Directional derivative — definition, formula: Question bank

Prerequisite ideas you'll lean on: Gradient vector, Partial derivatives, Dot product and cosine of angle, Level curves and level sets, Tangent planes and differentiability, and the Multivariable chain rule.

True or false — justify

A directional derivative is a vector pointing in the direction you walked.

False — it is a scalar, a single slope number. The vector that points "most uphill" is

\nabla f

, not

D_{u} f

∥ u ∥ \neq = 1

, the formula

\nabla f \cdot u

still gives the directional derivative.

False — it gives the rate scaled by

∥ u ∥

. Because

h

in the limit measures true distance, only a unit vector reports "change per unit distance."

D_{u} f

and

D_{- u} f

are always equal.

False — they are negatives of each other:

\nabla f \cdot (- u) = - (\nabla f \cdot u)

. Walking one way rises exactly as fast as the opposite way falls.

At a point where

\nabla f = 0

, every directional derivative is zero.

True —

0 \cdot u = 0

for all

u

. Such critical points (peaks, valleys, saddles) look flat to first order in every direction.

The maximum possible value of

D_{u} f

at a point equals

∥\nabla f ∥

True — since

D_{u} f = ∥\nabla f ∥ cos θ

and

cos θ \leq 1

, the largest slope is

∥\nabla f ∥

, reached when

u

aligns with

\nabla f

The directional derivative along

(0, 1)

equals the partial derivative

f_{y}

True —

\nabla f \cdot (0, 1) = f_{y}

. A directional derivative along a coordinate axis is exactly that partial derivative.

If all directional derivatives of

f

exist at

a

, then

f

is differentiable there.

False — existence of every directional derivative does not imply differentiability (classic counterexamples have a sharp ridge at the origin). Differentiability is a stronger condition.

Two functions with the same gradient at

a

have the same directional derivative in every direction at

a

True —

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

depends only on the gradient at that point, so equal gradients force equal directional derivatives.

Spot the error

"Direction is

(3, 4)

, gradient is

(4, 1)

, so

D_{u} f = (4, 1) \cdot (3, 4) = 16

The vector

(3, 4)

wasn't unit-ized (

∥ (3, 4) ∥ = 5

). The true answer is

16/5

; skipping normalisation inflated it fivefold.

D_{u} f = ∥\nabla f ∥∥ u ∥ cos θ = ∥\nabla f ∥ cos θ

, so I can drop

∥ u ∥

regardless."

You may drop

∥ u ∥

only because

∥ u ∥ = 1

was assumed. If

u

isn't unit, that step secretly loses the

∥ u ∥

factor.

"The gradient is the direction I'm walking, so

D_{u} f

is a vector."

Confuses two objects:

\nabla f

is a vector, but

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

is their dot product, hence a scalar slope.

f

has a corner at

a

but all directional derivatives exist, so the formula

\nabla f \cdot u

applies."

The formula needs differentiability, not merely existence of directional derivatives. At a corner

f

may fail to be differentiable and the shortcut can give wrong values.

"Since

D_{u} f = 0

along

u

, the function is constant along that whole line."

A zero directional derivative is an instantaneous fact at the point

a

only; it says you're tangent to a level curve there, not that

f

stays constant along the entire straight line.

"To find the steepest descent direction I maximise

\nabla f \cdot u

Steepest descent means minimising the dot product, giving

u = - \nabla f /∥\nabla f ∥

with value

- ∥\nabla f ∥

; maximising gives ascent instead.

Why questions

Why must

u

be a unit vector in the definition?

Because the parameter

h

measures actual distance travelled; if

∥ u ∥ \neq = 1

, a step of size

h

covers

h ∥ u ∥

real distance and the reported slope is stretched by

∥ u ∥

Why does the gradient point in the direction of steepest ascent?

Writing

D_{u} f = ∥\nabla f ∥ cos θ

, the slope is largest when

cos θ = 1

, i.e. when

u

points along

\nabla f

; every other heading only leans partway toward it and climbs less.

Why is the directional derivative zero when

u ⊥ \nabla f

Perpendicular means

cos θ = 0

, so

\nabla f \cdot u = 0

; geometrically you're walking along a level curve where height doesn't change to first order.

Why does the chain rule appear in deriving the formula?

Restricting to the line

g (h) = f (a + h u)

makes each coordinate a function of

h

; the Multivariable chain rule sums each partial times its coordinate's rate

u_{i}

, giving

\sum f_{x_{i}} u_{i} = \nabla f \cdot u

Why do we bother with a formula instead of always using the limit definition?

The limit must be recomputed for every new direction, while

\nabla f \cdot u

reuses one gradient for all directions — compute partials once, dot with any

u

Why is the maximum slope

∥\nabla f ∥

and not something involving

u

Because

∥ u ∥ = 1

is fixed, so

D_{u} f = ∥\nabla f ∥ cos θ

depends on the angle alone; its peak

∥\nabla f ∥

is a property of the gradient's length, independent of which unit direction you pick.

Why does a directional derivative along an axis reduce to a partial derivative?

The unit vector

(0, \dots, 1, \dots, 0)

picks out exactly one component of

\nabla f

in the dot product, which is precisely that partial derivative's definition.

Edge cases

What is

D_{u} f

at a point where

\nabla f = 0

Zero in every direction, since

0 \cdot u = 0

; to first order the surface is flat there, so slope information alone can't distinguish peak, pit, or saddle.

f

is constant everywhere, what are its directional derivatives?

All zero —

\nabla f = 0

at every point, so no direction of walking changes the height.

Along a level curve, what does

D_{u} f

equal for

u

tangent to that curve?

Zero, because the tangent to a level curve is perpendicular to

\nabla f

; this is why gradients are always normal to level curves.

For a linear function

f (x, y) = a x + b y

, how does

D_{u} f

vary from point to point?

It doesn't vary with position —

\nabla f = (a, b)

is constant, so along a fixed

u

the directional derivative is the same everywhere; only the direction changes the number.

Can two different directions give the same directional derivative value?

Yes — any two unit directions making the same angle

θ

with

\nabla f

share the value

∥\nabla f ∥ cos θ

(e.g. two symmetric directions straddling the gradient).

What happens to

D_{u} f

u

rotates a full circle around a point (in 2D)?

It traces

∥\nabla f ∥ cos θ

, oscillating smoothly from a maximum

+ ∥\nabla f ∥

(along

\nabla f

), through

0

(perpendicular), to

- ∥\nabla f ∥

(opposite), and back.

Recall One-line summary of the traps

Normalise the direction, remember the answer is a scalar, and never invoke $\nabla f \cdot u$ without differentiability. Reason ::: These three habits defuse most directional-derivative mistakes.