Visual walkthrough — Quotient rule — proof

Before we start, one promise: every letter is defined the first time it appears. Let us begin from what a "function" and a "slope" even are.

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Step 0 — What are we even asking? (the setup picture)

WHAT. We have two machines. Machine $u$ takes a number $x$ and outputs a number $u(x)$. Machine $v$ takes the same $x$ and outputs $v(x)$. We build a ratio machine $f(x)=\frac{u(x)}{v(x)}.$ Here the slash means ordinary division: "$u(x)$ shared among $v(x)$ parts". We require $v(x)\ne 0$ — you cannot share among zero parts.

WHY. We want the slope of $f$ — how fast its output changes when we nudge the input. The parent note calls this the derivative $f^{\prime }(x)$.

WHAT IT LOOKS LIKE. A slope is the steepness of a curve. If you stand on the graph of $f$ and take a tiny step to the right, the slope is how much you rise divided by how much you stepped.

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Step 1 — Write the rise as a difference of two fractions

WHAT. Put $f=u/v$ straight into the rise $f(x+h)-f(x)$: $f(x+h)-f(x)=\underset{\text{new value}}{\underbrace{\frac{u(x+h)}{v(x+h)}}}-\underset{\text{old value}}{\underbrace{\frac{u(x)}{v(x)}}}.$

WHY. This is literally the definition — nothing clever yet. Everything after this is just re-shaping until the limit becomes computable.

WHAT IT LOOKS LIKE. Two fractions with different denominators ($v(x+h)$ vs $v(x)$). Different bottoms means we cannot combine them yet — like $\frac{1}{2}-\frac{1}{3}$, we first need a common denominator.

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Step 2 — Combine over a common denominator

WHAT. Multiply each fraction so both sit over the same bottom $v(x+h)v(x)$: $=\frac{\stackrel{\text{new top}\times \text{old bottom}}{\overbrace{u(x+h)v(x)}}-\stackrel{\text{old top}\times \text{new bottom}}{\overbrace{u(x)v(x+h)}}}{\underset{\text{product of both bottoms}}{\underbrace{v(x+h)v(x)}}}.$

WHY. One single fraction is far easier to handle. Crucially, the denominator is now $v(x+h)v(x)$ — keep an eye on it, because when $h\to 0$ the two bottoms become equal and this turns into $v^2$. That is where the squared denominator is born.

WHAT IT LOOKS LIKE. A rectangle-area picture: to cross-multiply $\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$, we tile each fraction onto the shared grid $b\times d$.

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Step 3 — The "add zero" trick

WHAT. Look only at the numerator $u(x+h)v(x)-u(x)v(x+h)$. Slip in $-u(x)v(x)+u(x)v(x)$, which equals $0$ so it changes nothing: $u(x+h)v(x)\underset{=0,\text{added for free}}{\underbrace{-u(x)v(x)+u(x)v(x)}}-u(x)v(x+h).$ Now group the pieces: $=\underset{\text{how }u\text{ changed}}{\underbrace{[u(x+h)-u(x)]}}v(x)-u(x)\underset{\text{how }v\text{ changed}}{\underbrace{[v(x+h)-v(x)]}}.$

WHY. We want the differences $u(x+h)-u(x)$ and $v(x+h)-v(x)$ to appear, because those are the raw ingredients of $u^{\prime }$ and $v^{\prime }$. The add-zero trick manufactures them out of thin air. (This is the same manoeuvre used in the Product rule — proof.)

WHAT IT LOOKS LIKE. A bar splits into two coloured pieces: a magenta piece carrying "$u$ moved, weighted by the old bottom $v(x)$" and a violet piece carrying "$v$ moved, weighted by the old top $u(x)$" — with a minus already sitting in front of the violet piece.

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Step 4 — Divide by $h$ to expose two slopes

WHAT. The whole rise is now over $v(x+h)v(x)$; divide everything by $h$ as the definition demands: $\frac{f(x+h)-f(x)}{h}=\frac{1}{v(x+h)v(x)}\left[v(x)\underset{\to u^{\prime }(x)}{\underbrace{\frac{u(x+h)-u(x)}{h}}}-u(x)\underset{\to v^{\prime }(x)}{\underbrace{\frac{v(x+h)-v(x)}{h}}}\right].$

WHY. Dividing each bracketed difference by $h$ turns it into a difference quotient — the exact shape of the limit definition. So each bracket is secretly a derivative waiting to be revealed.

