Directional derivative — definition, formula: Visual walkthrough

Step 0 — The pictures we must be able to read first

Before anything, three plain-word ideas, each pinned to a picture.

Figure — Directional derivative — definition, formula

In the picture: the floor is the pale grid, your feet sit at the coral dot $a$ , the lavender arrow $u$ is your facing (drawn length 1), and the surface floating above is the hill $f$ . Everything below is about one number: how fast the hill rises as you step along that lavender arrow.

Step 1 — Freeze the whole hill down to a single straight walk

The spot we reach after walking a distance $h$ is $start a + how far (metres) h which way (length 1) u .$

Because $u$ has length 1, moving by the number $h$ really does move you $h$ metres — that is the whole reason we demanded a unit vector.

Step 2 — Name the height-along-the-walk: the helper $g$

Term by term:

$h$ — the input: distance walked so far (can be negative — that means walking backwards).
$a + h u$ — the floor spot under me at that moment.
$f (\dots)$ — the height there.
$g (h)$ — the single output number: my current altitude.

Step 3 — The slope at the start is the directional derivative

Every piece:

$g (h) - g (0)$ — the rise: altitude gained after walking $h$ metres.
the division by $h$ — makes it rise per metre, i.e. a slope.
$lim_{h \to 0}$ — shrink the step to an instant, so we get the instantaneous steepness right where I stand, not an average over a long walk.
$g^{'} (0)$ — the ordinary derivative of the plain curve $g$ , evaluated at the start $h = 0$ .

So the mission is now concrete: compute $g^{'} (0)$ . We don't want to grind the limit each time — so we use a tool that already knows how to differentiate a composed multivariable function.

Step 4 — Why the chain rule, and nothing else

Track how each floor-coordinate depends on $h$ . First write out the moving spot in components, recalling $u = (u_{1}, u_{2})$ : $a + h u = (a_{1} + h u_{1}, a_{2} + h u_{2}),$ so the two moving coordinates are $x_{1} (h) = a_{1} + h u_{1}, x_{2} (h) = a_{2} + h u_{2} .$ Each is a straight line in $h$ , so its speed is constant: $\frac{d x _{1}}{d h} = u_{1}, \frac{d x _{2}}{d h} = u_{2} .$ Term by term: $u_{1}$ is how fast my east-coordinate grows per metre walked; $u_{2}$ the same for north. Together $(u_{1}, u_{2}) = u$ — the components of the facing arrow are literally "east-speed" and "north-speed" of the walk.

Step 5 — Turn the crank: the chain rule sum

This is just the general chain rule from Step 4 with $x^{'} (h) = u_{1}$ and $y^{'} (h) = u_{2}$ plugged in. The recipe in words: (how much $f$ changes when this coordinate moves) $\times$ (how fast this coordinate moves), summed over every coordinate. The $\frac{\partial f}{\partial x _{i}}$ are the partial derivatives — pure east and pure north steepness, which the parent note already met.

Step 6 — Stand still at the start: evaluate at $h = 0$

$D_{u} f (a) = g^{'} (0) = \frac{\partial f}{\partial x _{1}} (a) u_{1} + \frac{\partial f}{\partial x _{2}} (a) u_{2} .$

Now bundle the two partials into one arrow — the gradient $\nabla f (a) = (\frac{\partial f}{\partial x _{1}}, \frac{\partial f}{\partial x _{2}})$ — and notice the right-hand side is precisely "matching components multiplied and added," which is the dot product:

This is general: for $f : R^{n} \to R$ the same argument gives a sum over $n$ coordinates, $D_{u} f = \sum_{i} \frac{\partial f}{\partial x _{i}} (a) u_{i} = \nabla f \cdot u$ .

Step 7 — Read the formula geometrically (the $cos θ$ face)

We bring in $cos θ$ because it is exactly the dial that measures "how much is my facing aligned with the steepest direction?" — value $1$ when aligned, $0$ when perpendicular, $- 1$ when opposite.

Step 8 — The degenerate cases (never leave a gap)

Intuition Every corner covered

(a) $\nabla f (a) = 0$ (flat spot / summit / saddle). Then $D_{u} f = 0 \cdot u = 0$ for every direction $u$ . Picture a hilltop or valley floor: step any way you like, the ground is momentarily level. No preferred direction exists.

(b) Walking backwards, $- u$ . $D_{- u} f = \nabla f \cdot (- u) = - D_{u} f$ . Facing the exact opposite way flips the sign: what was uphill is now downhill by the same amount. Consistent, no surprise.

(c) $f$ not differentiable at $a$ (a sharp ridge or crease). The Step 4 chain rule needs a genuine tangent plane at $a$ . If the surface has a kink there, no single flat tangent plane exists, so the neat sum can be wrong even though the raw limit $g^{'} (0)$ might still exist direction-by-direction. Rule: the boxed formula is licensed only when $f$ is differentiable (continuous partials guarantee it).

The one-picture summary

One panel chains the whole story: hill $\to$ straight slice $\to$ 1-D curve $g$ $\to$ tangent slope $g^{'} (0)$ $\to$ chain-rule sum $\to$ $\nabla f \cdot u$ $\to$ $∥\nabla f ∥ cos θ$ . Follow the arrows and you have re-derived the directional derivative from a single step on a hillside.

Recall Feynman retelling — the whole walkthrough in plain words

I'm standing on a hill and I face some way (a 1-metre arrow $u$ ). I don't try to understand the whole hill; I just walk straight that way and write down my height as a plain graph — distance walked on the bottom, altitude up the side. That graph is $g$ . "How steep is the ground the way I face?" is just the slope of $g$ at the start, which is $g^{'} (0)$ . To find that slope without grinding a limit, I notice my walk moves me a bit east and a bit north at speeds $u_{1}$ and $u_{2}$ . The chain rule says my climb rate is (east-steepness $\times$ east-speed) $+$ (north-steepness $\times$ north-speed). Bundling the two steepnesses into one arrow (the gradient) and the two speeds into $u$ , that sum is the dot product $\nabla f \cdot u$ . Finally, since $u$ is 1 metre long, this equals $∥\nabla f ∥ cos θ$ : I climb fastest facing straight up the gradient ( $cos θ = 1$ ), climb nothing walking sideways ( $cos θ = 0$ ), and descend fastest facing backwards ( $cos θ = - 1$ ). At a flat summit the gradient is zero, so every direction is level.

Recall Quick self-test

Why does reducing to $g (h) = f (a + h u)$ help? ::: It turns the 2-D hill into a 1-D curve, so ordinary single-variable calculus (a tangent slope) applies. In $g^{'} (0) = \sum_{i} f_{x_{i}} u_{i}$ , what does each $u_{i}$ represent? ::: The speed at which coordinate $x_{i}$ changes per metre walked along $u$ . Why is $D_{u} f = ∥\nabla f ∥ cos θ$ (no $∥ u ∥$ )? ::: Because $u$ is a unit vector, $∥ u ∥ = 1$ . If $\nabla f (a) = 0$ , what is $D_{u} f$ for every $u$ ? ::: Zero — a flat spot, level in all directions. When can the dot-product formula fail? ::: When $f$ is not differentiable at $a$ (a kink/ridge), so no tangent plane exists.

Parent: Directional derivative — definition, formula (index 4.4.8) · Prereqs: Gradient vector, Partial derivatives, Multivariable chain rule, Dot product and cosine of angle, Level curves and level sets, Tangent planes and differentiability