Let $P$ be any point in a domain of definition. Then we define a vector function $\mathbf{v}$ whose values are vectors, that is,

$$\mathbf{v}=\mathbf{v}(P)=[v_1(P),v_2(P),v_3(P)]$$

that depends on points $P$ in space. A vector function depends only on the point $P,$ not on the coordinate system chosen to represent its components.

We say that a vector function defines a vector field in a domain of definition.

Let $P$ be any point in a domain of definition. Then we define a scalar function $f,$ whose values are scalars, that is,

$$f=f(p)$$

that depends on $P.$

A scalar function depends only on the point $P,$ not on the coordinate system chosen to represent it.

Let $\mathbf{f}$ be a function that maps an open set $E\subset R^n$ into $R^m.$ Let $\{{\mathbf{e}}_1,\dots ,{\mathbf{e}}_n\}$ and $\{{\mathbf{u}}_1,\dots ,{\mathbf{u}}_n\}$ be the standard bases of $R^n$ and $R^m.$ The components of $\mathbf{f}$ are the real functions $f_1,\dots ,f_m$ defined by

$$\mathbf{f}(\mathbf{x})=\sum _{i=1}^m{f}_i(\mathbf{x}){\mathbf{u}}_i(\mathbf{x}\in E),$$

or, equivalently, by $f_i(\mathbf{x})=\mathbf{f}(\mathbf{x})\cdot {\mathbf{u}}_i,1\le i\le m.$

Using the setup from the definition of component, for $\mathbf{x}\in E,1\le i\le m,1\le j\le n,$ we define

$$(D_j{f}_i)(\mathbf{x})=\underset{t\to 0}{lim}\frac{f_i(\mathbf{x}+t{\mathbf{e}}_j)-f_i(\mathbf{x})}{t},$$

provided the limit exists. Writing $f_i(x_1,\dots ,x_n)$ in place of $f_i(\mathbf{x}),$ we see that $D_j{f}_i$ is the derivative of $f_i$ with respect to $x_j,$ keeping the other variables fixed. The notation

$$\frac{\partial f_i}{\partial x_j}$$

is therefore often used in place of $D_j{f}_i,$ and $D_j{f}_i$ is called a partial derivative.

For a space curve given parametrically by $\mathbf{r}(t),$ the tangent vector (or specifically, the unit tangent vector) at the point $\mathbf{r}(t)$ is the unit vector defined by:

$$\mathbf{T}(t)=\frac{{\mathbf{r}}^{\prime }(t)}{||{\mathbf{r}}^{\prime }(t)||}$$

Suppose $E$ is an open set in $R^n,$ $\mathbf{f}:E\to R^m,$ and $\mathbf{x}\in E.$ If there exists a linear transformation $A:R^n\to R^m$ such that

$$\begin{array}{cc}\underset{\mathbf{h}\to \mathbf{0}}{lim}\frac{|\mathbf{f}(\mathbf{x}+\mathbf{h})-\mathbf{f}(\mathbf{x})-A\mathbf{h}|}{|\mathbf{h}|}=0,& \text{(14)}\end{array}$$

then we say that $\mathbf{f}$ is differentiable at $\mathbf{x}$ and we write

$${\mathbf{f}}^{\prime }(\mathbf{x})=A.$$

If $\mathbf{f}$ is differentiable at every $\mathbf{x}\in E,$ we say that $\mathbf{f}$ is differentiable in $E.$

Note that in (14), $\mathbf{h}\in R^n.$ If $|\mathbf{h}|$ is small enough, then $\mathbf{x}+\mathbf{h}\in E,$ because $E$ is open. Therefore, $\mathbf{f}(\mathbf{x}+\mathbf{h})$ is defined, $\mathbf{f}(\mathbf{x}+\mathbf{h})\in R^m,$ and since $A\in L(R^n,R^m),$ $A\mathbf{h}\in R^m.$ Therefore,

$$\mathbf{f}(\mathbf{x}+\mathbf{h})-\mathbf{f}(\mathbf{x})-A\mathbf{h}\in R^m.$$

The norm in the numerator of (14) is that of $R^m,$ while the norm in the denominator is the $R^n$-norm.

