Directional derivative — definition, formula

WHAT is it?

Why must $\mathbf{u}$ be a unit vector? Because $h$ in the limit measures actual distance travelled. If $\|\mathbf{u}\|=2$ , then moving by $h$ in " $\mathbf{u}$ -steps" covers $2h$ real distance, and the slope would be doubled artificially. Forcing $\|\mathbf{u}\|=1$ makes "rate per unit distance" meaningful.

HOW do we get the working formula? (Derivation from scratch)

We don't want to compute that limit every time. Let's reduce it to something we already know — partial derivatives.

Step 1 — Restrict $f$ to the line through $\mathbf{a}$ in direction $\mathbf{u}$ . Define a single-variable helper $g(h) = f(\mathbf{a} + h\mathbf{u}).$ Why this step? Walking in a straight line turns the multivariable problem into a 1-D problem. The directional derivative is exactly $g'(0)$ : $D_{\mathbf{u}}f(\mathbf{a}) = \lim_{h\to0}\frac{g(h)-g(0)}{h} = g'(0).$

Step 2 — Differentiate $g$ with the chain rule. Write $\mathbf{a}+h\mathbf{u} = (a_1 + h u_1,\; \dots,\; a_n + h u_n)$ . Each coordinate is $x_i(h) = a_i + h u_i$ , so $\frac{dx_i}{dh}=u_i$ . By the multivariable chain rule: $g'(h) = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i}\Big(\mathbf{a}+h\mathbf{u}\Big)\cdot\frac{dx_i}{dh} = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i}\cdot u_i.$ Why this step? The chain rule sums up "how $f$ changes via each coordinate" $\times$ "how fast that coordinate moves."

Step 3 — Evaluate at $h=0$ (so the point is $\mathbf{a}$ ). $\boxed{\,D_{\mathbf{u}}f(\mathbf{a}) = \sum_{i=1}^{n}\frac{\partial f}{\partial x_i}(\mathbf{a})\,u_i = \nabla f(\mathbf{a})\cdot\mathbf{u}\,}$

Figure — Directional derivative — definition, formula

Geometric meaning via the dot product

Since $D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} = \|\nabla f\|\,\|\mathbf{u}\|\cos\theta = \|\nabla f\|\cos\theta$ (because $\|\mathbf{u}\|=1$ ):

Worked examples

Worked example 1 — Two variables, a clean direction

$f(x,y)=x^2 + xy$ , at $\mathbf{a}=(1,2)$ , direction toward $(4,6)$ .

Step A. Gradient. $\nabla f = (2x+y,\; x)$ . Why? Partials: $f_x = 2x+y$ , $f_y = x$ . At $(1,2)$ : $\nabla f(1,2) = (2\cdot1+2,\;1) = (4,1)$ .

Step B. Make a UNIT direction. Direction vector $= (4,6)-(1,2) = (3,4)$ , length $=\sqrt{3^2+4^2}=5$ . So $\mathbf{u} = (3/5,\,4/5)$ . Why? The raw vector has length 5; we must normalise or the answer scales wrongly.

Step C. Dot product. $D_{\mathbf{u}}f = (4,1)\cdot(3/5,4/5) = \frac{12 + 4}{5} = \frac{16}{5} = 3.2.$

Forecast-then-Verify

Recall Forecast before computing

For $f(x,y)=x^2+y^2$ at $(3,4)$ , predict $D_{\mathbf{u}}f$ along $\mathbf{u}=(0,1)$ , then check. Forecast: moving purely in $+y$ , slope should equal $f_y=2y=8$ . Verify: $\nabla f=(2x,2y)=(6,8)$ ; $D_{\mathbf{u}}f=(6,8)\cdot(0,1)=8.$ ✓ (A directional derivative along an axis = that partial derivative.)

Common mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a hilly field and you're standing somewhere. If you face exactly East and take a tiny step, you might go up. If you face North, you might go up faster, or even go down. The "directional derivative" is just a number that says: for the way I'm facing right now, how steeply does the ground rise as I step forward? There's one special direction — straight up the hill — where you climb fastest. That arrow is called the gradient. For any other facing, you climb less, by an amount equal to how much that facing leans toward the gradient.

