$\newcommand{\curl}{\operatorname{curl}}$ $\newcommand{\div}{\operatorname{div}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bfe}{\mathbf e}$ $\newcommand{\bft}{\mathbf t}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfF}{\mathbf F}$ $\newcommand{\bfk}{\mathbf k}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfv}{\mathbf v}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$
If $U$ is an open subset of $\R^n$ and $f:U\to \R$ is a function of class $C^1$, then $$ \mbox{ the gradient of }f = \grad f = \nabla f := (\partial_1 f, \ldots, \partial_n f). $$ We are already very familiar with this.
if $U$ is an open subset of $\R^n$ and $\bfF:U\to \R^n$ is a vector field of class $C^1$, then $$ \mbox{ the divergence of }\bfF = \div \bfF := \nabla \cdot \bfF = \partial_1 F_1+ \ldots +\partial_n F_n. $$
The definitions of $\grad$ and $\div$ make sense in $\R^n$ for any $n$. Our next definition only makes sense when $n=3$:
Whenever we refer to the curl, we are always implictly assuming that the number of dimensions is $\R^3$; we may not always say this explicitly.
We will talk about the meaning of the divergence and curl later on.
There are various ways of composing vector derivatives. Here are two of them: $$ \curl(\grad f) = {\bf 0} \mbox{ for all }C^2\mbox{ functions }f. $$ $$ \div(\curl \bfF) = 0 \mbox{ for all }C^2\mbox{ vector fields }\bfF. $$ Both of these are easy to verify, and both of them reduce to the fact that the mixed partial derivatives of a $C^2$ function are equal.
In other words, if we arrange $\div, \grad, \curl$ as indicated below, then following any two successive arrows yields $0$ (or $\bf 0$). $$ \mbox{functions} \quad\overset{\grad} { \xrightarrow{\hspace{1cm}}} \quad \mbox{vector fields} \quad \overset{\curl} {\xrightarrow{\hspace{1cm}}} \quad \mbox{vector fields} \quad \overset{\div} {\xrightarrow{\hspace{1cm}}} \quad \mbox{functions} . $$ Another way of composing vector derivatives is to take $\div(\grad f)$ for a scalar function $f$. It is easy to check that for $f:\R^n\to \R$ of class $C^2$, $$ \div(\grad f) = \sum_{j=1}^n \partial_{jj}f =: \mbox{ the Laplacian of }f. $$ The Laplacian of $f$ is usually denoted $\Delta f$ or $\nabla^2 f$. The former notation is used more often by mathematicians, and the latter by physicists and engineers. The Laplacian appears throughout mathematics, mathematical physics, and also in many areas of applied mathematics, including mathematical finance.
The other possible two combinations of vector derivatives are $\grad\div$ and $\curl\curl$. These are related to the so-called vector Laplacian. $$ \mbox{ for a }C^2\mbox{ vector field }\bfF, \quad \grad\div \bfF - \curl\curl \bfF = \Delta \bfF := (\Delta F_1, \Delta F_2, \Delta F_3). $$ where $(F_1,F_2,F_3)$ are the components of $\bfF$. The vector Laplacian too arises often in diverse areas of mathematics and the sciences.
One reason for the importance of both the Laplacian and the vector Laplacian is their relationship to the basic operations of $\div, \grad$, and $\curl$.
Since there are several ways of multiplying functions and vector fields and then several kinds of vector derivatives, there are a number of different product rules" for vector derivatives. $$ \begin{aligned} \grad (fg) &= f \grad g + g \grad f \\ \div (f\bfG) &= f\div \bfG + (\grad f) \cdot \bfG \\ \curl (f\bfG) &= f \curl \bfG +(\grad f) \times \bfG \\ \div(\bfF\times \bfG)&= \bfG\cdot \curl \bfF - \bfF\cdot \curl \bfG\\ \curl(\bfF\times \bfG)&= (\bfG\cdot \nabla)\bfF + (\div \bfG)\bfF+ (\bfF\cdot \nabla)\bfG + (\div \bfF) \bfG \\ \grad(\bfF\cdot \bfG) &= (\bfG\cdot \nabla)\bfF + \bfG\times \curl \bfF - (\bfF\cdot \nabla)\bfG - \bfF\times \curl \bfG \end{aligned} $$ where $$ (\bfF\cdot \nabla)\bfG = \sum_{j=1}^3 F_j \partial_j \bfG, \qquad (\bfG\cdot \nabla)\bfF = \sum_{j=1}^3 G_j \partial_j \bfF. $$ The first two identities, in which the curl does not appear, are valid in $\R^n$ for any $n$, whereas the others are restricted to $\R^3$.
Committing these product rules to memory may not be a good use of brainpower, But you should know that they exist, and if/when you come upon a situation when one or more of these is needed, you should be able to recognize that and look up the relevant identity.
In every case these can be proved by laboriously writing out both sides and checking to see that they are equal.
Be able to compute the curl or divergence of a vector field. For example, do some of the following. The linear examples are recommended -- they are both easy and instructive.
compute the divergence and the curl of $\bfF(x,y,z) = (x^2\cos y, zyx, e^{xy})$.
compute the divergence and the curl of $\bfF(x,y,z) = ( \sin yz, xz\cos yz, xy\cos yz)$.
Let $A$ be the (antisymmetric) matrix $$ A :=\left(\begin{array}{ccc} 0&-a&-b\\ a&0&-c\\ b&c&0 \end{array} \right) $$ and let $\bfF(\bfx) = A\bfx$. Compute $\curl \bfF$ and $\div \bfF$.
Let $S$ be the (symmetric) matrix $$ S :=\left(\begin{array}{ccc} a&b&c\\ b&d&e\\ c&e&f\end{array} \right) $$ and let $\bfF(\bfx) = S\bfx$. Compute $\curl \bfF$ and $\div \bfF$.
Let $M$ be an arbitrary matrix $$ M :=\left(\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{array} \right) $$ and let $\bfF(\bfx) = M\bfx$. Compute $\curl \bfF$ and $\div \bfF$.
Compute $\curl \bfF$ for a vector field of the form $\bfF = (0,0,f(x,y,z))$.
Compute $\curl \bfF$ and $\div \bfF$ for a vector field of the form $\bfF = (P(x,y), Q(x,y), 0)$.
(Note that $P$ and $Q$ depend only on $(x,y)$, not on $z$.)
Compute $\Delta f$ for $f:\R^3\setminus\{\bf 0\}\to \R$ defined by $f(\bfx) = \frac 1{|\bfx|}$.
Hint. $|\bfx| = (|\bfx|^2)^{1/2}$
prove one or more of the product rules for vector derivatives.
Prove that if $f$ and $g$ are $C^2$ functions of $3$ variables, then $\div(\grad f \times \grad g) = 0$