Gradient¶

Definition¶

Let $E\subseteq \mathbb{R}^n$ $R^{n}$ , and let $f \colon E \to \mathbb{R}$ be a real-valued function. If $\boldsymbol{x}_0$ is an interior point of $E$ and all first partial derivatives exist there, the gradient $\boldsymbol{\nabla} f (\boldsymbol{x}_0)$ of $f$ at $\boldsymbol{x}=\boldsymbol{x}_0$ is given as the $n$ -dimensional column vector:

\boldsymbol{\nabla} f (\boldsymbol{x}_0) = \left(\frac{\partial f}{\partial x_1}(\boldsymbol{x}_0),\, \frac{\partial f}{\partial x_2}(\boldsymbol{x}_0), \ldots,\, \frac{\partial f}{\partial x_n}(\boldsymbol{x}_0)\right)^{\top}.

Interpretation¶

The gradient collects the first partial derivatives of a real-valued function into one vector. For a totally differentiable $f$ it points in the direction in which $f$ increases fastest, and its length is that fastest rate of increase.

Elliptical level curves of f with gradient arrows at four points, each perpendicular to its level curve and pointing outwards, uphill.

The figure shows level curves of $f(x,y)=x^2+2y^2$ for the values $2$ , $4$ and $6$ , with $\boldsymbol{\nabla} f(x,y)=(2x,4y)^{\top}$ $(2 x, 4 y)^{⊤}$ . The arrows start at four points of the level curve $f=4$ $4$ , such as $(2,0)$ and $(0,\sqrt{2})$ ; each is perpendicular to that curve and points outwards, towards larger values of $f$ . The gradients there are $(4,0)^{\top}$ and $(0,4\sqrt{2})^{\top}$ : $f$ rises faster along the $y$ -axis, where the level curves lie closer together.

The gradient arrows point uphill, perpendicular to the level curves, and are longer where the curves are close together. Drag $\boldsymbol x_0$ to a stationary point.

The red path climbs the surface along $\boldsymbol{\nabla} f$ ; on the floor it crosses the level curves at right angles.

When $\boldsymbol{\nabla} f (\boldsymbol{x}_0)= \boldsymbol{0}$ $0$ , we say that $\boldsymbol{x}_0$ is a stationary point of $f$ (possibly a local extremum point). At an interior local extremum every partial derivative that exists is $0$ , but a stationary point need not be an extremum: $f(x,y)=x^2-y^2$ has $\boldsymbol{\nabla} f(0,0)=\boldsymbol{0}$ $0$ , yet takes both larger and smaller values than $f(0,0)$ near the origin.

When $f$ is totally differentiable at $\boldsymbol{x}_0$ and $\boldsymbol{\nabla} f (\boldsymbol{x}_0)\neq \boldsymbol{0}$ $0$ , the gradient gives two things: the direction $\boldsymbol{u}=\boldsymbol{\nabla} f(\boldsymbol{x}_0)/\|\boldsymbol{\nabla} f(\boldsymbol{x}_0)\|$ of steepest ascent, and the maximal rate of increase $\|\boldsymbol{\nabla} f(\boldsymbol{x}_0)\|=\max_{\|\boldsymbol{u}\|=1}\mathrm{D}_{\boldsymbol{u}}f(\boldsymbol{x}_0)\text{,}$ by the Cauchy-Schwarz inequality. Along a level curve $f$ does not change, so the directional derivative in a tangent direction is $0$ ; this is why the gradient is perpendicular to the level curves.

A common slip is to draw the gradient on the graph of $f$ . It is a vector in the domain $\mathbb{R}^n$ , drawn in the same plane as the level curves.

Key Results¶

If $f$ is totally differentiable at $\boldsymbol{x}_0\text{,}$ the gradient is the vector whose dot product with any unit vector $\boldsymbol{v}$ equals the directional derivative of $f$ at $\boldsymbol{x}=\boldsymbol{x}_0$ along $\boldsymbol{v}$ , i.e.,

\boldsymbol{\nabla} f (\boldsymbol{x}_0) \cdot \boldsymbol{v} = \mathrm{D}_{\boldsymbol{v}}f(\boldsymbol{x}_0).

The existence of the gradient (all partial derivatives at $\boldsymbol{x}_0$ ) alone does not guarantee this formula.

Examples¶

Let $f(x,y) = \mathrm{e}^{x+y}$ $e^{x + y}$ . Then $\displaystyle \boldsymbol{\nabla} f(x,y) = \begin{pmatrix} \mathrm{e}^{x+y} \\ \mathrm{e}^{x+y} \end{pmatrix}$ $(e^{x + y} e^{x + y})$ .
Let $f(x,y) = x^2 y + y^3$ $y^{3}$ . Then $\boldsymbol{\nabla} f(x,y) = \begin{pmatrix} 2xy \\ x^2 + 3y^2 \end{pmatrix}$ $(2 x y x^{2} + 3 y^{2})$ .

Prerequisites¶

Stationary Point

Sources¶

Math III: Analysis · Lecture 12

Aliases: gradient vector Symbols: nabla gradient, partial, real numbers, square root, subset

Last updated: 22 October 2026 19:04