Directional Derivatives and the Gradient

Partial derivatives measure rates of change along the coordinate axes. But a hiker on a mountain may walk in any direction — not just north–south or east–west. The directional derivative answers: how fast does $f$ change if we move in an arbitrary direction $\mathbf{u}$ ?

1. The question

Let $f(x,y)$ be a function and let $\mathbf{u} = \langle a, b \rangle$ be a unit vector ( $a^2 + b^2 = 1$ ). Starting at a point $(x,y)$ , we move a small distance $h$ in direction $\mathbf{u}$ to reach $(x+ha,\, y+hb)$ .

The average rate of change of $f$ over that step is

$\frac{f(x+ha,\, y+hb) - f(x,y)}{h}.$

What happens as $h \to 0$ ?

2. Definition via the limit

The directional derivative of $f$ at $(x,y)$ in the direction of the unit vector $\mathbf{u} = \langle a,b \rangle$ is

$\boxed{D_{\mathbf{u}} f(x,y) = \lim_{h \to 0} \frac{f(x+ha,\, y+hb) - f(x,y)}{h},}$

provided the limit exists.

Special cases. When $\mathbf{u} = \mathbf{i} = \langle 1,0 \rangle$ this is exactly $f_x$ ; when $\mathbf{u} = \mathbf{j} = \langle 0,1 \rangle$ it is $f_y$ . The partial derivatives are directional derivatives along the coordinate axes.

Note. The vector $\mathbf{u}$ must be a unit vector. If you are given a direction vector $\mathbf{v}$ that is not unit-length, normalize it first: $\mathbf{u} = \mathbf{v}/|\mathbf{v}|$ .

3. Geometric interpretation

Geometrically, $D_{\mathbf{u}} f$ is the slope of the surface $z = f(x,y)$ as seen by someone walking in direction $\mathbf{u}$ .

Use the interactive graph below to explore. The left panel shows a unit circle — click or drag to choose a direction $\mathbf{u}$ . The right panel shows the saddle surface $z = x^2 - y^2$ with the tangent plane at the base point $(1,0,1)$ and the red slice curve (the cross-section of the surface cut by a vertical plane through the base point in direction $\mathbf{u}$ ). The green dashed line is tangent to that slice; its slope equals $D_{\mathbf{u}} f$ .

Directional derivative on z = x² − y²

Choose direction u

θ =30°

u = (0.87, 0.50)

D_u f = 1.732

∇f = (2.0, 0.0) — gradient (dashed)

Click or drag on the circle to change direction. Red = slice curve, green dashed = tangent line, orange = tangent plane. Green dot on circle = gradient direction.

Notice how the slope of the tangent line changes as you rotate $\mathbf{u}$ : positive when climbing, negative when descending, zero when moving along a level direction.

4. Computing the directional derivative

Theorem. If $f$ is differentiable at $(x,y)$ and $\mathbf{u} = \langle a,b \rangle$ is a unit vector, then

$\boxed{D_{\mathbf{u}} f(x,y) = a\, f_x(x,y) + b\, f_y(x,y).}$

Define the single-variable function

$g(t) = f(x + ta,\; y + tb).$

Then $D_{\mathbf{u}} f(x,y) = g'(0)$ by definition. By the chain rule,

$g'(t) = f_x(x+ta,\, y+tb)\cdot a \;+\; f_y(x+ta,\, y+tb)\cdot b.$

Setting $t = 0$ gives

$D_{\mathbf{u}} f(x,y) = a\, f_x(x,y) + b\, f_y(x,y). \qquad \square$

Example. Let $f(x,y) = x^2 + 3xy$ and $\mathbf{u} = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$ . At the point $(2,1)$ :

$f_x = 2x + 3y = 7, \qquad f_y = 3x = 6.$

Therefore

$D_{\mathbf{u}} f(2,1) = \tfrac{1}{\sqrt{2}} \cdot 7 + \tfrac{1}{\sqrt{2}} \cdot 6 = \frac{13}{\sqrt{2}} \approx 9.19.$

5. The gradient vector

The formula $D_{\mathbf{u}} f = af_x + bf_y$ is a dot product. Define the gradient vector

$\nabla f(x,y) = \langle f_x(x,y),\; f_y(x,y) \rangle.$

Then the directional derivative formula becomes simply

$\boxed{D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}.}$

For a function of three variables, $\nabla f(x,y,z) = \langle f_x, f_y, f_z \rangle$ , and $D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}$ with $\mathbf{u}$ a unit vector in $\mathbb{R}^3$ .

6. The gradient direction maximizes the rate of change

By the Cauchy–Schwarz inequality,

$D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} \leq |\nabla f|\,|\mathbf{u}| = |\nabla f|,$

with equality exactly when $\mathbf{u}$ points in the same direction as $\nabla f$ .

Direction $\mathbf{u}$	Value of $D_{\mathbf{u}} f$
$\dfrac{\nabla f}{\lvert\nabla f\rvert}$	$\lvert\nabla f\rvert$ — maximum rate of increase
$-\dfrac{\nabla f}{\lvert\nabla f\rvert}$	$-\lvert\nabla f\rvert$ — maximum rate of decrease
Any $\mathbf{u} \perp \nabla f$	$0$ — no change

The gradient points in the direction of steepest ascent, and $|\nabla f|$ is the steepness.

