Math Core

Partial derivatives tell you the rate of change of $f$ when you move parallel to the $x$ -axis or the $y$ -axis. But on a real hillside you can walk in any direction. This lesson measures the rate of change in an arbitrary direction and packages all of that information into a single vector, the gradient, which points uphill and is perpendicular to the level curves.

Directional derivatives

A direction in the plane is described by a unit vector $\mathbf{u} = \langle a, b\rangle$ with $a^2 + b^2 = 1$ .

Definition

Directional derivative

The directional derivative of $f$ at $(x_0, y_0)$ in the direction of the unit vector $\mathbf{u} = \langle a, b\rangle$ is

D_{\mathbf{u}} f(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h},

if the limit exists.

With $\mathbf{u} = \mathbf{i} = \langle 1, 0\rangle$ this is exactly $f_x$ , and with $\mathbf{u} = \mathbf{j}$ it is $f_y$ . To compute it in general, notice that $g(h) = f(x_0 + ha, y_0 + hb)$ is $f$ evaluated along a line, so the chain rule gives $g'(0) = f_x\,a + f_y\,b$ . That is:

D_{\mathbf{u}} f(x_0, y_0) = f_x(x_0, y_0)\,a + f_y(x_0, y_0)\,b.

The gradient vector

The expression $f_x a + f_y b$ is a dot product. The vector of partial derivatives is important enough to have its own name.

The gradient

The gradient of $f(x, y)$ is the vector $\nabla f = \langle f_x, f_y\rangle$ . For $f(x, y, z)$ it is $\nabla f = \langle f_x, f_y, f_z\rangle$ . For any unit vector $\mathbf{u}$ ,

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

Worked example: A directional derivative

Let $f(x, y) = x^2y - 3y$ . Find the directional derivative at $(2, 1)$ in the direction of $\mathbf{v} = \langle 3, 4\rangle$ .

The gradient is $\nabla f = \langle 2xy,\ x^2 - 3\rangle$ , so $\nabla f(2, 1) = \langle 4, 1\rangle$ .

The vector $\mathbf{v}$ is not a unit vector: $\lvert\mathbf{v}\rvert = 5$ . Normalize it: $\mathbf{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle$ . Then

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

Common mistake

Always convert the direction to a unit vector before taking the dot product. Using $\mathbf{v} = \langle 3, 4\rangle$ directly would give $16$ , five times too large. If a direction is given as an angle $\theta$ , the unit vector is $\langle \cos\theta, \sin\theta\rangle$ .

Steepest ascent

Write the dot product with the angle $\theta$ between $\nabla f$ and $\mathbf{u}$ :

D_{\mathbf{u}} f = \lvert\nabla f\rvert\,\lvert\mathbf{u}\rvert\cos\theta = \lvert\nabla f\rvert\cos\theta.

Since $\cos\theta$ ranges from $-1$ to $1$ , this single line answers three questions at once.

What the gradient tells you

At a point where $\nabla f \ne \mathbf{0}$ :

$f$ increases fastest in the direction of $\nabla f$ ( $\theta = 0$ ), and the maximum rate is $\lvert\nabla f\rvert$ .
$f$ decreases fastest in the direction of $-\nabla f$ ( $\theta = \pi$ ), at rate $-\lvert\nabla f\rvert$ .
The rate of change is $0$ in directions perpendicular to $\nabla f$ ( $\theta = \pi/2$ ).

Worked example: Heat-seeking

The temperature on a metal plate is $T(x, y) = 100 - x^2 - 2y^2$ . A bug sits at $(3, 1)$ . In which direction should it move to warm up as fast as possible, and how fast does the temperature rise in that direction?

$\nabla T = \langle -2x, -4y\rangle$ , so $\nabla T(3, 1) = \langle -6, -4\rangle$ . The bug should head in the direction of $\langle -6, -4\rangle$ , or equivalently the unit vector $\left\langle -\tfrac{3}{\sqrt{13}}, -\tfrac{2}{\sqrt{13}}\right\rangle$ . The maximum rate of increase is

\lvert\nabla T(3, 1)\rvert = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21 \text{ degrees per unit distance}.

The gradient is perpendicular to level curves

Moving along a level curve keeps $f$ constant, so the rate of change in the direction tangent to the level curve is $0$ . By the last bullet above, that tangent direction is perpendicular to $\nabla f$ . So the gradient at a point is perpendicular to the level curve through that point, and it points toward higher values.

Level curves of f = x² + 2y². At (2, 1.5), the gradient ⟨4, 6⟩ (drawn scaled down) is perpendicular to the level curve x² + 2y² = 8.5 through that point.Open in grapher →

This is the same fact a hiker uses: the steepest path up a mountain crosses the contour lines at right angles.

Tangent planes to level surfaces

The same reasoning works in three dimensions. If $F(x, y, z) = k$ is a level surface and $P = (x_0, y_0, z_0)$ lies on it, then $\nabla F(P)$ is perpendicular to every curve on the surface through $P$ . So $\nabla F(P)$ is a normal vector to the surface, and the tangent plane is

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

This handles surfaces, like spheres and ellipsoids, that are not graphs of a single function $z = f(x, y)$ .

Worked example: Tangent plane to an ellipsoid

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

With $F = x^2 + 2y^2 + 3z^2$ , $\nabla F = \langle 2x, 4y, 6z\rangle$ and $\nabla F(1, 1, 1) = \langle 2, 4, 6\rangle$ . The tangent plane is

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

Tip

The tangent plane formula from the previous lessons is a special case. The graph $z = f(x, y)$ is the level surface $F(x, y, z) = f(x, y) - z = 0$ , whose gradient is $\langle f_x, f_y, -1\rangle$ . Plugging that into the formula above gives $z - z_0 = f_x(x - x_0) + f_y(y - y_0)$ .

Practice

Practice 1

Find $\nabla f(1, -1)$ for $f(x, y) = x^3y + 2y^2$ .

Enter a point like (2, -3)

Practice 2

Find $\nabla f\left(0, \tfrac{\pi}{2}\right)$ for $f(x, y) = e^x\sin y$ .

Enter a point like (2, -3)

Practice 3

Find the directional derivative of $f(x, y) = x^2 + xy$ at $(1, 2)$ in the direction of $\mathbf{v} = \langle 4, -3\rangle$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Find the directional derivative of $f(x, y) = xy^2$ at $(2, 1)$ in the direction making an angle $\theta = \tfrac{\pi}{3}$ with the positive $x$ -axis.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the maximum rate of change of $f(x, y) = x^2 + 4y^2$ at the point $(3, 1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the unit vector in the direction in which $f(x, y) = x^2y$ decreases most rapidly at $(1, 2)$ .

Enter a point like (2, -3)

Practice 7

The tangent plane to the sphere $x^2 + y^2 + z^2 = 14$ at $(1, 2, 3)$ can be written as $x + 2y + 3z = d$ . Find $d$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

At a point $P$ , $\nabla f(P) = \langle 3, 4\rangle$ . In which unit direction is the directional derivative of $f$ at $P$ equal to $0$ ?