Flashcards

Define the directional derivative of

f

\mathbf{a}

in direction

\mathbf{u}

(limit form).

D_{\mathbf{u}}f(\mathbf{a})=\lim_{h\to0}\frac{f(\mathbf{a}+h\mathbf{u})-f(\mathbf{a})}{h}

, with

\|\mathbf{u}\|=1

Why must

\mathbf{u}

be a unit vector?

So that

h

measures actual distance travelled; otherwise the slope is scaled by

\|\mathbf{u}\|

Working formula for

D_{\mathbf{u}}f

(differentiable

f

D_{\mathbf{u}}f(\mathbf{a}) = \nabla f(\mathbf{a})\cdot\mathbf{u}

How is the formula derived?

Set

g(h)=f(\mathbf{a}+h\mathbf{u})

; then

D_{\mathbf{u}}f=g'(0)

and the chain rule gives

g'(0)=\sum f_{x_i}u_i=\nabla f\cdot\mathbf{u}

In which direction is

D_{\mathbf{u}}f

largest, and what is its value?

Along

\mathbf{u}=\nabla f/\|\nabla f\|

; max value

=\|\nabla f\|

What is

D_{\mathbf{u}}f

when

\mathbf{u}\perp\nabla f

Zero — you move along a level curve.

Is the directional derivative a scalar or a vector?

A scalar (a single slope number).

When can the dot-product formula fail despite the limit existing?

When

f

is not differentiable at

\mathbf{a}

(e.g. a sharp point), even if all directional derivatives exist.

Relation between

D_{\mathbf{u}}f

and partial derivatives along axes?

D_{\mathbf{e}_i}f = \partial f/\partial x_i

(axis direction recovers the partial).

Connections

Gradient vector — the direction of steepest ascent; $D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}$ .
Partial derivatives — special case when $\mathbf{u}$ is a coordinate axis.
Multivariable chain rule — the engine behind the derivation.
Level curves and level sets — where $D_{\mathbf{u}}f=0$ .
Tangent planes and differentiability — the condition that validates the formula.
Dot product and cosine of angle — gives the $\|\nabla f\|\cos\theta$ interpretation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, partial derivative sirf ye batata hai ki agar tum exactly $x$ -axis ke direction me chalo to function kitna fast change hota hai. Lekin zindagi me sirf East-West thodi chalna hai — tum kisi bhi direction me ja sakte ho. Directional derivative wahi cheez hai: tum ek point $\mathbf{a}$ pe khade ho, ek direction $\mathbf{u}$ choose karte ho, aur poochhte ho "is taraf chaloon to height kitni tezi se badhegi?" Ye answer ek number hota hai (slope), vector nahi.

Formula bahut simple hai: $D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}$ — yaani gradient ka dot product unit direction ke saath. Yaad rakhne ka trick: GUN — Gradient nikaalo, direction ko Unit banao (length 1 karo), phir Number nikaalo dot product se. Sabse important galti jo log karte hain: direction vector ko normalise karna bhool jaate hain. Agar direction $(3,4)$ di hai to uski length 5 hai, to pehle $5$ se divide karke $\mathbf{u}=(3/5,4/5)$ banao, warna answer 5 guna galat aayega.

Ek aur khaas baat: $\nabla f\cdot\mathbf{u} = \|\nabla f\|\cos\theta$ . Iska matlab — jab tum gradient ke saath-saath chalo ( $\theta=0$ ), slope maximum hota hai aur uski value $\|\nabla f\|$ hoti hai. Yahi reason hai ki gradient ko "steepest ascent direction" kehte hain. Aur agar tum gradient ke perpendicular chalo, slope zero — tum level curve pe ghoom rahe ho, height constant. Ye intuition hill-climbing aur optimization (gradient descent) me directly kaam aati hai, isliye ise rat-ta na maaro, samajh ke yaad rakho.

Directional derivative — definition, formula

WHAT is it?

HOW do we get the working formula? (Derivation from scratch)

Geometric meaning via the dot product

Worked examples

Forecast-then-Verify

Common mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections