4.1.14 · D2Calculus I — Limits & Derivatives
Visual walkthrough — Product rule — proof
Before we start, three plain words we will lean on the whole way:
Everything below is a story about what happens to the rectangle's area when we push by .
Step 1 — Draw the area as a rectangle
WHAT. Put the two changing quantities on perpendicular sides. Width runs along the bottom, height runs up the side. The grey box they enclose has area — and that product is the function we want to differentiate.
WHY. Multiplication of two positive numbers is the area of a rectangle. That is not a metaphor — it is the definition of area. So if we want to understand how a product changes, the honest thing is to watch how a rectangle's area changes.
PICTURE.
The bottom edge is labelled , the left edge , and the shaded region is annotated . Nothing has moved yet — this is the "before" frame.
Step 2 — Nudge ; both sides stretch
WHAT. Advance to . The width does not stay put: it becomes . The height becomes . Draw the new, bigger rectangle on top of the old one so we can compare.
WHY. The derivative asks "what does a tiny push to do?" A tiny push to moves both and , because both depend on . This is the whole reason the answer is not just — two things move at once, and we must account for both.
PICTURE.
The original rectangle is the inner solid box; the new outline extends right by (blue arrow) and up by (pink arrow). The new area is .
Step 3 — Cut the growth into four honest pieces
WHAT. The new area minus the old area is the L-shaped rim we just added. Slice that rim into three tiles:
\;=\; \underbrace{v\,\Delta u}_{\text{right strip}} \;+\; \underbrace{u\,\Delta v}_{\text{top strip}} \;+\; \underbrace{\Delta u\,\Delta v}_{\text{corner}} $$ Reading the tiles right where they sit: - $v\,\Delta u$ — the **tall thin strip on the right**: it is $\Delta u$ wide and $v$ tall. - $u\,\Delta v$ — the **flat thin strip on top**: it is $u$ wide and $\Delta v$ tall. - $\Delta u\,\Delta v$ — the **little corner square** where the two strips overlap: $\Delta u$ by $\Delta v$. **WHY.** We want the *change* in area, $\Delta(uv)$. Expanding $(u+\Delta u)(v+\Delta v)$ and cancelling the $uv$ that was already there leaves exactly these three tiles. Geometry and algebra agree — the multiplied-out formula and the sliced picture are the same statement. **PICTURE.** ![[deepdives/dd-maths-4.1.14-d2-s03.png]] Each tile is coloured and labelled with its area. Notice the corner is genuinely tiny — a small number times a small number. > [!formula] The change in area, exactly > $$ \Delta(uv) = v\,\Delta u + u\,\Delta v + \Delta u\,\Delta v $$ > Every term is a real area on the picture — nothing has been approximated yet. --- ## Step 4 — Why the corner is negligible (the key idea) **WHAT.** Compare the three tiles as $h\to 0$. The strips shrink in **one** direction ($\Delta u$ or $\Delta v$ heads to $0$). The corner shrinks in **both** directions at once — it is a product of two things going to zero, so it shrinks *far faster*. **WHY.** Divide everything by $h$ (we are about to build a rate of change, and rate = change per unit $h$): $$ \frac{\Delta(uv)}{h} = v\,\underbrace{\frac{\Delta u}{h}}_{\to\, u'} + u\,\underbrace{\frac{\Delta v}{h}}_{\to\, v'} + \underbrace{\frac{\Delta u\,\Delta v}{h}}_{\to\, 0} $$ Look at the last term. Write it as $\dfrac{\Delta u}{h}\cdot \Delta v$. The first factor tends to the finite number $u'$; the second factor $\Delta v$ tends to **$0$** (a hair of $x$ makes a hair of height). Finite $\times$ zero $=0$. **The corner dies.** **PICTURE.