Directional derivative — definition, formula: Worked examples

This is a companion to the parent note. There we built the formula $D_{u} f (a) = \nabla f (a) \cdot u, ∥ u ∥ = 1.$ Here we drill it. The plan: first lay out a scenario matrix — a checklist of every kind of case this topic can throw at you — then work examples that together tick off every cell.

The scenario matrix

Each row is a "case class" — a situation with its own trap. The last column names the example that covers it.

#	Case class	What's tricky about it	Covered by
C1	Direction given as two points	must subtract, then normalise	Ex 1
C2	Direction given as an angle	already unit — don't re-normalise	Ex 2
C3	Negative directional derivative (walking downhill)	sign of the answer is meaningful	Ex 3
C4	Zero directional derivative ( $u ⊥ \nabla f$ )	you're on a level curve	Ex 4
C5	Three variables, mixed signs in $u$	normalising with a minus sign	Ex 5
C6	Maximum / minimum rate, and the special directions	$\pm ∥\nabla f ∥$	Ex 6
C7	Degenerate: $\nabla f = 0$ at the point	every direction gives $0$	Ex 7
C8	Word problem (temperature / hill), with units	translate words → vectors	Ex 8
C9	Exam twist: given $D_{u} f$ in two directions, find $\nabla f$	reverse the dot product	Ex 9

Prerequisites you may want open: Gradient vector, Partial derivatives, Dot product and cosine of angle, Level curves and level sets.

C1 — Direction from two points

Worked example Ex 1 — subtract, then unit-ize

$f (x, y) = x^{2} + x y$ at $a = (1, 2)$ , walking from $(1, 2)$ toward $(4, 6)$ .

Forecast: the target is up-and-right, and the gradient (we'll see) points strongly rightward — so expect a solidly positive number, a few units in size.

Gradient. $f_{x} = 2 x + y, f_{y} = x$ , so $\nabla f = (2 x + y, x)$ . At $(1, 2)$ : $\nabla f = (4, 1)$ . Why this step? The formula is $\nabla f \cdot u$ ; we always need the gradient at the specific point first.
Direction vector. $(4, 6) - (1, 2) = (3, 4)$ . Why this step? "Toward a point" means the arrow from here to there — subtract start from end.
Unit-ize. $∥ (3, 4) ∥ = 9 + 16 = 5$ (nonzero, so this is legal), so $u = (3/5, 4/5)$ . Why this step? The directional derivative measures change per unit distance, so the heading must have length $1$ ; otherwise the slope is scaled by $5$ .
Dot. $D_{u} f = (4, 1) \cdot (3/5, 4/5) = \frac{12 + 4}{5} = \frac{16}{5} = 3.2$ . Why this step? $\nabla f \cdot u$ multiplies matching components and adds — this collapses "how $f$ changes per axis" into one slope for our heading.

Verify: $∥ u ∥ = (3/5)^{2} + (4/5)^{2} = 9/25 + 16/25 = 1$ ✓. Positive, moderate — matches the forecast.

C2 — Direction from an angle

Worked example Ex 2 — the angle is already a unit vector

$f (x, y) = sin (x y)$ at $(1, 0)$ , direction $3 0^{\circ}$ above the $+ x$ -axis.

Forecast: at $(1, 0)$ the " $x y$ " inside sine is $0$ , so things are near their fastest change — but which way? Guess before reading on.

Turn the angle into $u$ . $u = (cos 3 0^{\circ}, sin 3 0^{\circ}) = (\frac{3}{2}, \frac{1}{2})$ . Why this step? A point on the unit circle at angle $θ$ is $(cos θ, sin θ)$ — automatically length $1$ , so there is nothing to normalise. In Figure 1 this arrow is a radius of the unit circle: its tip sits on the circle, which is the visual meaning of "length $1$ ".
Gradient. $f_{x} = y cos (x y), f_{y} = x cos (x y)$ , so $\nabla f = (y cos (x y), x cos (x y))$ . At $(1, 0)$ : $cos (0) = 1$ , so $\nabla f = (0, 1)$ . Why this step? We need the gradient at the given point before dotting; the chain-rule factor $cos (x y)$ appears in both partials because it comes from differentiating the outer $sin$ .
Dot. $D_{u} f = (0, 1) \cdot (\frac{3}{2}, \frac{1}{2}) = \frac{1}{2}$ . Why this step? Dotting selects the overlap between heading and gradient. Here the $x$ -rate is $0$ , so only the vertical lean of the walk ( $\frac{1}{2}$ ) contributes.

