The del function is not commutative.
$$ \textrm{nabla/del function: }\nabla V=\begin{bmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y} \\ \frac{\partial }{\partial z} \end{bmatrix} V $$
$$ \textrm{Divergence theorem: }\iiint_V (\nabla \cdot \mathbf v )\, \mathrm{d}V=\oiint_S \mathbf v \cdot \mathrm{d}\mathbf{S} $$
$$ \textrm{Curl theorem: }\iint_S (\nabla \times \mathbf v) \cdot \mathrm{d}\mathbf{S}=\oint_l \mathbf v \cdot \mathrm{d}\mathbf{l} $$
$$ \textrm{Gradient theorem: }\int_l\nabla f \, \cdot \mathrm{d}\mathbf{l}=f \vert_l $$
There are 6 product rules, since there are 3 ways to use del (gradient, divergence, curl) and two ways to make a scalar or vector.
$$ \nabla \cdot (\nabla \times \mathbf v)=0 \quad \nabla \times (\nabla f)=0 $$
$$ \nabla\times(\nabla \times \mathbf v)=\nabla(\nabla \cdot \mathbf v)- \nabla^2 \mathbf v $$
The Dirac delta function is a function with a finite integral but blows up at 0.
$$ \delta (x) = \begin{cases}\infty , \ x=0 \\ 0, \ x \neq 0 \end{cases}, \quad \int^\infty_{-\infty} \delta(x) dx = 1 $$
The delta function effectively “picks out” the value at the origin:
$$ \int^\infty_{-\infty} f(x)\delta(x)dx = f(0) $$
The delta function isn’t a real function, as it is undefined at $x=0$. However, we can still compare delta functions by using it in an integral. We can say that two expressions with delta functions $D_1$ and $D_2$ are considered equal if their integrals over all space are equal.
$$ \int^\infty_{-\infty} f(x) D_1(x) dx = \int^\infty_{-\infty} f(x) D_2(x) dx \quad \textrm{for all functions } f(x) $$
The equivalence of $D_1$ and $D_2$ yields useful identities which can be used to simplify functions involving the Dirac delta.