Quotient rule — proof: Question bank — Maths deep dive

Throughout, we write the rule for $f = \frac{u}{v}$ (top $u$ over bottom $v$ , with $v (x) \neq = 0$ ) as $f^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}} .$ Here $u^{'}$ means "how fast the top changes" and $v^{'}$ means "how fast the bottom changes" — nothing more exotic than that.

The proof this bank interrogates has five moves; the "Spot the error" items refer to them by name:

Recall The five proof steps (reference for the traps)

Step 1 — write $f (x + h) - f (x) = \frac{u ( x + h )}{v ( x + h )} - \frac{u ( x )}{v ( x )}$ .
Step 2 — combine over a common denominator: top $= u (x + h) v (x) - u (x) v (x + h)$ , bottom $= v (x + h) v (x)$ .
Step 3 — add-and-subtract zero ( $- u (x) v (x) + u (x) v (x)$ ) inside the top.
Step 4 — factor and divide by $h$ , exposing the difference quotients for $u$ and $v$ .
Step 5 — let $h \to 0$ (using continuity of $v$ ) to reach the formula.

True or false — justify

The quotient rule is just the product rule applied to

u

and

1/ v

True — writing

\frac{u}{v} = u \cdot v^{- 1}

and using the product rule plus

\frac{d}{d x} v^{- 1} = - v^{'} / v^{2}

reproduces the same formula. See Product rule — proof and Chain rule.

Swapping the two numerator terms,

\frac{u v ^{'} - u ^{'} v}{v ^{2}}

, gives the same answer because subtraction of two things "looks symmetric".

False — it flips the sign of the entire derivative, turning increasing into decreasing. The order Lo dHi first is essential.

v (x) = 0

at some point, the quotient rule still applies there as long as

u

and

v

are differentiable.

False — the formula divides by

v^{2}

, so

v (x) \neq = 0

is required; at a zero of

v

the function

u / v

typically isn't even defined.

Applying the quotient rule requires you to separately assume

v

is continuous, on top of assuming it is differentiable.

False — you never assume continuity separately, because differentiability already guarantees continuity; the proof simply uses that free continuity in the

h \to 0

step. See Continuity.

You can derive

\frac{d}{d x} tan x = sec^{2} x

without ever knowing

cos^{2} x + sin^{2} x = 1

False — the Pythagorean identity is what collapses the numerator

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

1

; without it you only reach

\frac{1}{c o s ^{2} x}

by that identity anyway.

The reciprocal rule

\frac{d}{d x} \frac{1}{v} = - \frac{v ^{'}}{v ^{2}}

is a separate theorem you must memorise on its own.

False — it is the quotient rule with

u = 1

u^{'} = 0

; the rule already contains it as a special case.

The "add and subtract zero" step changes the value of the numerator.

False — it adds

- u (x) v (x) + u (x) v (x) = 0

, so the value is untouched; only the grouping changes, revealing two difference quotients.

If both

u

and

v

are constants, the quotient rule correctly gives

f^{'} = 0

True —

u^{'} = v^{'} = 0

makes the numerator

0 \cdot v - u \cdot 0 = 0

, matching that a constant ratio has zero slope.

Spot the error

(\frac{u}{v})^{'} = \frac{u ^{'}}{v ^{'}}

— differentiate top and bottom."

Division is not term-wise. Test

f = \frac{x ^{2}}{x} = x

: truth is

1

, but

\frac{2 x}{1} = 2 x

. The shortcut is simply false.

f^{'} = \frac{u ^{'} v - u v ^{'}}{v}

— the bottom keeps a single

v

like the original fraction."

Wrong — the denominator must be

v^{2}

. It arises from the common-denominator product

v (x + h) v (x) \to v^{2}

in Step 2 of the proof.

"In Step 2 I combined over a common denominator and got

\frac{u ( x + h ) v ( x + h ) - u ( x ) v ( x )}{v ( x + h ) v ( x )}

The cross-multiplication is wrong: Step 2 should give

u (x + h) v (x) - u (x) v (x + h)

on top. Each numerator term pairs the opposite denominator (recall

h

is the small step:

x + h

x

"Since

v

is differentiable, in the limit

v (x + h) \to v (x + h)

, so the bottom becomes

v (x + h)^{2}

h \to 0

v (x + h) \to v (x)

, not

v (x + h)

; the bottom tends to

v (x)^{2}

. Letting

h \to 0

must be applied everywhere.

\frac{d}{d x} tan x = \frac{c o s x c o s x - s i n x s i n x}{c o s ^{2} x}

Sign error:

v^{'} = - sin x

, so the second term is

- sin x \cdot (- sin x) = + sin^{2} x

. The correct numerator is

cos^{2} x + sin^{2} x = 1

, giving

sec^{2} x

"For

\frac{1}{v}

I used the quotient rule and got

\frac{0 \cdot v - 1 \cdot v ^{'}}{v} = - \frac{v ^{'}}{v}

The denominator must be squared:

- \frac{v ^{'}}{v ^{2}}

. Forgetting the square is the classic slip.