Verify: $∥ u ∥^{2} = \frac{3}{4} + \frac{1}{4} = 1$ ✓. Answer $\frac{1}{2} = 0.5$ , positive and less than the max $∥\nabla f ∥ = 1$ — consistent.

Figure — Directional derivative — definition, formula

Reading Figure 1. The plum circle is the unit circle. The orange arrow is $u$ at $3 0^{\circ}$ — notice its tip lands exactly on the circle, confirming length $1$ with no re-normalising. The teal arrow is $\nabla f = (0, 1)$ , pointing straight up; the directional derivative $\frac{1}{2}$ is precisely how much $u$ leans toward that teal arrow.

C3 — A negative answer (walking downhill)

Worked example Ex 3 — the sign carries the geometry

$f (x, y) = x^{2} + y^{2}$ at $(3, 4)$ , in direction $v = (- 1, 0)$ (due west).

Forecast: this bowl rises away from the origin. Standing at $(3, 4)$ and heading west means heading back toward the centre — so downhill — expect a negative slope.

Gradient. $f_{x} = 2 x, f_{y} = 2 y$ , so $\nabla f = (2 x, 2 y) = (6, 8)$ at $(3, 4)$ . Why this step? We need the local uphill direction and its steepness before we can measure any heading against it.
Unit-ize $v$ . $∥ (- 1, 0) ∥ = 1$ (nonzero, and already length $1$ ), so $u = (- 1, 0)$ . Why this step? We must check the length rather than assume it — here it happens to be $1$ , so no division is needed.
Dot. $D_{u} f = (6, 8) \cdot (- 1, 0) = - 6$ . Why this step? The dot product picks out how much the heading aligns with $\nabla f$ ; the heading points opposite to the $+ x$ -part of the gradient, producing the minus sign.

Verify: A directional derivative along an axis equals $\pm$ the partial: heading in $- x$ gives $- f_{x} = - 6$ ✓. The minus sign literally means " $f$ decreases as you step west" — you're descending into the bowl.

C4 — A zero answer (on a level curve)

Worked example Ex 4 — perpendicular to the gradient

Same $f = x^{2} + y^{2}$ at $(3, 4)$ , but head in $v = (- 4, 3)$ .

Forecast: $\nabla f = (6, 8)$ points straight out from the origin. The direction $(- 4, 3)$ — is it perpendicular? $(6) (- 4) + (8) (3) = - 24 + 24 = 0$ . So expect exactly zero: you're sliding around the circle of constant height.

Gradient. $\nabla f = (2 x, 2 y) = (6, 8)$ at $(3, 4)$ . Why this step? Zero slope happens exactly when the heading is perpendicular to $\nabla f$ , so we need $\nabla f$ to test perpendicularity.
Unit-ize. $∥ (- 4, 3) ∥ = 16 + 9 = 5$ (nonzero), so $u = (- 4/5, 3/5)$ . Why this step? Normalising doesn't change perpendicularity, but a unit heading is required for the rate to mean "per unit distance."
Dot. $D_{u} f = (6, 8) \cdot (- 4/5, 3/5) = \frac{- 24 + 24}{5} = 0$ . Why this step? A zero dot product is the algebraic signature of a right angle — heading and gradient are perpendicular.

Verify: The level curve of $x^{2} + y^{2}$ through $(3, 4)$ is the circle of radius $5$ . The tangent to that circle at $(3, 4)$ is perpendicular to the radius $(3, 4)$ — and $(- 4, 3)$ is exactly that tangent. Zero slope = walking along constant height.

Reading Figure 2. The plum circles are level curves of $f$ (constant height). At $(3, 4)$ the orange arrow $\nabla f$ points radially outward — the uphill direction. The teal arrow is our heading $u = (- 4/5, 3/5)$ : it lies along the plum circle (tangent), at a right angle to orange. Because it never leaves a single circle, the height doesn't change — that is the geometric reason the answer is $0$ .

C5 — Three variables, mixed signs

Worked example Ex 5 — a minus sign inside the unit vector

$f (x, y, z) = x^{2} y + z^{3}$ at $(1, 1, 1)$ , direction $v = (2, - 1, 2)$ .

