Chain rule — proof, composite function derivatives: Visual walkthrough

Everything rests on one earlier idea: the derivative as a limit of a difference quotient. If that phrase means nothing yet, Step 1 rebuilds it from zero.

Step 1 — What a "rate" even is (the slope of a tiny secant)

WHAT. Pick a function — a machine that eats a number $x$ and outputs a number. Nudge the input by a tiny amount we'll call $Δ x$ ("delta-x", read as "a small change in $x$ "). The output changes by some amount. The rate is

"rise over run" \frac{change in output}{change in input} = \frac{f ( x + Δ x ) - f ( x )}{Δ x} .

Term by term: $x + Δ x$ is the nudged input; $f (x + Δ x) - f (x)$ is how much the output moved (the rise); $Δ x$ is how much the input moved (the run). The fraction is the slope of the secant line joining the two dots.

WHY. We need one honest number for "how fast the output responds to the input". Slope = rise/run is exactly that. As $Δ x$ shrinks to $0$ , the secant snuggles onto the curve and its slope becomes the derivative $f^{'} (x)$ — the instantaneous rate. That shrinking limit is written $lim_{Δ x \to 0}$ .

PICTURE. The blue curve; two dots a distance $Δ x$ apart; the orange secant; the run and rise labelled. Watch the secant tilt toward the green tangent as the dots close in.

Figure — Chain rule — proof, composite function derivatives

Step 2 — A composite is TWO machines in a row

WHAT. Now chain two machines. The inner machine $g$ eats $x$ and outputs a middle value we name $u = g (x)$ (" $u$ " is just a nickname for the inner output). The outer machine $f$ eats that $u$ and outputs the final number $y = f (u)$ . The whole pipeline is

h (x) = f (= u g (x)) .

Term by term: $g (x)$ runs first and lands at $u$ ; $f$ acts on that $u$ ; $h$ is the name for the two-step journey $x \to u \to y$ .

WHY. Most real functions are secretly stacked like this — $sin (x^{2})$ , $e^{s i n x}$ , $cos 4 x$ . If we can differentiate a stack, we can differentiate almost anything. We must first name the middle station $u$ , because the whole proof is about how a wiggle travels through it.

PICTURE. Two boxes on a conveyor belt: $x$ enters the $g$ -box, exits as $u$ on the middle rail, enters the $f$ -box, exits as $y$ .

Step 3 — Wiggle the input; watch the wiggle get relayed

WHAT. Nudge the input: $x \to x + Δ x$ . This forces the middle station to move by an amount we name the inner increment

Δ u = g (x + Δ x) - g (x) .

Term by term: $Δ u$ is how far the middle rail slid because we pushed $x$ . Then the outer machine turns that $Δ u$ into a final move $Δ y = f (u + Δ u) - f (u)$ .

WHY. This is the heart of the "gear-ratio" story. One push at the input ( $Δ x$ ) becomes a push at the middle ( $Δ u$ ), which becomes a push at the output ( $Δ y$ ). We want the total ratio $Δ y /Δ x$ , and the natural move is to route it through the middle.

PICTURE. Three number-lines stacked: the $x$ -line with a small arrow $Δ x$ ; the $u$ -line with a bigger arrow $Δ u$ ; the $y$ -line with an even bigger arrow $Δ y$ . Each amplification is one machine.

Step 4 — The naïve split (route through the middle)

WHAT. Write the rate we want and multiply-and-divide by $Δ u$ :

\frac{Δ y}{Δ x} = outer rate \to f^{'} (u) \frac{f ( u + Δ u ) - f ( u )}{Δ u} \cdot inner rate \to g^{'} (x) \frac{Δ u}{Δ x} .

Term by term: the left fraction is "final move per middle move" — as $Δ u \to 0$ it becomes $f^{'} (u)$ . The right fraction is "middle move per input move" — as $Δ x \to 0$ it becomes $g^{'} (x)$ . Their product is the total stretch.

WHY. This is the gear-ratio intuition made algebraic: total speed-up $=$ (outer gear ratio) $\times$ (inner gear ratio). It gives the correct answer $h^{'} (x) = f^{'} (u) g^{'} (x)$ .

PICTURE. Two gears meshed: small input gear (ratio $g^{'} (x)$ ) drives a big output gear (ratio $f^{'} (u)$ ); a caption reads "total $=$ product". A red warning tag hangs off the $Δ u$ in the denominator.

Step 5 — The hole: when the middle rail does NOT move

WHAT. The split in Step 4 divides by $Δ u$ . But $Δ u = g (x + Δ x) - g (x)$ can be exactly $0$ even when $Δ x \neq = 0$ . Two ways this happens:

Flat patch: if $g$ is constant on a stretch, every nudge gives $Δ u = 0$ .
Wiggly zero-crosser: $g (x) = x^{2} sin (1/ x)$ near $0$ hits $Δ u = 0$ infinitely often.

Dividing by $0$ is undefined, so the "one-line proof" is not yet a proof.

WHY. A theorem must hold in every case, not just the friendly ones. We can't wave away $Δ u = 0$ ; we must build machinery that survives it. That's the whole reason for Step 6.

PICTURE. Left panel: a flat plateau of $g$ — two dots at different $x$ but identical height, so $Δ u = 0$ . Right panel: the spiky $x^{2} sin (1/ x)$ crossing zero over and over near the origin, each crossing a place where $Δ u = 0$ .

Step 6 — The repair: an "error function" that multiplies instead of divides

WHAT. Because $f$ is differentiable at $u$ , define one bookkeeping function $E$ that measures how far the secant slope sits from the true tangent slope $f^{'} (u)$ :

E (k) = ⎩ ⎨ ⎧ \frac{f ( u + k ) - f ( u )}{k} - f^{'} (u), 0, k \neq = 0, k = 0.