"The proof factors the numerator as

u (x) [u (x + h) - u (x)] - v (x) [v (x + h) - v (x)]

The factors are swapped: Step 4 gives

v (x) [u (x + h) - u (x)] - u (x) [v (x + h) - v (x)]

. The constant-in-the-limit factor multiplying each change is the other function.

Why questions

Why is there a minus sign, in plain physical terms?

Because a fraction shrinks when its bottom grows; so

v

increasing (

v^{'} > 0

) must push $f^{'}$ down, which is exactly the subtracted

- u v^{'}

term.

Why do we divide by

v^{2}

specifically, and not

v

v^{3}

Differentiating

1/ v

by the Chain rule gives

- v^{'} / v^{2}

; one power of

v

stays from the original fraction, one more comes from the chain-rule derivative of

v^{- 1}

Why does the proof start from the Limit definition of derivative instead of just quoting the product rule?

To prove the rule from first principles we need an argument that doesn't assume other differentiation rules; the limit definition is the common engine.

Why must we insert exactly

- u (x) v (x) + u (x) v (x)

, and not some other zero?

This particular pair lets us factor out

v (x)

from one bracket and

u (x)

from the other, manufacturing the difference quotients that become

u^{'}

and

v^{'}

Why does continuity of

v

(and not of

u

) get singled out in the last step?

Only

v

appears in the denominator

v (x + h) v (x)

; we need

v (x + h) \to v (x)

so that bottom tends to a nonzero

v (x)^{2}

. See Continuity.

Why can we pull

u^{'}

and

v^{'}

out as separate limits at the end?

Each difference quotient converges (both functions are differentiable) and the surrounding factors converge to finite numbers, so the limit of the product/sum is the product/sum of the limits.

Why is the mnemonic order "Lo dHi minus Hi dLo" and not "Hi dLo minus Lo dHi"?

Because the correct sign has

u^{'} v

(Lo·dHi) positive; reversing it negates the whole derivative, giving increasing/decreasing backwards.

Edge cases

What happens to the quotient rule when

u = v

, i.e.

f = \frac{v}{v} = 1

The numerator becomes

v^{'} v - v v^{'} = 0

, so

f^{'} = 0

— correctly matching that the constant

1

has zero derivative wherever

v \neq = 0

At a point where

v^{'} (x) = 0

but

u^{'} (x) \neq = 0

, does

f^{'}

have to be zero?

No — the formula gives

f^{'} = \frac{u ^{'} v - 0}{v ^{2}} = \frac{u ^{'}}{v}

, generally nonzero; a flat denominator doesn't flatten the fraction.

What is

f^{'}

where the numerator

u^{'} v - u v^{'}

vanishes but

v \neq = 0

There

f^{'} = 0

: a critical point (horizontal tangent) of the ratio, e.g.

x = 0, - 2

\frac{x ^{2}}{x + 1}

v

has a zero of odd order (sign change) at

x = a

, what happens near there?

f = u / v

blows up with a vertical asymptote and the fraction flips sign across

a

(opposite signs on the two sides); the rule simply doesn't apply at

a

since

v (a) = 0

v

has a zero of even order (no sign change) at

x = a

, like

v = (x - a)^{2}

, how does that differ?

Still a vertical asymptote, but

v > 0

on both sides so

f

blows up to the same sign either side (both

+ \infty

or both

- \infty

); again the rule fails at

a

because

v (a) = 0

Can the quotient rule produce a wrong answer if

u / v

can be simplified first (like

\frac{x ^{2}}{x} = x

No — it gives the same correct result off the removable point, but at the cancelled point (

x = 0

) the original

u / v

is undefined, so agreement holds only where the ratio is defined.

What does the rule give when

u^{'} = 0

(constant top)?

It reduces to

- \frac{u v ^{'}}{v ^{2}}

, the reciprocal rule scaled by the constant

u

— a built-in sanity check.

Does the rule care whether

u

and

v

are trig, polynomial, or exponential functions?

No — it only uses that both are differentiable at

x

with

v (x) \neq = 0

; the specific formulas for

u^{'}, v^{'}

come from elsewhere, e.g. Derivatives of trig functions.

Recall Self-test after the bank

Without looking: (1) why $v^{2}$ and not $v$ , (2) why the minus, (3) what zero we add, (4) why continuity of $v$ matters. Any miss → revisit that trap.

Connections

Quotient rule — proof — the parent this bank interrogates.
Product rule — proof — the twin "add-zero" trick.
Chain rule — source of the $- v^{'} / v^{2}$ and the $v^{2}$ denominator.
Limit definition of derivative — the engine behind every "why" here.
Continuity — the quiet hero of the last limit step.
Derivatives of trig functions — where $tan, sec, csc$ traps come from.