7. Gradient direction — interactive illustration

The graph below uses $z = x^2 + y^2$ at the base point $(1,1,2)$ , where $\nabla f = \langle 2,2 \rangle$ . Drag the direction selector and watch $D_{\mathbf{u}} f = 2\cos\theta + 2\sin\theta$ update live. Click "Jump to gradient direction" to confirm that $\theta = 45°$ achieves the maximum $2\sqrt{2} \approx 2.83$ .

Gradient direction maximizes D_u f on z = x² + y²

Choose direction u

θ =30°

u = (0.87, 0.50)

D_u f = 2.732

∇f = (2.0, 2.0) — gradient (dashed)

Click or drag on the circle to change direction. Red = slice curve, green dashed = tangent line, orange = tangent plane. Green dot on circle = gradient direction.

8. Level sets and the gradient

A level surface of $F(x,y,z)$ is the set of all points satisfying $F(x,y,z) = k$ for some constant $k$ . For a function $f(x,y)$ the level surfaces are the familiar level curves $f(x,y) = c$ .

The key geometric fact is:

$\boxed{\nabla F \text{ is perpendicular to the level surface } F = k.}$

The graph below shows the paraboloid $z = x^2 + y^2$ together with a family of level curves. The gradient at any point on a level curve points outward — perpendicular to the curve.

Level curves of z = x² + y²

c = 3.00

c =

The blue surface is z = x² + y². The colored circles are a family of five level curves, and the matching circles below are their projections onto the xy-plane.

Let $\mathbf{r}(t) = (x(t), y(t), z(t))$ be any smooth curve that lies on the level surface $F = k$ . Then

$F(\mathbf{r}(t)) = k \quad \text{for all } t.$

Differentiating both sides with respect to $t$ using the chain rule:

$\nabla F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \frac{dk}{dt} = 0.$

So $\nabla F$ is perpendicular to $\mathbf{r}'(t)$ — the tangent vector of every curve on the surface. The tangent vectors of all such curves span the tangent plane, so $\nabla F$ is a normal vector to the surface. $\square$

Consequence. The tangent plane to $F(x,y,z) = k$ at $(x_0,y_0,z_0)$ is

$F_x(x_0,y_0,z_0)(x-x_0) + F_y(x_0,y_0,z_0)(y-y_0) + F_z(x_0,y_0,z_0)(z-z_0) = 0.$

9. Exercises

Let $f(x,y) = x^2 + xy$ and $\mathbf{v} = \langle 3, 4 \rangle$ . Find $D_{\mathbf{u}} f(1,2)$ where $\mathbf{u} = \mathbf{v}/|\mathbf{v}|$ .

Solution.

First normalize: $|\mathbf{v}| = 5$ , so $\mathbf{u} = \langle \tfrac{3}{5}, \tfrac{4}{5} \rangle$ .

Partial derivatives at $(1,2)$ :

$f_x = 2x + y = 4, \qquad f_y = x = 1.$

Therefore

$D_{\mathbf{u}} f(1,2) = \tfrac{3}{5}\cdot 4 + \tfrac{4}{5}\cdot 1 = \tfrac{12}{5} + \tfrac{4}{5} = \frac{16}{5}.$

Let $f(x,y) = xe^y$ . Compute $\nabla f(x,y)$ and evaluate at $(2,0)$ . Find the direction of steepest ascent and the maximum rate of increase at that point.

Solution.

$f_x = e^y, \qquad f_y = xe^y, \qquad \nabla f(x,y) = \langle e^y,\; xe^y \rangle.$

At $(2,0)$ : $\nabla f(2,0) = \langle 1, 2 \rangle$ .

Direction of steepest ascent: $\mathbf{u} = \langle 1,2 \rangle/\sqrt{5}$ .

Maximum rate: $|\nabla f(2,0)| = \sqrt{5}$ .

Find the direction of steepest ascent of $f(x,y) = x^2 - y^2$ at the point $(1,1)$ .

Solution.

$\nabla f(x,y) = \langle 2x,\; -2y \rangle, \qquad \nabla f(1,1) = \langle 2, -2 \rangle.$

Direction of steepest ascent:

$\mathbf{u} = \frac{\langle 2,-2 \rangle}{2\sqrt{2}} = \left\langle \frac{1}{\sqrt{2}},\; -\frac{1}{\sqrt{2}} \right\rangle.$

Maximum rate of increase: $|\nabla f(1,1)| = 2\sqrt{2}$ .

Find the tangent plane to the ellipsoid $x^2 + 2y^2 + 3z^2 = 6$ at the point $(1, 1, 1)$ .

Solution.

Let $F(x,y,z) = x^2 + 2y^2 + 3z^2$ . Then

$\nabla F = \langle 2x,\; 4y,\; 6z \rangle, \qquad \nabla F(1,1,1) = \langle 2, 4, 6 \rangle.$

The tangent plane is

$2(x-1) + 4(y-1) + 6(z-1) = 0 \implies x + 2y + 3z = 6.$