** ![[deepdives/dd-maths-4.1.14-d2-s04.png]] Three bars show the sizes of the tiles as $h$ shrinks (left to right): the two strips settle to steady heights $v\,u'$ and $u\,v'$, while the corner bar collapses to the floor. > [!mistake] "But we *divided* the corner by $h$ — doesn't that keep it alive?" > **Why it feels risky:** dividing a tiny thing by a tiny $h$ could give something finite (like > $\Delta u/h \to u'$). > **The fix:** yes — but the corner has *two* tiny factors and only *one* $h$ underneath. After > $\Delta u/h$ uses up its tininess to become $u'$, there is still a leftover $\Delta v \to 0$ with > no $h$ to rescue it. So the corner truly vanishes. --- ## Step 5 — Do the same thing with the limit definition (the algebra twin) **WHAT.** The picture is convincing, but let us confirm it with the [[Limit definition of the derivative]]. By definition, $$ f'(x) = \lim_{h\to 0} \frac{u(x+h)\,v(x+h) - u(x)\,v(x)}{h} $$ Now the algebraic version of "slice into strips": add and subtract the middle term $u(x+h)\,v(x)$ (it equals zero net, so it is legal): $$ u(x{+}h)v(x{+}h) - u(x)v(x) = \underbrace{u(x{+}h)\big[v(x{+}h)-v(x)\big]}_{\text{top strip}} + \underbrace{v(x)\big[u(x{+}h)-u(x)\big]}_{\text{right strip}} $$ Term by term: $u(x{+}h)$ is the **new width**; $[v(x{+}h)-v(x)]=\Delta v$ is the **extra height** — their product is the top strip. And $v(x)$ is the **old height**; $[u(x{+}h)-u(x)]=\Delta u$ is the **extra width** — their product is the right strip. *This is Step 3 written in symbols.* **WHY.** The clever $\pm u(x+h)v(x)$ is exactly how you force one $\Delta v$ and one $\Delta u$ to appear so each becomes a difference quotient. Where did the corner go? It is hidden inside $u(x{+}h)\,\Delta v$ — because $u(x{+}h) = u + \Delta u$, that term secretly contains both $u\,\Delta v$ (top strip) **and** $\Delta u\,\Delta v$ (corner) at once. **PICTURE.** ![[deepdives/dd-maths-4.1.14-d2-s05.png]] The two coloured brackets in the picture point from the algebra terms to the matching tiles in the rectangle — a dictionary between symbols and shapes. --- ## Step 6 — Take the limit, piece by piece **WHAT.** Divide by $h$ and let $h\to 0$: $$ f'(x) = \lim_{h\to 0}\left[\, u(x{+}h)\,\frac{v(x{+}h)-v(x)}{h} \;+\; v(x)\,\frac{u(x{+}h)-u(x)}{h}\,\right] $$ Now read each limit: - $\dfrac{v(x{+}h)-v(x)}{h} \to v'(x)$ — the definition of $v'$. - $\dfrac{u(x{+}h)-u(x)}{h} \to u'(x)$ — the definition of $u'$. - $u(x{+}h) \to u(x)$ — the **new width slides back to the old width** as the nudge vanishes. - $v(x)$ does not contain $h$ at all, so it just sits there. **WHY that third limit needs care.** Saying $u(x{+}h)\to u(x)$ is exactly the statement that $u$ is **continuous** — no sudden jumps. We are allowed to say it *because* $u$ is differentiable, and [[Differentiability implies continuity]]. This is the quiet lemma the whole proof leans on. **PICTURE.** ![[deepdives/dd-maths-4.1.14-d2-s06.png]] The new-width marker (blue dot at $u(x{+}h)$) is shown creeping leftward onto the old-width marker (yellow dot at $u(x)$) as $h\to 0$ — continuity in one frame. Assembling the survivors: $$ f'(x) = \underbrace{u(x)}_{\text{width slid back}}\,\underbrace{v'(x)}_{\text{height's speed}} + \underbrace{v(x)}_{\text{old height}}\,\underbrace{u'(x)}_{\text{width's speed}} \qquad\blacksquare $$ which we tidy into the headline: $(uv)' = u'v + uv'$. --- ## Step 7 — Edge cases: the picture never breaks We promised every scenario. A rectangle argument seems to assume $u,v>0$ and both growing. What if they are not? **WHAT / WHY / PICTURE for each case:** - **$u'=0$ (width frozen).