Derivatives and rules (product, quotient, chain): Visual walkthrough

Step 1 — What "slope" even means (a picture, not a formula)

WHAT. A function $f$ is a machine: feed it a number $x$ , it hands back a number $f (x)$ . Draw $x$ on the horizontal line, $f (x)$ on the vertical line, and mark the dot. As $x$ slides, the dot traces a curve.

WHY. Before any rule, we need one honest question: if I push $x$ right by a tiny step, how far does the dot rise? That "rise per push" is the entire subject.

PICTURE. In the figure, the small horizontal push is called $h$ (the width of our step). The matching vertical jump is $f (x + h) - f (x)$ (the height gained). Their ratio is the steepness of the little staircase step.

This is only the average over a window of width $h$ . To get the slope at the point we shrink the window. See Limits and Continuity for what "shrink to zero" formally means.

Step 2 — Shrinking the step to a single point

WHAT. Slide the right edge of the step toward the left edge: let $h$ get smaller and smaller.

WHY. A wide step averages over a stretch of curve. We want the steepness exactly here, at one dot. So we let the run $h \to 0$ and read off where the ratio settles.

PICTURE. As the step narrows, the little staircase line pivots until it just kisses the curve — that kissing line is the tangent. Its slope is the derivative.

Everything below is just this one idea, applied to a machine made of two machines.

Step 3 — Stacking two machines (composition)

WHAT. Now build a bigger machine by feeding one into another. Call the inner machine $g$ and the outer machine $f$ . Input $x$ goes into $g$ , out comes a middle number we name $u = g (x)$ , and $u$ goes into $f$ , out comes $y = f (u)$ .

WHY. Real formulas are nested: $(3 x^{2} + 1)^{5}$ is "square-and-add-one" wrapped inside "raise-to-the-fifth"; the sigmoid $σ (x) = \frac{1}{1 + e ^{- x}}$ is a stack too. We need the slope of the whole stack, not the pieces. This nesting is exactly a neural-network layer feeding the next — see Backpropagation.

PICTURE. Three number lines in a row: the $x$ -line, the middle $u$ -line, and the $y$ -line, with arrows showing $x g u f y$ .

Step 4 — Push the input, watch the ripple

WHAT. Nudge $x$ by a tiny $h$ . This forces $u$ to move by some small amount we call $Δ u$ (Greek "delta u" = "a small change in $u$ "). That $Δ u$ in turn forces $y$ to move by $Δ y$ .

WHY. A change can only reach $y$ by first passing through $u$ . So the effect travels in two hops: $x \to u$ , then $u \to y$ . If we understand each hop's steepness, we can chain them.

PICTURE. The same three-line diagram, now with a small red bracket $h$ on the $x$ -line, a blue bracket $Δ u$ on the middle line, and a pink bracket $Δ y$ on the $y$ -line. The push propagates left to right.

Step 5 — Why the rates MULTIPLY (the gear picture)

WHAT. We assemble the total steepness $\frac{Δ y}{Δ x}$ from the two hop-steepnesses.

WHY. Watch a purely algebraic magic trick: the $Δ u$ that comes out of hop 1 is the very same $Δ u$ that goes into hop 2. So we can wedge it in and cancel it:

$\frac{Δ y}{Δ x} = \frac{Δ y}{Δ u} \cdot \frac{Δ u}{Δ x}$

$\frac{Δ y}{Δ x}$ ::: total steepness of the whole stack.
$\frac{Δ y}{Δ u}$ ::: how strongly the outer machine amplifies a wiggle (hop 2 gear ratio).
$\frac{Δ u}{Δ x}$ ::: how strongly the inner machine amplifies a wiggle (hop 1 gear ratio).
the $Δ u$ ::: appears once on top, once on bottom — it cancels, which is why the equation is even allowed.

Why multiply and not add? Gears. If a small gear turns a middle gear 3× as fast, and the middle turns a big gear 2× as fast, the big gear spins $2 \times 3 = 6$ × as fast — you compound the ratios, you don't add them. Adding would say " $5 \times$ ", which is plainly wrong: nothing added, everything passed through and got amplified twice.

