$
• $\nabla \times \vec{b}f(\vec{a}\cdot\vec{r})=\vec{a}\times\vec{b}f'(\vec{a}\cdot\vec{r})$

where $\vec{a}$ and $\vec{b}$ are arbitrary vectors independent of position.

Is this generally true? How to prove this (I can see how it works in a certain coordinate system, but how to prove it in general?).

Is this a symptom of something that is more generally true (how nabla operations on functions that depend on $\vec{r}$ correspond with normal 'scalar differentiation' w.r.t. $\vec{r}$)

Answer 1

If $\vec a$ is independent of $\vec r$, then we have from the chain rule

$$\begin{align} \nabla f(\vec a\cdot \vec r)&=\sum_{i=1}^3\hat x_i \frac{\partial f(a_1x_1+a_2x_2+a_3x_3)}{\partial x_i}\\\\&=\sum_{i=1}^3\hat x_i \,a_i f'(\vec a\cdot \vec r)\\\\&=\vec a f(\vec a\cdot \vec r) \end{align}$$

If $\vec a$ and $\vec b$ are independent of $\vec r$, then we have

$$\begin{align} \require{cancelto} \nabla \cdot \left(\vec b\cdot f(\vec a\cdot \vec r)\right)&=\vec b\cdot\nabla \left(f(\vec a\cdot \vec r)\right)+f(\vec a\cdot \vec r) \cancelto0{\nabla \cdot \left(\vec b\right)}\\\\ &=\vec b\cdot\nabla \left(f(\vec a\cdot \vec r)\right)\\\\ &=\vec b \cdot \vec a f'\left(\vec a\cdot \vec r\right) \end{align}$$

and

$$\begin{align} \nabla \times \left(\vec b f(\vec a\cdot \vec r)\right)&=\nabla \left(f(\vec a\cdot \vec r)\right)\times \vec b+f(\vec a\cdot \vec r)\cancelto0{\nabla \times \left(\vec b\right)}\\\\ &=-\vec b\times \nabla \left(f(\vec a\cdot \vec r)\right)\\\\ &=\vec a\times \vec b \, f'\left(\vec a\cdot \vec r\right) \end{align}$$

Answer 2

Intuitively this is because the derivative we're used to seeing in single variable calculus is really best thought of as the one dimensional gradient, which it has more in common with than partial derivatives. In a (very) heuristic way

$$\nabla "=" \frac{d}{d\vec{r}} "=" \frac{d}{(dx^1,\cdots,dx^n)} "=" \left(\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^n}\right)$$

This notion is made more rigorous in tensor analysis, in understanding the difference between a partial and total derivative on a space.