We can also rewrite (14) as

$$\begin{array}{cc}\mathbf{f}(\mathbf{x}+\mathbf{h})-\mathbf{f}(\mathbf{x})={\mathbf{f}}^{\prime }(\mathbf{x})\mathbf{h}+\mathbf{r}(\mathbf{h}),& \text{(17)}\end{array}$$

where the remainder $\mathbf{r}(\mathbf{h})$ satisfies

$$\underset{\mathbf{h}\to \mathbf{0}}{lim}\frac{|\mathbf{r}(\mathbf{h})|}{|\mathbf{h}|}=0.$$

This means that for a fixed $\mathbf{x}$ and a small $\mathbf{h},$ the left side of (17) is approximately equal to ${\mathbf{f}}^{\prime }(\mathbf{x})\mathbf{h},$ that is, to the value of a linear transformation applied to $\mathbf{h}.$

The derivative defined above in differentiable is called the total derivative of $\mathbf{f}$ at $\mathbf{f},$ or the differential of $\mathbf{f}$ at $\mathbf{x}.$

A differentiable function $\mathbf{f}$ of an open set $E\subset R^n$ into $R^m$ is said to be continuously differentiable in $E$ if ${\mathbf{f}}^{\prime }$ is a continuous function of $E$ into $L(R^n,R^m).$ More explicitly, it is required that to every $\mathbf{x}\in E$ and to every corresponds a such that

if $\mathbf{y}\in E$ and

If this is the case, we also say that $\mathbf{f}$ is a ${\mathcal{C}}^{\prime }$-mapping or that $\mathbf{f}\in {\mathcal{C}}^{\prime }(E).$

In Euclidean space ${\mathbb{R}}^n$ with coordinates $(x_1,\dots ,x_n)$ and standard basis $({\mathbf{e}}_1,\dots ,{\mathbf{e}}_n),$ del is a vector operator whose $x_1,\dots ,x_n$ components are the partial derivative operators $\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n};$ that is

$$\mathrm{\nabla }=\sum _{i=1}^n{\mathbf{e}}_i\frac{\partial }{\partial x_i}=(\frac{\partial }{\partial x_1},\dots ,\frac{\partial }{\partial x_n}).$$

Given a scalar function $f:{\mathbb{R}}^n\to \mathbb{R}$, the gradient of $f$, denoted as $\mathrm{\nabla }f$, is defined as the vector of its partial derivatives. Specifically, for a function $f(x_1,x_2,\cdots ,x_n)$, the gradient is given by

$$\mathrm{\nabla }f=\left[\begin{array}{c}\frac{\partial f}{\partial x_1}\\ \frac{\partial f}{\partial x_2}\\ ⋮\\ \frac{\partial f}{\partial x_n}\end{array}\right]$$

Using the setup from the definition of component, let us fix an $\mathbf{x}\in E$ (with $E\subset R^n$,) and let $\mathbf{u}\in R^n$ be a unit vector. Then,

$$\underset{t\to 0}{lim}\frac{f(\mathbf{x}+t\mathbf{u})-f(\mathbf{x})}{t}=(\mathrm{\nabla }f)(\mathbf{x})\cdot \mathbf{u}$$

is called the directional derivative of $f$ at $\mathbf{x}$ in the direction of the unit vector $u$ and is denoted as $D_{\mathbf{u}}f(\mathbf{x}).$

Let $S$ be a surface represented by $f(x,y,z)=c,$ where $c$ is constant and $f$ is differentiable. Such a surface is called a level surface of $f,$ and for different $c,$ we get different level surfaces.

If $S$ is a level surface of a function $f,$ and $P$ is a point of $S,$ then the set of all tangent vectors of all curves passing through $P$ will generally form a plane, called the tangent plane of $S$ at $P.$

Given a tangent plane to a surface $S$ at $P,$ the normal to this plane (the straight line through $P$ perpendicular to the tangent plane) is called the surface normal to $S$ at $P.$

A vector in the direction of the surface normal of a surface $S$ at point $P$ is called a surface normal vector of $S$ at $P.$

A potential function is a scalar function whose gradient is a vector field. They allow representing certain vector fields in a simpler, more fundamental form.