Term by term: $k$ is a stand-in for "the outer machine's input nudge" (later we'll plug $k = Δ u$ ). The first branch is secant slope minus tangent slope — the error. The second branch patches the hole: we simply define the error to be $0$ when $k = 0$ .

Because $f^{'} (u)$ is precisely the limit of that secant slope, $lim_{k \to 0} E (k) = 0 = E (0)$ , so $E$ is continuous at $0$ (no jump). Now rearrange the first branch to a form valid for all $k$ (including $k = 0$ , where both sides are $0$ ):

f (u + k) - f (u) = (f^{'} (u) + E (k)) k .

WHY. Look: $k$ now sits on top, multiplied — never on the bottom. This is the magic. The identity holds even when $k = 0$ (it reads $0 = 0$ ). So we can safely substitute the troublesome $k = Δ u$ .

PICTURE. The tangent line (green, slope $f^{'} (u)$ ) and a secant (orange). The vertical gap between them, scaled per unit of $k$ , is exactly $E (k)$ ; as $k \to 0$ the gap collapses to zero — $E$ is continuous at $0$ .

Step 7 — Substitute, divide by $Δ x$ (legal!), take the limit

WHAT. Put $k = Δ u = g (x + Δ x) - g (x)$ into the multiply-form of Step 6:

f (g (x + Δ x)) - f (g (x)) = (f^{'} (u) + E (Δ u)) Δ u .

Now divide by $Δ x$ — allowed, since in the limit $Δ x \neq = 0$ :

\frac{h ( x + Δ x ) - h ( x )}{Δ x} = (f^{'} (u) + E (Δ u)) \cdot \frac{Δ u}{Δ x} .

Let $Δ x \to 0$ and track each piece:

$g$ differentiable $\Rightarrow$ $g$ continuous $\Rightarrow$ $Δ u \to 0$ , so $E (Δ u) \to E (0) = 0$ . (This is the differentiable ⇒ continuous fact.)
$\frac{Δ u}{Δ x} \to g^{'} (x)$ — the inner rate.

So the whole thing tends to $(f^{'} (u) + 0) g^{'} (x)$ .

WHY. We only ever divided by $Δ x$ (never by $Δ u$ ), so no step was illegal. The error term $E (Δ u)$ politely vanishes because $g$ drags $Δ u$ to $0$ . What survives is precisely the product of the two rates.

PICTURE. A collapsing bar chart: three bars for $f^{'} (u)$ , $E (Δ u)$ , $Δ u /Δ x$ ; as $Δ x \to 0$ the middle bar $E$ shrinks to nothing and the surviving product settles at $f^{'} (u) g^{'} (x)$ .

Step 8 — See it on a real example: $h (x) = sin (x^{2})$

WHAT. Inner $g (x) = x^{2}$ so $g^{'} (x) = 2 x$ ; outer $f (u) = sin u$ so $f^{'} (u) = cos u$ . Assemble:

h^{'} (x) = f^{'} (g (x)) cos (x^{2}) \cdot g^{'} (x) 2 x = 2 x cos (x^{2}) .

WHY. This makes the abstract product concrete and exposes the classic trap: it is $cos (x^{2})$ (inside kept intact), not $cos (2 x)$ . The inner rate $2 x$ is a separate multiplied factor, never smuggled inside the cosine.

PICTURE. Top: $h (x) = sin (x^{2})$ speeding up its oscillations as $x$ grows. Bottom: the computed $h^{'} (x) = 2 x cos (x^{2})$ overlaid on a numerical slope of $h$ — they land on top of each other, confirming the formula.

The one-picture summary

Everything on one canvas: input nudge $Δ x$ enters the inner gear (ratio $g^{'} (x)$ ), becomes the middle nudge $Δ u$ , enters the outer gear (ratio $f^{'} (u)$ ), becomes the output nudge $Δ y$ — and the total ratio is the product. The little error term $E$ that made the proof honest is drawn fading to zero.

Recall Feynman retelling — the whole walkthrough in plain words

A composite is two machines on a belt. Push the input a hair (Step 1's rate idea). That push reaches a middle rail $u = g (x)$ and slides it by $Δ u$ (Step 3). The middle push then reaches the output and slides it by $Δ y$ . To get "output-per-input" you route through the middle: (output-per-middle) × (middle-per-input) (Step 4) — gear ratios multiply. But there's a trap: the middle rail can stand still ( $Δ u = 0$ ) even when you pushed the input, and you can't divide by a still rail (Step 5). Fix: instead of dividing by $Δ u$ , write the outer machine's response as $(f^{'} (u) + E) Δ u$ where $E$ is a tiny "error" that we define to be $0$ when nothing moves (Step 6). Now $Δ u$ is multiplied, never in a denominator, so it's safe. Divide only by $Δ x$ , shrink it, and $E$ fades away because a differentiable $g$ can't jump — so $Δ u$ eases to zero too (Step 7). What's left is exactly $f^{'} (g (x)) g^{'} (x)$ . Test it on $sin (x^{2})$ : $cos (x^{2}) \cdot 2 x$ , keeping the inside intact (Step 8).

Connections

Derivative — limit definition — Step 1 is literally this; the whole proof is a limit of a difference quotient.
Continuity implies differentiability fails converse — Step 7 uses "differentiable $g$ ⇒ continuous $g$ " to force $Δ u \to 0$ .
Product rule and Quotient rule — sibling rules often stacked with the chain rule.
Implicit differentiation — the chain rule applied to $y (x)$ .
Related rates — the same $\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}$ multiplication in a time setting.
Inverse function derivative — chain-rule $f (f^{- 1} (x)) = x$ .