Forecast: the $z$ -slope is big ( $3 z^{2} = 3$ ) and the $x$ -slope is up too, but $u$ has a $- y$ component fighting the $+ y$ -slope. Net should be positive but reduced.

Gradient. $f_{x} = 2 x y, f_{y} = x^{2}, f_{z} = 3 z^{2}$ , so $\nabla f = (2 x y, x^{2}, 3 z^{2}) = (2, 1, 3)$ at $(1, 1, 1)$ . Why this step? Each partial answers "how fast does $f$ change if I nudge just this one variable?"; stacking them gives the full gradient we dot against.
Unit-ize. $∥ (2, - 1, 2) ∥ = 4 + 1 + 4 = 3$ (nonzero), so $u = (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})$ . Why this step? The minus stays with its component; normalising divides all three by the same length $3$ , preserving heading but fixing length to $1$ .
Dot. $D_{u} f = (2, 1, 3) \cdot (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3}) = \frac{4}{3} - \frac{1}{3} + 2 = 1 + 2 = 3$ . Why this step? Same recipe as in 2-D — multiply matching components and add; the $- \frac{1}{3}$ is the $y$ -slope working against us.

Verify: $∥ u ∥^{2} = \frac{4}{9} + \frac{1}{9} + \frac{4}{9} = \frac{9}{9} = 1$ ✓. The $- \frac{1}{3}$ term shows the $y$ -component pulling the total down; still solidly positive, as forecast.

C6 — Maximum and minimum rate

Worked example Ex 6 — the two extreme directions

For $f = x^{2} + x y$ at $(1, 2)$ (so $\nabla f = (4, 1)$ ), find the greatest and least possible directional derivatives, and the directions achieving them.

Forecast: the biggest slope is the length of the gradient; the most negative is minus that.

Max value. $D_{u} f = ∥\nabla f ∥ cos θ$ , maximised at $θ = 0$ : value $= ∥\nabla f ∥ = 4^{2} + 1^{2} = 17$ . Why this step? With $θ$ the angle between $\nabla f$ and $u$ , the only free part of $∥\nabla f ∥ cos θ$ is $cos θ$ , which peaks at $1$ when $θ = 0$ (walk exactly along the gradient).
Direction of max. $u = \frac{1}{17} (4, 1)$ . Why this step? $cos θ = 1$ means $u$ points the same way as $\nabla f$ ; unit-izing $\nabla f$ gives that heading.
Min value. At $θ = π$ , $cos θ = - 1$ : value $= - 17$ , direction $- \frac{1}{17} (4, 1)$ . Why this step? The most negative slope is achieved by facing exactly opposite the gradient — steepest descent.

Verify: $17 \approx 4.123$ . Check Ex 1's answer $3.2$ is below this cap ( $3.2 < 4.123$ ) ✓ — no direction can beat the gradient's length.

C7 — Degenerate: the gradient is zero

Worked example Ex 7 — a flat spot, every direction gives

$0$ $f (x, y) = x^{2} + y^{2}$ at the origin $(0, 0)$ , any (nonzero) direction $u$ .

Forecast: the origin is the bottom of the bowl — perfectly flat to first order. Whatever way you face, the instantaneous slope should be $0$ .

Gradient. $f_{x} = 2 x, f_{y} = 2 y$ , so $\nabla f = (2 x, 2 y) = (0, 0)$ at the origin. Why this step? The gradient's length is the maximum possible slope; if that length is $0$ , no direction can produce any first-order rise.
Dot. $D_{u} f = (0, 0) \cdot u = 0$ for every unit $u$ . Why this step? The zero vector dotted with anything is $0$ — the heading becomes irrelevant, which is exactly what "flat spot" means.

Verify: Directly from the parent note's limit definition, $f (h u) = h^{2} (u_{1}^{2} + u_{2}^{2}) = h^{2}$ , so $\frac{f ( h u ) - f ( 0 )}{h} = \frac{h ^{2}}{h} = h \to 0$ as $h \to 0$ ✓. The surface curves up in every direction, but the first-order rate is $0$ — a genuine flat spot (a minimum).

C8 — Word problem with units

Worked example Ex 8 — temperature on a metal plate

The temperature (in $^{\circ} C$ ) at point $(x, y)$ on a plate is $T (x, y) = 100 - x^{2} - 2 y^{2}$ , with $x, y$ in metres. An ant sits at $(2, 1)$ and marches in the direction of the vector $(3, 4)$ . How fast does it feel the temperature change, per metre walked?

