I'm working through an introductory electrodynamics text (Griffiths), and I encountered a pair of questions asking me to show that:

1. the divergence transforms as a scalar under rotations
2. the gradient transforms as a vector under rotations

I can see how to show these things mathematically, but I'd like to gain some intuition about what it means to "transform as a" vector or scalar. I have found definitions, but none using notation consistent with the Griffiths book, so I was hoping for some confirmation.

My guess is that "transforms as a scalar" applies to a scalar field, e.g. $T(y,z)$ (working in two dimensions since the questions in the book are limited to two dimensions). It says that if you relabel all of the coordinates in the coordinate system using: $$\begin{pmatrix}\bar{y} \\ \bar{z}\end{pmatrix} = \begin{pmatrix}\cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix} \begin{pmatrix}y \\ z\end{pmatrix}$$ so $(\bar{y},\bar{z})$ gives the relabeled coordinates for point $(y,z)$, then: $$\bar{T}(\bar{y},\bar{z}) = T(y,z)$$ for all y, z in the coordinate system, where $\bar{T}$ is the rotated scalar field. Then I thought perhaps I'm trying to show something like this? $$\overline{(\nabla \cdot T)}(\bar{y},\bar{z})=(\nabla \cdot T)(y,z) $$ where $\overline{(\nabla \cdot T)}$ is the rotated gradient of $T$.

The notation above looks strange to me, so I'm wondering if it's correct. I'm also quite curious what the analogous formalization of "transforms as a vector field" would look like.