Partial derivatives — notation, calculation, geometric meaning: Visual walkthrough

Throughout we use the running surface $z = f (x, y) = x^{2} + y^{2}$ (a round bowl) at the point $(a, b) = (1, 2)$ . It is concrete enough to draw and simple enough to check every number by hand.

Step 1 — What "height over a floor" even means

WHAT. A function of two variables $f (x, y)$ takes a point on a flat floor — the $x y$ -plane — and returns a single number: a height. Stack all those heights up and you get a landscape floating over the floor: the surface $z = f (x, y)$ .

WHY. Before we can talk about slopes we must fix the picture: two horizontal inputs $x, y$ (where you stand) and one vertical output $z$ (how high the ground is). Everything else is bookkeeping on top of this one image.

PICTURE. The floor is the grid; the arrow rising off the point $(1, 2)$ shows the height $z = f (1, 2) = 1^{2} + 2^{2} = 5$ .

Figure — Partial derivatives — notation, calculation, geometric meaning

Read the equation $z = x^{2} + y^{2}$ term by term: $x^{2}$ says "the further East–West from centre, the higher"; $y^{2}$ says the same North–South; their sum is the total height. At $(1, 2)$ that height is $1 + 4 = 5$ .

Step 2 — Freeze $y$ : cut the surface with a wall

WHAT. Hold $y$ fixed at $b = 2$ . Every point with $y = 2$ forms a vertical wall — the plane $y = 2$ — slicing straight through the bowl. Where wall meets surface, we get a single curve.

WHY. A surface has infinitely many directions of slope at a point, so "the slope" is meaningless. To recover an ordinary one-slope situation we throw away one degree of freedom: freeze $y$ . Now only $x$ can move, and the surface becomes a plain curve we already know how to differentiate — this is the whole trick of single-variable thinking reused.

PICTURE. The cyan wall is $y = 2$ ; the amber curve is where it carves the bowl.

Notice: freezing $y$ did not delete $y$ ; it turned $y^{2}$ into the number $4$ . That constant $4$ raises the curve but adds $0$ to any slope.

Step 3 — The $x$ -slope is an ordinary derivative

WHAT. On the slice-curve $g (x) = x^{2} + 4$ we take a plain single-variable derivative and evaluate at $x = a = 1$ . This number is defined to be the partial derivative $f_{x} (1, 2)$ .

WHY. We spent Step 2 collapsing the surface to a curve precisely so that this step needs no new machinery — the limit definition of a derivative you already own does all the work. Term by term the limit is: $f_{x} (a, b) = lim_{h \to 0} \frac{f ( a + h , b ) - f ( a , b )}{h}$

$f (a + h, b)$ ::: height one small East step $h$ away, same $y = b$ .
$f (a, b)$ ::: height where you stand.
divide by $h$ ::: rise over run — a slope.
$lim_{h \to 0}$ ::: shrink the step to a point so "run" becomes "the tangent at that exact spot".

PICTURE. The amber tangent line kisses the slice-curve at $x = 1$ ; its rise-over-run is the red slope triangle.

$g^{'} (x) = 2 x \Rightarrow f_{x} (1, 2) = g^{'} (1) = 2 \cdot 1 = 2 .$

The frozen $4$ contributed $0$ , exactly as promised. The $x$ -slope of the bowl at $(1, 2)$ is $2$ .

Step 4 — Freeze $x$ instead: the other wall

WHAT. Now do the mirror move — hold $x$ fixed at $a = 1$ , sweep $y$ . The plane $x = 1$ is a wall running the other way, and it carves a different curve.

WHY. The bowl's steepness East–West and North–South need not agree, so a second, independent measurement is required. Symmetry of the setup means we repeat Steps 2–3 verbatim with the roles of $x$ and $y$ swapped.

PICTURE. The second cyan wall $x = 1$ and its amber slice-curve, side by side with the first for contrast.

The $y$ -slope is $4$ — steeper than the $x$ -slope $2$ , because at $(1, 2)$ we are further out along $y$ , and the bowl climbs faster the further you are from centre. This matches Example 4 in the parent note.