WHAT IT LOOKS LIKE. Each coloured piece from Step 4 becomes a little right-triangle slope: rise $u(x+h)-u(x)$ over run $h$ (magenta), and rise $v(x+h)-v(x)$ over run $h$ (violet).

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Step 5 — Take the limit (where continuity does its job)

WHAT. Let $h\to 0$. Four things happen simultaneously: $\frac{u(x+h)-u(x)}{h}\to u^{\prime }(x),\frac{v(x+h)-v(x)}{h}\to v^{\prime }(x),v(x+h)\to v(x).$ The third one uses that $v$ is differentiable, hence continuous: a hair-step in $x$ makes only a hair-step in $v$. So the denominator $v(x+h)v(x)\to v(x)v(x)=(v(x){)}^2$. Assemble: $f^{\prime }(x)=\frac{v(x)u^{\prime }(x)-u(x)v^{\prime }(x)}{(v(x){)}^2}=\frac{u^{\prime }(x)v(x)-u(x)v^{\prime }(x)}{(v(x){)}^2}.■$

WHY continuity is essential. If $v$ jumped (was not continuous), $v(x+h)$ would not approach $v(x)$, the denominator would not become $v^2$, and the proof would collapse.

WHAT IT LOOKS LIKE. The two shrinking triangles freeze into exact tangent slopes $u^{\prime }(x)$ and $v^{\prime }(x)$, and the two nearby bottoms $v(x+h),v(x)$ slide together into one value $v(x)$, squared.

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Step 6 — Degenerate case: constant numerator ($u=1$)

WHAT. Set $u(x)=1$, a flat machine, so $u^{\prime }(x)=0$. The rule collapses: $\frac{d}{dx}\frac{1}{v}=\frac{\stackrel{u^{\prime }v}{\overbrace{0\cdot v}}-1\cdot v^{\prime }}{v^2}=-\frac{v^{\prime }}{v^2}.$

WHY. This is the reciprocal rule hiding inside. A good sanity check: the quotient rule must contain it. It also confirms the minus — nothing on top can grow, so the only thing moving the fraction is the bottom, always pushing the opposite way.

WHAT IT LOOKS LIKE. With a flat top, $y=1/v$ falls when $v$ rises and rises when $v$ falls — the curve is a mirror-flip of $v$'s motion.

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The one-picture summary

Read it left to right: the limit definition feeds in $f=u/v$; common denominator plants the future $v^2$; add-zero splits the numerator into a magenta "$u$-moved" piece and a violet "$v$-moved" piece (minus in front); dividing by $h$ turns both into slopes; the limit freezes them into $u^{\prime }v-u{v}^{\prime }$ over $v^2$.

Recall Feynman retelling — the whole walkthrough in plain words

You are cutting a pizza. Top $u$ = slices you own, bottom $v$ = friends sharing. You want to know how each person's share changes when things shift a tiny bit. First (Step 1–2) you write "new share minus old share" and put both over one common crowd-size. Then (Step 3) the magic: you add and subtract the exact same amount, which is like adding zero — free, harmless — but it lets you separate "slices changed" from "crowd changed". Divide by the tiny time-step (Step 4) and each of those becomes a rate: how fast slices grow ($u^{\prime }$), how fast the crowd grows ($v^{\prime }$). Let the step shrink (Step 5): the crowd barely moves in an instant (that's continuity), so the two crowd sizes merge into one, squared. Final law: share-change = (crowd × slice-growth − slices × crowd-growth) ÷ crowd². More friends (bottom grows) shrinks everyone's share — that's the minus. And the bigger the crowd already, the less one more matters — that's the squared bottom.

Connections

• Limit definition of derivative — the engine of every step.
• Product rule — proof — same add-zero trick.
• Chain rule — gives $\frac{d}{dx}{v}^{-1}=-v^{\prime }/v^2$ directly.
• Continuity — makes $v(x+h)\to v(x)$ in Step 5.
• Derivatives of trig functions — the $\tan x$ payoff.