Visual walkthrough — Chain rule for multivariable functions — all cases

Before line one, three plain words:

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Step 1 — One walker, one clock

WHAT. Instead of standing still on the landscape, imagine a walker whose position moves as time $t$ ticks. Their east coordinate is a function $x=x(t)$ and their north coordinate is $y=y(t)$. As $t$ changes, the point $(x,y)$ traces a path across the ground, and the height $z$ under their feet rises and falls.

WHY. This is the entire question of Case 1: how fast does the height change as the clock ticks? We want one number, $\frac{dz}{dt}$ — an ordinary derivative ($d$, not $\partial$), because in the end $z$ depends on the single variable $t$.

PICTURE. The curve below is the walker's trail on the map. The height is the color of the ground. The whole rest of the page zooms into ONE tiny green segment of this trail.

• $t$ ::: the clock; the only truly independent variable in Case 1.
• $(x(t),y(t))$ ::: the walker's position — both coordinates move as $t$ moves.

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Step 2 — Take one tiny step in time

WHAT. Advance the clock by a small amount, written $\Delta t$ (the symbol $\Delta$, capital Greek "D", means "a small change in"). The walker moves. Their east coordinate changes by $\Delta x$, their north coordinate by $\Delta y$, and the height changes by $\Delta z$.

WHY. Derivatives are born from tiny changes. If we understand how one small $\Delta t$ produces $\Delta z$, we can divide and shrink to get the exact rate. This is the same idea as the Single-variable Chain Rule, just with two coordinates moving at once.

PICTURE. Zoom into the green box from Step 1. The step in time splits into a horizontal nudge $\Delta x$ and a vertical nudge $\Delta y$ on the map.

• $\Delta t$ ::: tiny tick of the clock (the cause).
• $\Delta x,\Delta y$ ::: the east and north nudges it produces (the two "doors").
• $\Delta z$ ::: the resulting change in height (the effect we want).

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Step 3 — How height responds to a nudge: the two ramps

WHAT. Freeze time for a moment and ask: given the nudges $\Delta x$ and $\Delta y$, how much does the height change? Because $f$ is differentiable, near our point the curved ground looks flat — a tilted plane. On a plane the height change is just:

$\Delta z\approx \underset{\text{east steepness}}{\underbrace{\frac{\partial z}{\partial x}}}\underset{\text{east nudge}}{\underbrace{\Delta x}}+\underset{\text{north steepness}}{\underbrace{\frac{\partial z}{\partial y}}}\underset{\text{north nudge}}{\underbrace{\Delta y}}.$

WHY THIS TOOL. Why a plane, and why do the two effects add? Because "differentiable" is precisely the promise that up close the surface is well-approximated by a flat tilted sheet — this sheet is the Total Differential. On a flat sheet, moving east then north changes height by (east-steepness × east-distance) plus (north-steepness × north-distance). Independent directions contribute independently, so we sum. This is why the chain rule adds — the addition is inherited from the flatness of the local plane.

PICTURE. The tilted plane over the point. Walking east climbs one ramp; walking north climbs the other; the total rise is the sum of the two ramp-rises.

• $\frac{\partial z}{\partial x}\Delta x$ ::: rise from the east door = (east steepness) × (east nudge).
• $\frac{\partial z}{\partial y}\Delta y$ ::: rise from the north door = (north steepness) × (north nudge).

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Step 4 — The honest version: name the leftover error

WHAT. The $\approx$ hides a small lie: the ground is not exactly a plane. Differentiability lets us write the truth exactly by adding leftover terms:

$\Delta z=\frac{\partial z}{\partial x}\Delta x+\frac{\partial z}{\partial y}\Delta y+\underset{\text{the leftover curvature}}{\underbrace{{\epsilon }_1\Delta x+{\epsilon }_2\Delta y}},$

where ${\epsilon }_1,{\epsilon }_2$ (Greek "epsilon", meaning "a vanishingly small number") both shrink to $0$ as the nudges shrink to $0$.

WHY. We must not cheat. Keeping the error terms explicitly means our final rule will be exact, not approximate. The whole trick will be to show these leftovers die in the limit.

PICTURE. The gap between the true curved surface and the flat plane is the leftover. As the step shrinks, the gap shrinks faster than the step.

• ${\epsilon }_1,{\epsilon }_2$ ::: the curvature error per unit nudge — both $\to 0$ as $(\Delta x,\Delta y)\to (0,0)$.

