The gradient of the dot product of a point and a vector? This is a list of vector identities from the first page of the text on Classical Electrodynamics by Jackson. There is not much explanation provided.
https://www.physics.rutgers.edu/~shapiro/504/lects/vecidents.pdf


*is the usual gradient of a dot product of two vectors identity.


Then: 15. $$ \nabla(\vec A \cdot \vec x) $$
Is $\vec x$ a point? Never encountered this I think? Can someone give me an explanation? I don't even know what's the proper name for this identity.
Thanks
 A: Hint
Let us see the $j^{th}$ component :
$$(\vec{\nabla}(\vec{A}•\vec{x}))_j=$$
$$\frac{\partial \vec{A}•\vec{x}}{\partial x_j}=$$
$$\frac{\partial \vec{A}}{\partial x_j}•\vec{x}+\vec{A}•\frac{\partial \vec{x}}{\partial x_j}$$
A: The position vector $\vec x \in \Bbb R^3$ is such that the gradient operator is
$$
\vec\nabla \psi = \frac{\partial}{\partial \vec x} \psi \, ,
$$
for any field $ \psi$. Therefore, one has (12)
\begin{aligned}
\vec\nabla\vec x &= \text{id} \, , \\
\vec\nabla\cdot\vec x &= \frac{\partial}{\partial \vec x}\cdot\vec x = \text{tr}(\text{id}) = 3\, ,\\
\vec\nabla \times \vec x &= \frac{\partial}{\partial \vec x}\times\vec x = 0 \, .
\end{aligned}
If we plug $\vec B = \vec x$ in (9), we have
\begin{aligned}
\vec\nabla (\vec x\cdot\vec \Lambda) &= (\vec\Lambda\cdot\vec \nabla) \vec x + (\vec x \cdot\vec\nabla)\vec\Lambda + \vec x\times (\vec\nabla\times\vec\Lambda) \\
&= \vec\Lambda\cdot (\vec\nabla\vec x) + \vec x\cdot (\vec\nabla\vec \Lambda) \, .
\end{aligned}
There may be a way to obtain (15)-(16) out of it, but I doubt that this is very general and of wide practical use.
