Derivatives and rules (product, quotient, chain): Question bank

A quick reminder of the three tools, so every symbol below is earned:

A derivative is the local slope: nudge the input, measure how fast the output moves.
Product rule: $(uv)^{'} = u^{'} v + u v^{'}$ — imagine a rectangle of sides $u, v$ ; growing each side adds one area strip.
Quotient rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ — same as the product rule solved backwards.
Chain rule: for $f (g (x))$ , $f^{'} (g (x)) \cdot g^{'} (x)$ — nested functions are like gears, so their rates multiply.
Sigmoid: $σ (x) = \frac{1}{1 + e ^{- x}}$ , an S-shaped function used in ML; its derivative is $σ (x) (1 - σ (x))$ . We use it several times below.

True or false — justify

The derivative of a product equals the product of the derivatives.

False. Linearity only works for addition/scaling; multiplication needs two terms

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

. Test

u = v = x

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

, but

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

u

and

v

are both increasing, then

(uv)^{'}

must be positive.

False.

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

is positive only when both functions are also positive; if

v < 0

the term

u^{'} v

can be negative and drag the sum below zero, so "both increasing" is not enough.

The chain rule and the product rule are the same underlying idea.

False. The chain rule multiplies rates through a composition (nesting), while the product rule adds strips from a multiplication. Different constructions, different formulas.

The quotient rule is a brand-new rule you must memorise separately.

False. Writing

u = q v

and applying the product rule, then solving for

q^{'}

, gives the quotient rule — it's the product rule in disguise.

\frac{d}{d x} (uv)^{2} = 2 (uv) (uv)^{'}

requires the chain rule.

True. The outer function is "square" and the inner is the product

uv

; chain rule handles the square, and

(uv)^{'}

still needs the product rule inside.

Swapping

u

and

v

in the product rule changes the answer.

False.

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

is symmetric — renaming gives

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

, the same sum. (The quotient rule is not symmetric because division isn't.)

Swapping numerator and denominator in the quotient rule just flips the sign.

False. It gives a different function entirely,

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

, with a different denominator

u^{2}

, not a sign flip of the original.

f (g (x))

has

g^{'} (x) = 0

at a point, then

f^{'}

of the whole thing is

0

there.

True. By the chain rule the total rate is

f^{'} (g) \cdot g^{'}

; a zero gear ratio

g^{'} = 0

freezes the output no matter how steep the outer

f^{'} (g)

is.

The sigmoid derivative

σ (1 - σ)

can be negative.

False. Since

σ (x) = \frac{1}{1 + e ^{- x}} \in (0, 1)

, both

σ

and

1 - σ

are positive; the product is always positive — the sigmoid is always increasing.

Spot the error

\frac{d}{d x} [x^{2} sin x] = 2 x cos x

Wrong — this treats it like a chain rule and only differentiates each piece once. It's a product:

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

\frac{d}{d x} (3 x^{2} + 1)^{5} = 5 (3 x^{2} + 1)^{4}

Missing the inner derivative. The inside also moves, so multiply by

\frac{d}{d x} (3 x^{2} + 1) = 6 x

, giving

30 x (3 x^{2} + 1)^{4}

(\frac{u}{v})^{'} = \frac{u v ^{'} - u ^{'} v}{v ^{2}}

Sign flipped. The first numerator term must carry

u^{'}

(the top's own rate): correct is

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

. Check

\frac{x}{1} = x

: right rule gives

1

, this gives

- 1

\frac{d}{d x} e^{- x} = e^{- x}

Forgot the chain rule on the exponent. Inner is

- x

with derivative

- 1

, so

\frac{d}{d x} e^{- x} = - e^{- x}

"For

\frac{u}{v}

, the denominator of the derivative is

v

It's

v^{2}

. Dividing by

v

once for the fraction and once for the sensitivity squares the scale — the

v^{2}

is not optional.

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

No — extend the product rule term by term:

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

, one strip per factor. Products of derivatives is never how it works.