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Step 5 — Divide by the clock tick

WHAT. We want a rate per unit time, so divide every term of the exact equation by $\Delta t$:

$\frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial x}\frac{\Delta x}{\Delta t}+\frac{\partial z}{\partial y}\frac{\Delta y}{\Delta t}+{\epsilon }_1\frac{\Delta x}{\Delta t}+{\epsilon }_2\frac{\Delta y}{\Delta t}.$

WHY. $\frac{dz}{dt}$ literally means "change in $z$ per change in $t$." So we measure everything per tick $\Delta t$. Notice $\frac{\Delta x}{\Delta t}$ = "east-speed of the walker" and $\frac{\Delta y}{\Delta t}$ = "north-speed."

PICTURE. Each nudge becomes a speed: divide the map nudges by the same clock tick.

• $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$ ::: how fast the walker goes east.
• $\frac{\Delta y}{\Delta t}\to \frac{dy}{dt}$ ::: how fast the walker goes north.

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Step 6 — Shrink the tick: the leftovers vanish, the rule appears

WHAT. Now let $\Delta t\to 0$. As the tick shrinks, the nudges $\Delta x,\Delta y\to 0$ too (the walker's path is continuous), so ${\epsilon }_1,{\epsilon }_2\to 0$. Meanwhile $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$ stays finite. So the two leftover terms are (something $\to 0$) × (something finite) $\to 0$ and die:

WHY the limit. The limit is what turns the approximate plane into the exact rate — the curvature error only mattered at finite step size, and it evaporates faster than $\Delta t$.

PICTURE. Two supply-roads flow into the output. Each road is a multiplication (steepness × speed); the meeting point is the sum.

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Step 7 — Edge cases: don't get caught out

Every scenario the reader could hit:

(a) A frozen coordinate. If the walker moves only east, then $\frac{dy}{dt}=0$ and the north path contributes nothing — we recover the Single-variable Chain Rule $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}$. The formula contains the 1-D case; it never contradicts it.

(b) Zero steepness but moving fast. If you walk along a level contour (height constant in that direction), the matching steepness is $0$, so that path adds $0$ no matter how fast you go. Speed alone doesn't change height — only speed times steepness does.

(c) Two ultimate variables → use $\partial$ (Case 2). If $x=x(s,t)$ and $y=y(s,t)$, then $z$ ends up depending on two variables $s,t$. Freeze $t$, wiggle $s$: this is exactly Step 6 applied to a slice, so $\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}.$ We now write $\partial$ (not $d$) because a companion variable $t$ survives.

(d) The sneaky direct path (implicit reappearance). If $w=f(x,y,t)$ and $x,y$ depend on $t$, there is a third road straight to $t$: $\frac{dw}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}.$ The tree literally shows three branches — miss the direct one and your answer is wrong. This is the engine behind Implicit Differentiation (multivariable).

PICTURE. Four mini-trees, one per case, showing which branches survive.

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The one-picture summary

The whole derivation on a single map: a tiny wiggle $\Delta t$ splits into two nudges, each nudge climbs its own ramp, the two rises add, we divide by $\Delta t$, shrink, and the error washes out — leaving the boxed rule. In matrix language this compresses to a Jacobian product $J_{z\circ x}=J_z{J}_x$ (see Gradient and Jacobian).

Recall Feynman retelling of the whole walkthrough

You're hiking on a hilly landscape, but you don't control where you stand directly — a clock does. Each tick, the clock pushes you a little east and a little north at once. To find how fast your altitude changes per tick: figure out how steep the hill is toward the east and multiply by your eastward speed — that's how much the east-move lifts you. Do the same for north. The hill isn't perfectly flat, so there's a tiny curvature fudge, but when the tick gets small the fudge shrinks away to nothing. Add the two lifts and you have your altitude's rate of change. More doors (more intermediate variables) → more roads to add; if the clock also changes the landscape itself, add one more straight road. That's the entire multivariable chain rule: sum the paths, multiply the links.

Recall Quick self-check

Why do the error terms ${\epsilon }_1\Delta x/\Delta t$ vanish? ::: Because ${\epsilon }_1\to 0$ as the step shrinks while $\Delta x/\Delta t\to dx/dt$ stays finite; zero times finite is zero. In Case 2 why is it $\partial z/\partial s$, not $dz/ds$? ::: Because a second ultimate variable $t$ still exists, so $z$ depends on more than one variable. Ex-check: for $z=x^2+y^2$, $x=r\cos \theta$, $y=r\sin \theta$, what is $\partial z/\partial r$? ::: $2r$ — the chain rule matches the direct answer $z=r^2$.

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Connections

• Parent topic — the full catalogue of cases.
• Partial Derivatives — the "steepness" ramps $\partial z/\partial x$.
• Total Differential — the tilted plane of Step 3.
• Directional Derivative — Case 2c is a directional derivative in disguise.
• Gradient and Jacobian — the one-picture matrix compression.
• Implicit Differentiation (multivariable) — the sneaky direct path (7d).
• Single-variable Chain Rule — the one-road special case (7a).