Product rule — proof: Question bank — Maths deep dive

Vocabulary this page uses (build it once, use it everywhere)

Before any trap, we fix the symbols so nothing appears un-defined.

True or false — justify

The product rule is $(uv)^{'} = u^{'} v + u v^{'}$ ; do not answer "true/false" alone — say why.

Every item below asks you to defend the verdict, not just state it.

(uv)^{'} = u^{'} v^{'}

for all differentiable

u, v

False — linearity respects addition only. Test

u = v = x

(x^{2})^{'} = 2 x

but

u^{'} v^{'} = 1

. The missing cross-terms

u^{'} v + u v^{'}

are exactly what makes products different from sums.

u

and

v

are both differentiable, then

uv

is automatically differentiable.

True — the proof constructs

h \to 0 lim \frac{uv ( x + h ) - uv ( x )}{h}

from the four pieces above and shows it exists, equal to

u^{'} v + u v^{'}

; existence of the limit is differentiability.

The product rule needs

u

and

v

to be continuous, but not differentiable.

False — it needs both differentiable; continuity alone leaves

u^{'}

and

v^{'}

undefined, so

u^{'} v + u v^{'}

is meaningless. (Differentiability does give continuity for free, see Differentiability implies continuity.)

u

is constant, the product rule collapses to

(c v)^{'} = c v^{'}

True — with

u = c

u^{'} = 0

, so

u^{'} v + u v^{'} = 0 \cdot v + c v^{'} = c v^{'}

. The constant-multiple rule is a special case of the product rule.

The corner term

Δ u Δ v

vanishes because

Δ u

and

Δ v

are literally zero.

False — they are small but nonzero; the term vanishes because

\frac{Δ u Δ v}{h} = Δ u \cdot \frac{Δ v}{h} \to 0 \cdot v^{'} = 0

(the spare

Δ u \to 0

kills it), a second-order effect, not an exact zero.

(uv)^{'} = u^{'} v + u v^{'}

still holds if

v (x) = 0

at the point of interest.

True — the rule makes no assumption that

u

v

be nonzero; at a point where

v = 0

it simply reads

(uv)^{'} = u^{'} \cdot 0 + u v^{'} = u v^{'}

The product rule applied to

u \cdot u = u^{2}

gives

(u^{2})^{'} = 2 u u^{'}

True — set

v = u

(uu)^{'} = u^{'} u + u u^{'} = 2 u u^{'}

. (The Chain rule

\frac{d}{d x} u^{2} = 2 u \cdot u^{'}

and the Power rule on

x^{2}

give the same number — they must, since all three describe one function — but here we derived it purely from the product rule.)

For three factors,

(uv w)^{'} = u^{'} v^{'} w^{'}

False — group

(uv) \cdot w

and apply twice:

(uv w)^{'} = u^{'} v w + u v^{'} w + uv w^{'}

. Each factor takes exactly one turn being differentiated while the rest stay put.

Spot the error

Each line contains a flawed statement or step; the reveal names the flaw.

"In the proof we write

lim_{h \to 0} u (x + h) = u (x)

with no justification — it's obvious."

The flaw is skipping why: this step uses that

u

is continuous at

x

, i.e.

Δ u \to 0

, which holds only because

u

is differentiable there. Naming this dependence is the whole subtle point of the proof.

"To prove the product rule, add and subtract

u (x) v (x)

in the numerator."

Wrong middle term — you must add/subtract

u (x + h) v (x)

(or

u (x) v (x + h)

). Using

u (x) v (x)

just re-inserts the term already there and produces no difference quotients

Δ u / h, Δ v / h

"Since

(u + v)^{'} = u^{'} + v^{'}

, by the same logic

(uv)^{'} = u^{'} v^{'}

The analogy is illegitimate: the sum rule follows from limits distributing over addition; multiplication does not distribute over the limit that way, and the counterexample

u = v = x

kills it instantly.

Δ u Δ v

is dropped because we're allowed to ignore small numbers."