In other words, given a vector field $\mathbf{F},$ a potential function $\varphi$ is a scalar function such that

$$\mathbf{F}=-\mathrm{grad}\varphi \text{or}\mathbf{F}=\mathrm{grad}\varphi .$$

The normal derivative is the directional derivative in the direction of the @normal-vector.

For a vector field $\mathbf{F}=(F_1,F_2,F_3)$ defined in three-dimensional space ${\mathbb{R}}^3$, the divergence is defined as

$$\mathrm{div}\mathbf{F}=\mathrm{\nabla }\cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}.$$

Let $\mathbf{v}$ be the velocity vector of of the motion of particles in a fluid. If $\mathrm{div}\mathbf{v}=0,$ then the fluid has constant density and is said to be incompressible.

For a vector field $\mathbf{F}=(F_1,F_2,F_3)$ defined in three-dimensional space ${\mathbb{R}}^3$, with each component function $F_i$ depending on the variables $x$, $y$, and $z$, the curl of $\mathbf{F}$ is defined as

$$\mathrm{curl}\mathbf{F}=\mathrm{\nabla }\times \mathbf{F}=(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z})\mathbf{i}+(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x})\mathbf{j}+(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y})\mathbf{k}$$

If the curl of a vector field is $\mathbf{0},$ i.e. if $\mathrm{curl}\mathbf{v}=0,$ the field is said to be irrotational.

The Laplace operator is a second-order differential operator in the $n$-dimensional Euclidean space, defined as the divergence $(\mathrm{\nabla }\cdot )$ of the gradient $(\mathrm{\nabla }f$). Thus, if $f$ is a twice differentiable real-valued function, then the Laplacian of $f$ is the real-valued function defined by

$$\mathrm{\Delta }f={\mathrm{\nabla }}^{2}f=\mathrm{\nabla }\cdot \mathrm{\nabla }f.$$

The Laplacian of $f$ is the sum of all the unmixed second partial derivatives in the Cartesian coordinates $x_i$:

$${\mathrm{\nabla }}^{2}f=\sum _{i=1}^n\frac{{\partial }^{2}f}{\partial x_i^2}.$$

In two dimensions, using Cartesian coordinates, the Laplace operator is given by

$${\mathrm{\nabla }}^{2}f=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}$$

and in three dimensions by

$${\mathrm{\nabla }}^{2}f=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}$$

Suppose $f:{\mathbb{R}}^n\to \mathbb{R}$ is a function taking as input a vector $\mathbf{x}\in {\mathbb{R}}^n$ and outputting a scalar $f(\mathbf{x})\in \mathbb{R}.$ If all second-order partial derivatives of $f$ exist, then the Hessian matrix $\mathbf{H}$ of $f$ is a square $n\times n$ @matrix, usually defined and arranged as

That is, the @entry of the $i$th row and the $j$th column is

$$({\mathbf{H}}_f{)}_{i,j}=\frac{{\partial }^{2}f}{\partial x_i\partial x_j}.$$

For a function $f:{\mathbb{R}}^3\to \mathbb{R},$ this is

Let $\mathbf{f}:{\mathbb{R}}^n\to {\mathbb{R}}^m$ be a function such that each of its first-order partial derivatives exists on ${\mathbb{R}}^n.$ This function takes a point $\mathbf{x}=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ as input and produces the vector $\mathbf{f}(\mathbf{x})=(f_1(\mathbf{x}),\dots ,f_m(\mathbf{x}))\in {\mathbb{R}}^m$ as output. Then the Jacobian matrix of $\mathbf{f},$ denoted ${\mathbf{J}}_{\mathbf{f}},$ is the $m\times n$ @matrix whose $(i,j)$ @entry is $\frac{\partial f_i}{\partial x_j};$ explicitly

where ${\mathrm{\nabla }}^{\mathsf{T}}{f}_i$ is the @transpose (@row-vector) of the gradient of the $i$-th component.