# 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

1. 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

Let us see the $j^{th}$ component :
$$(\vec{\nabla}(\vec{A}•\vec{x}))_j=$$
$$\frac{\partial \vec{A}•\vec{x}}{\partial x_j}=$$
$$\frac{\partial \vec{A}}{\partial x_j}•\vec{x}+\vec{A}•\frac{\partial \vec{x}}{\partial x_j}$$
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.