** The right strip $v\,\Delta u$ has zero width, so it contributes nothing; only the top strip survives: $(uv)' = u\,v'$. Correct — a constant width times a changing height grows only through the height. - **$v$ constant ($v'=0$).** Symmetric: $(uv)' = u'\,v$. This is the **constant-multiple rule**, $(c\,u)' = c\,u'$, falling out for free. - **A side shrinks ($\Delta u<0$).** The "strip" is *removed* from the right rather than added, so its signed area $v\,\Delta u$ is negative. The algebra of Step 5 never assumed a sign — $\Delta u$ can be any real number. Every conclusion holds unchanged. - **A negative value ($u<0$ or $v<0$).** Then $uv$ is a *signed* area, but the limit derivation in Steps 5–6 is pure algebra on real numbers and never used positivity. The rectangle is a **thinking aid**; the proof is airtight for all signs. - **$h<0$ (nudging backward).** The limit $h\to 0$ includes both sides. The same three quotients appear with $h$ negative and converge to the same $u',v'$. No case is missed. ![[deepdives/dd-maths-4.1.14-d2-s07.png]] Four mini-panels: frozen width, constant height, a shrinking side (strip subtracted), and a negative-valued (signed) rectangle — each landing on the same rule. --- ## The one-picture summary ![[deepdives/dd-maths-4.1.14-d2-s08.png]] One frame holds the entire argument: the growing rectangle, its three tiles labelled $v\,\Delta u$, $u\,\Delta v$, and the doomed corner $\Delta u\,\Delta v$, an arrow marked "$\div h$, then $h\to 0$", and the two surviving strips flowing into the final line $(uv)' = u'v + uv'$. > [!recall]- Feynman: tell the whole walkthrough in plain words > Picture a rectangle: so wide, so tall, area = width times height. Now nudge time forward a > whisker. The width gets a bit wider and the height gets a bit taller — both at once. That adds a > thin strip on the right (height × extra-width), a thin strip on top (width × extra-height), and a > tiny corner square where the strips overlap. To get a *rate*, divide the added area by the size of > the nudge and shrink the nudge to nothing. The two strips settle into steady rates — "height > times how fast width grows" plus "width times how fast height grows." The corner had two tiny > factors but only one nudge to divide by, so it starves and vanishes. What's left is the product > rule: $(uv)' = u'v + uv'$. And it doesn't care whether the sides grow, shrink, freeze, or go > negative — the algebra behind the picture never used a single sign. > [!mnemonic] Chant it: "**speed of first** × second **plus** first × **speed of second**." > $(uv)' = u'v + uv'$. --- ## Connections - [[Limit definition of the derivative]] — the engine behind Steps 5–6. - [[Differentiability implies continuity]] — why $u(x{+}h)\to u(x)$ in Step 6. - [[Power rule]] — pops out of the $v'=0$ edge case and the parent's induction. - [[Quotient rule]] — the same rectangle idea with a division twist. - [[Chain rule]] — its partner for compositions. - [[Leibniz rule (nth derivative of a product)]] — this walkthrough, iterated $n$ times. ## 🖼️ Concept Map ```mermaid flowchart TD R["Rectangle area A = u times v"] -->|"nudge x by h"| G["New area, both sides grow"] G --> RS["Right strip v times du"] G --> TS["Top strip u times dv"] G --> C["Corner du times dv"] RS -->|"divide by h"| L["Difference quotients"] TS -->|"divide by h"| L C -->|"two tinies, one h"| V["Corner vanishes"] L -->|"needs u continuous"| CO["Differentiable implies continuous"] L --> RULE["Product rule uv prime = u prime v + u v prime"] V --> RULE ```