Forecast: $(3, 4)$ points outward, where $- x^{2} - 2 y^{2}$ makes $T$ smaller — expect a negative rate (getting colder), a few $^{\circ} C$ per metre.

Gradient. $T_{x} = - 2 x, T_{y} = - 4 y$ , so $\nabla T = (- 2 x, - 4 y) = (- 4, - 4)$ at $(2, 1)$ . Why this step? The gradient collects the temperature's rate along each axis (in $^{\circ} C$ per metre) — the raw material for any heading.
Unit-ize. $∥ (3, 4) ∥ = 9 + 16 = 5$ (nonzero), so $u = (3/5, 4/5)$ . Why this step? The question asks per metre, so the heading must itself be one metre long — i.e. length $1$ .
Dot. $D_{u} T = (- 4, - 4) \cdot (3/5, 4/5) = \frac{- 12 - 16}{5} = \frac{- 28}{5} = - 5.6$ . Why this step? Dotting combines both cooling rates, weighted by how much the ant's path leans along each axis.

Verify: Units: $\nabla T$ is $^{\circ} C / m$ , $u$ is dimensionless, so the answer is $^{\circ} C / m$ . Result $- 5.6^{\circ} C / m$ — negative (cooling), as forecast. The ant cools by $5. 6^{\circ}$ for each metre it walks along $(3, 4)$ .

C9 — Exam twist: recover the gradient

Worked example Ex 9 — reverse the dot product

A differentiable $f (x, y)$ satisfies $D_{u} f = 1$ in direction $u_{1} = (1, 0)$ and $D_{u} f = 2$ in direction $u_{2} = (0, 1)$ , both at the same point $a$ . Find $\nabla f (a)$ , then the directional derivative along $u_{3} = (3/5, 4/5)$ .

Forecast: directional derivatives along the axes are the partials, so $\nabla f$ should just be $(1, 2)$ .

Read off the partials. $D_{(1, 0)} f = \nabla f \cdot (1, 0) = f_{x} = 1$ and $D_{(0, 1)} f = \nabla f \cdot (0, 1) = f_{y} = 2$ . Why this step? Dotting the gradient with $(1, 0)$ selects its first component; with $(0, 1)$ , its second — so these two special headings hand us the partials directly.
Assemble. $\nabla f (a) = (1, 2)$ . Why this step? The gradient is defined as the vector of partials, so we just stack the numbers from step 1.
Use it. $D_{u_{3}} f = (1, 2) \cdot (3/5, 4/5) = \frac{3 + 8}{5} = \frac{11}{5} = 2.2$ . Why this step? With $\nabla f$ known and $f$ differentiable, the dot-product formula works for any unit heading.

Verify: $∥ u_{3} ∥ = 9/25 + 16/25 = 1$ ✓. And $2.2 < ∥\nabla f ∥ = 1^{2} + 2^{2} = 5 \approx 2.236$ ✓ — below the max, as it must be. This works because $f$ is differentiable, so the linear (dot-product) formula holds in every direction.

Wrapping up — every cell is ticked

Recall

Recall Which cell is which?

C3 asks for the sign of $D_{u} f$ when walking toward the origin in a bowl. What sign, and why? ::: Negative — the height decreases as you approach the minimum. In C4, why is the answer exactly $0$ ? ::: The direction is perpendicular to $\nabla f$ , i.e. tangent to the level curve, so height is momentarily constant. In C7 the gradient is $0$ . What is $D_{u} f$ for a general $u$ ? ::: $0$ for every direction — the zero vector dotted with anything is $0$ . In C9, how do you recover $\nabla f$ from two axis-direction derivatives? ::: They are the partials: $D_{(1, 0)} f = f_{x}$ , $D_{(0, 1)} f = f_{y}$ , so $\nabla f = (f_{x}, f_{y})$ . What does $θ$ mean in $D_{u} f = ∥\nabla f ∥ cos θ$ ? ::: The angle between the gradient $\nabla f$ and the heading $u$ . Why can't we take a directional derivative "in direction $0$ "? ::: The zero vector has no heading and $∥ 0 ∥ = 0$ makes unit-izing divide by zero — the question is undefined.