Product rule — proof — Maths notes

WHAT — the statement

WHY the corner term vanishes (Dual Coding)

The picture shows area $u v$ as a rectangle. Increasing $x\to x+h$ makes $u\to u+\Delta u$ and $v\to v+\Delta v$ . The new area splits into 4 pieces:

Piece	Area
original	$uv$
right strip (wider)	$v\,\Delta u$
top strip (taller)	$u\,\Delta v$
corner (tiny)	$\Delta u\,\Delta v$

The change in area is $\Delta(uv) = v\,\Delta u + u\,\Delta v + \Delta u\,\Delta v$ . The corner is a product of two small things → second-order tiny → vanishes in the limit.

HOW — Derivation from scratch (the limit definition)

Formula Step-by-step from first principles

Start from the definition of the derivative of $f=uv$ : $f'(x) = \lim_{h\to 0}\frac{u(x+h)v(x+h) - u(x)v(x)}{h}$

Step 1 — Add and subtract a clever middle term. Insert $-\,u(x+h)v(x) + u(x+h)v(x) = 0$ in the numerator: $u(x{+}h)v(x{+}h) - u(x)v(x) = \underbrace{u(x{+}h)\big[v(x{+}h)-v(x)\big]}_{\text{height changes}} + \underbrace{v(x)\big[u(x{+}h)-u(x)\big]}_{\text{width changes}}$ Why this step? This is the algebra version of "split the strips". We force a $\Delta v$ to appear in one group and a $\Delta u$ in the other so each turns into a known derivative.

Step 2 — Divide by $h$ and split the limit. $f'(x) = \lim_{h\to 0}\left[ u(x{+}h)\,\frac{v(x{+}h)-v(x)}{h} + v(x)\,\frac{u(x{+}h)-u(x)}{h} \right]$ Why this step? Each fraction is exactly a difference quotient — the seed of a derivative.

Step 3 — Take limits piece by piece.

$\displaystyle\lim_{h\to0}\frac{v(x{+}h)-v(x)}{h}=v'(x)$ (definition).
$\displaystyle\lim_{h\to0}\frac{u(x{+}h)-u(x)}{h}=u'(x)$ (definition).
$\displaystyle\lim_{h\to0}u(x{+}h)=u(x)$ — this needs that $u$ is continuous at $x$ , which is guaranteed because $u$ is differentiable (differentiable ⇒ continuous).
$v(x)$ is constant in $h$ .

Putting them together: $f'(x) = u(x)\,v'(x) + v(x)\,u'(x) \qquad\blacksquare$

Worked examples

Flashcards

State the product rule.

(uv)' = u'v + uv'

What trick starts the limit proof of the product rule?

Add and subtract

u(x+h)v(x)

in the numerator.

Why does

\lim_{h\to0}u(x+h)=u(x)

hold in the proof?

Because

u

is differentiable, hence continuous.

Why is

(uv)'=u'v'

wrong (give a counterexample)?

With

u=v=x

(x^2)'=2x

but

u'v'=1

Geometrically, what does the product rule's "corner term"

\Delta u\,\Delta v

represent?

A second-order tiny area that vanishes in the limit.

Differentiate

x^2\sin x

2x\sin x + x^2\cos x

What's

(uvw)'

u'vw + uv'w + uvw'

Recall Feynman: explain to a 12-year-old

Imagine a rectangle of stickers — so many stickers wide, so many tall. Now both the width and the height start growing. After a tiny moment you've added a thin strip on the side and a thin strip on top. There's also a teeny corner square where the two strips overlap, but it's so small we ignore it. So how fast the total number of stickers grows = (how fast it widens)×height

width×(how fast it gets taller). That's the product rule!

Connections

Limit definition of the derivative — the engine of this proof.
Differentiability implies continuity — the hidden lemma in Step 3.
Quotient rule — derived from product rule + chain rule.
Chain rule — partners with this for compositions.
Power rule — bootstrapped from product rule (Example 2).
Leibniz rule (nth derivative of a product) — the grand generalisation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, product rule ka idea bilkul simple hai. Socho ek rectangle hai jiski width $u(x)$ hai aur height $v(x)$ hai — area to $u\cdot v$ hoga. Ab jab $x$ thoda sa change hota hai, to width bhi thodi badhti hai ( $\Delta u$ ) aur height bhi thodi badhti hai ( $\Delta v$ ). Naya area do patli strips add karta hai: ek side wali ( $v\,\Delta u$ ) aur ek upar wali ( $u\,\Delta v$ ). Ek chhota sa corner square bhi banta hai ( $\Delta u\,\Delta v$ ), lekin wo do choti cheezon ka product hai, itna chhota ki limit mein zero ho jaata hai. Isliye area ki growth rate banti hai $u'v + uv'$ .

Proof ka asli trick limit definition mein hai. Numerator $u(x+h)v(x+h)-u(x)v(x)$ ko hum $u(x+h)v(x)$ ko add-and-subtract karke do parts mein todte hain — ek part mein $\Delta v$ aata hai, doosre mein $\Delta u$ . Phir $h$ se divide karke limit lete hain, aur dono difference quotients $v'$ aur $u'$ ban jaate hain. Ek important point: $u(x+h)\to u(x)$ tabhi hota hai jab $u$ continuous hai — aur yeh free mein milta hai kyunki differentiable function hamesha continuous hota hai.

Sabse common galti yeh hai ki students soch lete hain $(uv)' = u'v'$ . Yeh galat hai! Addition mein derivative split hota hai, multiplication mein nahi. Test karo: $u=v=x$ , to $(x^2)'=2x$ , par $u'v'=1$ . Sahi answer $u'v+uv'=2x$ deta hai. Yaad rakhne ke liye chant karo: "first d-second plus second d-first". Bas har factor ko baari-baari differentiate karo, baaki ko constant rakho, aur sum kar do — chahe do factor ho ya teen.

Product rule — proof

WHAT — the statement

WHY the corner term vanishes (Dual Coding)

HOW — Derivation from scratch (the limit definition)

Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections