Directional Derivatives and the Gradient

Quick answer

The directional derivative gives the rate of change of f in any direction u. By the chain rule it equals ∇f · u, the dot product of the unit vector u with the gradient ∇f = (f_x, f_y). Because of that dot product, the gradient points in the direction of fastest increase, its length is that fastest rate, and it is perpendicular to the level curve through the point. In three variables the gradient is normal to level surfaces, which gives their tangent planes.

What you'll learn

Compute the gradient of a function
Find a directional derivative as a dot product
Find the direction and rate of fastest increase
Use the gradient as a normal to level curves and surfaces

Rates in every direction

The partial derivatives measure how $f$ changes moving east or north. A hiker can walk any compass direction, though. Given a unit vector $\mathbf{u} = (u_1, u_2)$ , the directional derivative $D_{\mathbf{u}}f(a, b)$ is the rate of change of $f$ at $(a, b)$ moving in the direction $\mathbf{u}$ .

Collect the partial derivatives into one vector, the gradient:

\nabla f = \left(f_x, f_y\right)

Then every directional derivative is a dot product with it:

D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = f_x u_1 + f_y u_2

Why a dot product, and why the gradient points uphill

Walk from $(a, b)$ along the line $x = a + tu_1$ , $y = b + tu_2$ . The chain rule gives the rate at $t = 0$ :

\frac{d}{dt}f(a + tu_1,\ b + tu_2) = f_x u_1 + f_y u_2 = \nabla f \cdot \mathbf{u}

A dot product with a unit vector is $\|\nabla f\|\cos\theta$ , where $\theta$ is the angle between $\mathbf{u}$ and $\nabla f$ . It is largest when $\theta = 0$ , zero when $\theta = 90°$ , and most negative when $\theta = 180°$ . So the gradient points in the direction of fastest increase, its length is that fastest rate, and moving perpendicular to it leaves $f$ unchanged. Moving perpendicular to the gradient is walking along a level curve, so the gradient is perpendicular to the level curves.

f = 8
f = 16

The gradient of f(x, y) = x² + 4y² at (2, 1)

Worked examples

Common mistakes

Practice problems

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

Answer

$(4, 1)$

Full solution

$f_x = 2xy = 4$ and $f_y = x^2 = 1$ at $(1, 2)$ .
For the same function and point, find the directional derivative in the direction of $(1, 1)$ .

Answer

$\tfrac{5}{\sqrt{2}} \approx 3.54$

Full solution

The unit vector is $\tfrac{1}{\sqrt{2}}(1, 1)$ , and $(4, 1) \cdot (1, 1) = 5$ .
What are the directional derivatives of that function at $(1, 2)$ in the directions $\mathbf{i}$ and $\mathbf{j}$ ?

Answer

$4$ and $1$

Full solution

Along the axes, the directional derivatives are the partial derivatives themselves.
Find the maximum rate of change of $f(x, y) = e^x\sin y$ at $\left(0, \tfrac{\pi}{2}\right)$ , and its direction.

Answer

$1$ , in the direction $\mathbf{i}$

Full solution

$\nabla f = (e^x\sin y,\ e^x\cos y) = (1, 0)$ at that point.
Find $\nabla f$ for $f(x, y) = \sqrt{x^2 + y^2}$ at $(3, 4)$ . What is its length?

Answer

$(0.6, 0.8)$ ; length $1$

Full solution

$f_x = \tfrac{x}{\sqrt{x^2 + y^2}}$ and $f_y = \tfrac{y}{\sqrt{x^2 + y^2}}$ . The distance from the origin grows at rate $1$ moving straight away from the origin.
Find a vector perpendicular to the level curve $x^2 + y^2 = 25$ at $(3, 4)$ .

Answer

$(6, 8)$ , or any multiple

Full solution

The gradient of $x^2 + y^2$ is $(2x, 2y) = (6, 8)$ , pointing along the radius.
In which direction does $T(x, y) = 100 - x^2 - 3y^2$ fall fastest at $(2, 1)$ ?

Answer

The direction of $(4, 6)$

Full solution

The fastest decrease is opposite the gradient $(-4, -6)$ , away from the center, at rate $\sqrt{52}$ .
Find the tangent plane to the surface $z = x^2 + y^2$ at $(1, 1, 2)$ by treating it as a level surface.

Answer

$2x + 2y - z = 2$

Full solution

Write $F = x^2 + y^2 - z = 0$ . Then $\nabla F = (2, 2, -1)$ at the point, and $2(x - 1) + 2(y - 1) - (z - 2) = 0$ simplifies to the answer.
A function has $\nabla f(a, b) = (3, -4)$ . What are the largest and smallest directional derivatives at $(a, b)$ ?

Answer

$5$ and $-5$

Full solution

The largest is $\|\nabla f\| = 5$ , in the gradient’s direction; the smallest is $-5$ , in the opposite direction.
A student finds the rate of change of $f$ at a point in the direction $(3, 4)$ as $\nabla f \cdot (3, 4)$ . What went wrong?

Hint

What is the length of $(3, 4)$ ?

Answer

The direction must be a unit vector: use $\left(\tfrac{3}{5}, \tfrac{4}{5}\right)$ .

Full solution

$(3, 4)$ has length $5$ , so the student’s answer is five times the true rate. The directional derivative measures change per unit of distance, which requires $\|\mathbf{u}\| = 1$ .

Frequently asked questions

What is the gradient?

The vector of partial derivatives, ∇f = (f_x, f_y), or (f_x, f_y, f_z) in three variables.

How do you compute a directional derivative?

Take the dot product of the gradient with a unit vector u in the chosen direction: D_u f = ∇f · u. If the direction is given by a vector that is not a unit vector, divide it by its length first.

Which direction gives the fastest increase?

The direction of the gradient itself, where the rate is |∇f|. The fastest decrease is in the opposite direction, at rate −|∇f|.

Why is the gradient perpendicular to level curves?

Along a level curve f does not change, so the directional derivative along the curve is 0. A zero dot product with the tangent direction means the gradient is perpendicular to it.

How does the gradient give a tangent plane to a surface F(x, y, z) = k?

∇F at the point is a normal vector to the level surface, so the tangent plane is ∇F · (x − x₀, y − y₀, z − z₀) = 0.

What to learn next

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