PICTURE. Three interlocking gears labelled with ratios $\frac{Δ u}{Δ x} = 3$ and $\frac{Δ y}{Δ u} = 2$ , the far gear spinning at $6$ .

Step 6 — Take the limit: the Chain Rule

WHAT. Shrink the push $h \to 0$ . Each hop-ratio settles into a genuine derivative.

WHY. By Step 2, $\frac{Δ u}{Δ x} \to g^{'} (x)$ and $\frac{Δ y}{Δ u} \to f^{'} (u) = f^{'} (g (x))$ . The product of two settling numbers settles to the product of the limits.

PICTURE. The gear diagram with each ratio replaced by its derivative label, and a banner equation underneath.

Step 7 — Watch it run: $(3 x^{2} + 1)^{5}$ term by term

WHAT. Outer $f (u) = u^{5}$ , inner $u = g (x) = 3 x^{2} + 1$ .

WHY. This is the smallest example where forgetting the inner slope changes the answer dramatically — a stress test for the rule.

$\frac{d y}{d x} = f^{'} (u) 5 u^{4} \cdot g^{'} (x) 6 x = 5 (3 x^{2} + 1)^{4} \cdot 6 x = 30 x (3 x^{2} + 1)^{4}$

$5 u^{4}$ ::: outer slope — power rule on $u^{5}$ ; see Power Rule and Basic Derivatives.
then $u = 3 x^{2} + 1$ ::: put the middle value back so the outer slope is measured where we are.
$6 x$ ::: inner slope — power rule on $3 x^{2} + 1$ .
product ::: the two gear ratios compounded.

PICTURE. The stack drawn concretely with both slopes labelled on their own hop.

Step 8 — The degenerate cases (so you never get surprised)

WHAT. Cover the boundary situations the smooth formula quietly handles.

WHY. The contract: the reader must never hit a scenario we didn't show. Three matter.

Case	What happens	Why it's fine
$g^{'} (x) = 0$ (inner is momentarily flat)	Whole product $= 0$	The first gear isn't turning, so nothing downstream moves — $\frac{d y}{d x} = f^{'} (g (x)) \cdot 0 = 0$ .
$f^{'} (g (x)) = 0$ (outer flat at the landed spot)	Whole product $= 0$	The middle gear turns but the last gear is at a flat plateau — motion arrives but produces no output rise. This is the vanishing-gradient seed in Backpropagation.
Constant inner, $g (x) = c$	$g^{'} (x) = 0$ , so $\frac{d y}{d x} = 0$	If the middle value never changes, the outer machine's output never changes either.

PICTURE. Two mini-curves: left, a flat inner (gear stuck); right, a flat outer plateau (last gear on a flat).

The one-picture summary

Everything collapses to one image: a chain of gears. Push $x$ , the wiggle rides through each machine, and every machine multiplies it by its local slope. Read left to right, multiply as you go.

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} (rates compound, they never add)$

Recall Feynman retelling — the whole walkthrough in plain words

A derivative is just how steep — push the input a hair, see how much the output climbs. To find the steepness at a single spot, we shrink the push until our little step-line pivots into the tangent that kisses the curve. Now stack two machines. A push on $x$ can only reach the answer by first passing through a middle value $u$ . So it takes two hops. Hop one has a steepness; hop two has a steepness. Because the same middle wiggle leaves hop one and enters hop two, we can cancel it — and what's left is: total steepness = (second steepness) × (first steepness). Why times and not plus? Gears. A small gear spinning a big gear spinning a huge gear — the huge one's speed is all the ratios multiplied. Add nothing; pass through and amplify twice. Edge cases: if any gear is stuck (a slope is zero), the whole chain's slope is zero — nothing gets through. In a deep neural net you have hundreds of these gears in a row, and many slopes are small; multiply many small numbers and you get almost nothing. That vanishing product is the deep-learning headache the whole chain-rule story predicts.

Connections

Parent: Derivatives and rules
Limits and Continuity — the "shrink $h$ to zero" that makes slope exact.
Power Rule and Basic Derivatives — the outer/inner slopes we plugged in.
Gradient and Partial Derivatives — the multi-input version of this same push-and-watch idea.
Backpropagation — the chain rule run backwards through many layers.
Gradient Descent / Activation Functions — why multiplied small slopes make deep training hard.