Vague and wrong reasoning — we do not "ignore" it, we compute its limit:

\frac{Δ u Δ v}{h} = Δ u \cdot \frac{Δ v}{h} \to 0 \cdot v^{'} = 0

. Rigor, not hand-waving, sends it away.

(uv w)^{'} = u^{'} v w

because you differentiate the product like one big function."

Only one term of three — a product is not a single black box; each of the three factors must be differentiated in turn, giving

u^{'} v w + u v^{'} w + uv w^{'}

(x^{2} sin x)^{'} = 2 x cos x

Only the second half is right, and even it is misplaced. Correct:

2 x sin x + x^{2} cos x

. The writer differentiated both factors at once, the classic product-rule blunder.

"The proof splits the limit into two pieces before checking each piece's limit exists."

Splitting a limit into a sum/product is only valid because each piece's limit exists (and the product

u (x + h) \cdot \frac{Δ v}{h}

splits since both factors converge). Order matters: existence justifies the split.

Edge cases

v^{'} (x) = 0

at a point (v momentarily flat), what does the rule give?

(uv)^{'} = u^{'} v + u \cdot 0 = u^{'} v

— the height isn't stretching this instant, so all area growth comes from the widening strip.

At a point where both

u = 0

and

v = 0

, what is

(uv)^{'}

(uv)^{'} = u^{'} \cdot 0 + 0 \cdot v^{'} = 0

— even if

u^{'}, v^{'} \neq = 0

, the product's rate of change is zero there because both "sides" of the rectangle have length zero.

Does the product rule still hold if

u

is differentiable but

v

is only continuous (not differentiable) at

x

No — the formula

u^{'} v + u v^{'}

requires

v^{'} (x)

to exist. If

v

has a corner there,

uv

may or may not be differentiable, and the product rule does not apply.

What happens to

(uv)^{'}

as one factor tends to a constant, say

u \to c

u^{'} \to 0

, so

(uv)^{'} \to 0 \cdot v + c v^{'} = c v^{'}

— smoothly recovering the constant-multiple rule as a limiting/boundary case.

For a product of

n

factors, how many terms appear and why don't higher-order pieces survive?

Exactly

n

terms, each differentiating one factor:

\sum_{k = 1}^{n} f_{1} \dots f_{k}^{'} \dots f_{n}

. Expanding

\prod (f_{k} + Δ f_{k}) - \prod f_{k}

, every term is a product of some

Δ f_{k}

's; after dividing by

h

, a term with two or more

Δ

's reads (finite)

\times (Δ f_{j} \to 0)

, so it dies — only terms with a single

Δ

(one derivative, rest undisturbed) survive, exactly like the single corner in the two-factor case.

(uv)^{'}

defined at a point where

u

v

jumps (a discontinuity)?

No — a jump means that factor isn't even continuous, hence not differentiable, so neither the factor's derivative nor the product's derivative exists there.

Recall One-line summary of every trap

The product rule's cross-terms ( $u^{'} v + u v^{'}$ , never $u^{'} v^{'}$ ) come from two growing sides of a rectangle; each strip is a difference quotient $Δ u / h$ or $Δ v / h$ , while the corner $Δ u Δ v$ dies because one spare $Δ \to 0$ . The proof leans on differentiability ⇒ continuity; every extra factor adds one term (differentiate one factor at a time, higher- $Δ$ terms vanish), and it degenerates gracefully at zeros, flats, and constants.

Connections

Product rule — proof — the parent every trap here refers back to.
Limit definition of the derivative — where the difference quotients $Δ u / h$ live.
Differentiability implies continuity — the hidden lemma several traps target.
Quotient rule — a "why" item: product rule + chain rule.
Chain rule — needed to turn $v^{- 1}$ into the quotient rule.
Power rule — bootstrapped, not replaced, by the product rule.
Leibniz rule (nth derivative of a product) — the large- $n$ edge case generalised.

Question bank — Product rule — proof