"Since

\frac{d}{d x} ln x = \frac{1}{x}

, we have

\frac{d}{d x} ln (5 x) = \frac{1}{5 x}

Chain rule needed: inner

5 x

has derivative

5

, so

\frac{1}{5 x} \cdot 5 = \frac{1}{x}

. (It happens to simplify, but the reasoning must include the inner rate.)

Why questions

Why does the product rule have exactly two terms, not one and not three?

A rectangle of sides

u, v

grows on two sides; each growing side adds one area strip. The tiny corner square is second-order (

Δ u Δ v

) and vanishes in the limit.

Why is there a minus sign in the quotient rule?

When the denominator

v

grows, the fraction shrinks — that opposing effect on the

u v^{'}

term shows up as a subtraction.

Why do rates multiply in the chain rule instead of adding?

Nested functions act like gears driving gears: if

y

moves 2× as fast as

u

and

u

moves 3× as fast as

x

, then

y

moves

2 \times 3 = 6

× as fast as

x

. Speed-ups compound.

Why is the chain rule "exactly backpropagation"?

A neural net is a composition of layers; the loss's sensitivity to an early weight is the product of each layer's local derivative — the chain rule applied layer by layer.

Why does the sigmoid have the tidy derivative

σ (1 - σ)

Because

\frac{e ^{- x}}{1 + e ^{- x}} = 1 - σ

, so the chain-rule result

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

collapses to

σ (1 - σ)

— a self-referential form that's cheap to compute during training.

Why can't we differentiate

(3 x^{2} + 1)^{5}

with the product rule directly?

It's a composition (a power of an inner function), not a product of two separate factors. Expanding it into a product first would work but is far more painful than one chain step.

Why must a symbol like

tan

e^{- x}

be broken into outer/inner before differentiating?

The chain rule only sees "outer shell times inner rate"; identifying which part is the shell and which is nested tells you what to differentiate and what to multiply by.

Edge cases

What is

(uv)^{'}

when

v

is a constant

c

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

. The product rule reduces to the constant-multiple rule — consistent, not a special case.

What happens to the quotient rule when the numerator

u

is constant?

u^{'} = 0

, so

(\frac{c}{v})^{'} = \frac{0 \cdot v - c v ^{'}}{v ^{2}} = - \frac{c v ^{'}}{v ^{2}}

. Only the denominator's motion matters, and it opposes the fraction — hence the minus.

Is the quotient rule valid where

v (x) = 0

No — the derivative isn't defined there because

\frac{u}{v}

itself blows up (division by zero). The

v^{2}

in the denominator also vanishes, so the formula is meaningless at those points.

For

f (g (x))

, what if the outer function is the identity

f (u) = u

Then

f^{'} = 1

and the chain rule gives

1 \cdot g^{'} (x) = g^{'} (x)

— composing with "do nothing" leaves the rate unchanged, as it should.

What is the chain-rule result when the inner function is constant,

g (x) = c

g^{'} = 0

, so the whole derivative is

0

. A constant input never moves, so no matter how steep

f

is, the output is frozen.

Does

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

still hold if one factor is zero at a point?

Yes. If

u (a) = 0

the formula gives

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

; the rule holds everywhere the factors are differentiable, zeros included.

At the peak of

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

, why is the derivative zero even though both

x

and

x^{2} + 1

are changing?

The numerator of the quotient derivative,

1 - x^{2}

, hits zero at

x = \pm 1

; the top's growth and the bottom's growth exactly cancel there, so the net slope is

0

Recall Self-test

For each rule, can you state (a) the formula, (b) the picture (rectangle strips / gears), and (c) one trap it invites? If yes for all three rules, you own this topic.

Connections

Power Rule and Basic Derivatives — the raw pieces these rules combine.
Limits and Continuity — where "shrink $h$ to zero" comes from.
Activation Functions — why the sigmoid derivative form matters.
Backpropagation — the chain rule at industrial scale.
Gradient Descent — where these derivatives get used.
Gradient and Partial Derivatives — the multi-input generalisation.