Step 5 — Two tangent lines, one flat plane

WHAT. Each partial gave a tangent line living inside its wall. Both lines pass through the same surface point $P = (1, 2, 5)$ . Two lines crossing at a point pin down exactly one flat plane — the tangent plane.

WHY. A plane is the flattest object that can touch a curved surface and share its direction. To be tangent it must reproduce the surface's slope in the $x$ -direction and in the $y$ -direction. Those two slopes are precisely $f_{x}$ and $f_{y}$ — so the plane is forced; there is no freedom left.

PICTURE. The amber $x$ -tangent and cyan $y$ -tangent through $P$ , with the translucent plane they span.

Build the plane's equation piece by piece: $z = height at P f (a, b) + x -slope f_{x} (a, b) (x - a) + y -slope f_{y} (a, b) (y - b)$

$f (a, b)$ ::: start at the right height, $5$ , so the plane touches $P$ .
$f_{x} (a, b) (x - a)$ ::: for every step $(x - a)$ East, climb by the East-slope $2$ .
$f_{y} (a, b) (y - b)$ ::: for every step $(y - b)$ North, climb by the North-slope $4$ .

Plugging our numbers: $z = 5 + 2 (x - 1) + 4 (y - 2) .$

Check the two slices agree. Set $y = 2$ : $z = 5 + 2 (x - 1)$ , slope $2$ ✔ (matches Step 3). Set $x = 1$ : $z = 5 + 4 (y - 2)$ , slope $4$ ✔ (matches Step 4). The plane reproduces both partials — that is what "tangent" means.

Step 6 — Degenerate & edge cases (never let the reader fall through)

Real surfaces are not always tame. Here is what each odd situation looks like.

Case A — flat spot ( $f_{x} = f_{y} = 0$ ). At the bottom of the bowl, $(0, 0)$ : $f_{x} = 2 \cdot 0 = 0$ , $f_{y} = 2 \cdot 0 = 0$ . Both slice-tangents are horizontal, so the tangent plane is the flat lid $z = 0$ . This is a minimum — the surface curves up away from a level plane.

Case B — one slope zero, one not. On the trough of a valley like $f = y^{2}$ (no $x$ at all): $f_{x} = 0$ everywhere but $f_{y} = 2 y$ . The plane tilts in $y$ only; it stays level East–West. A valid tangent plane still exists.

Case C — a corner/kink (no partial at all). For $f = x^{2} + y^{2}$ (an ice-cream cone) at the tip $(0, 0)$ , the East step and West step give slopes $+ 1$ and $- 1$ : the limit disagrees left vs right, so $f_{x} (0, 0)$ does not exist. No tangent line ⇒ no tangent plane. Partials require the slice-curve to be smooth at that point.

PICTURE. Bowl bottom (flat lid), valley (one-way tilt), cone tip (kink) — three miniatures.

The one-picture summary

Everything at once: the bowl, the two orthogonal cutting walls, the two amber/cyan tangent lines meeting at $P = (1, 2, 5)$ , and the flat tangent plane they span — annotated with the slopes $2$ and $4$ .

Recall Feynman retelling — the whole walk in plain words

Stand on a hilly field at one spot; your height there is $5$ . You cannot say "the slope" because you could face any direction. So you cheat honestly: face straight East and take a tiny step — the ground rose with steepness $2$ ; that is $f_{x}$ . Come back, face North, tiny step — steepness $4$ ; that is $f_{y}$ . Those two little slope measurements are two straight sticks laid on the hill, one pointing East, one North, both touching your feet. A flat board resting on those two sticks is the tangent plane. Its recipe is "start at height $5$ , add $2$ for every step East, add $4$ for every step North." If the ground had a sharp point under your feet (a cone tip), the East step up and West step down would disagree — no single slope, no stick, no board. Otherwise the two partials are the plane.

Connections

Tangent plane and linear approximation — this page is its geometric derivation.
Gradient vector — bundles the two slopes $f_{x}, f_{y}$ into one arrow $\nabla f$ .
Directional derivative — the slope when you face a diagonal, not just E or N.
Total differential — the plane written as $d z = f_{x} d x + f_{y} d y$ .
Chain rule (multivariable) — how these slices compose.
Single-variable derivative — each slice-curve reuses it wholesale.
Clairaut's theorem — symmetry of the second-order slopes.