Chain rule — proof, composite function derivatives: Question bank

Before the traps, a refresher so nothing is used unexplained:

True or false — justify

TF1. "

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

is true because the

d u

's cancel like fractions."

False — the equation is true, but the reason is not.

d y / d u

and

d u / d x

are single limits, not fractions with a shared factor; cancellation is only a memory aid, proven properly by the error-function argument.

TF2. "If

f

and

g

are both differentiable everywhere, then

f \circ g

is differentiable everywhere."

True — the chain rule's hypotheses (

g

differentiable at

x

f

differentiable at

g (x)

) are met at every

x

, so the composite is differentiable there with the product formula.

TF3. "The chain rule can fail even if the answer

f^{'} (g (x)) g^{'} (x)

still exists as a number."

True — differentiability of

f

must hold at the point $g (x)$ , not just somewhere. If

f

is not differentiable at that particular

u

, the formula's value is meaningless even if it computes to a number.

TF4. "

\frac{d}{d x} sin (x^{2}) = cos (2 x)

False — you must keep the inside intact and multiply by its rate:

cos (x^{2}) \cdot 2 x

cos (2 x)

wrongly doubles the angle instead of keeping

x^{2}

and multiplying by

g^{'} = 2 x

TF5. "In the rigorous proof, the error function

E

divides by

Δ u

, so it still breaks when

Δ u = 0

False —

E

is defined to equal

0

s = 0

, and the key identity

f (u + s) - f (u) = (f^{'} (u) + E (s)) s

multiplies by

s

. Nothing is ever divided by

Δ u

TF6. "Differentiability of

g

is needed only to make

g^{'} (x)

appear; continuity of

g

plays no role."

False — continuity of

g

is what guarantees

Δ u \to 0

Δ x \to 0

, which is exactly what lets

E (Δ u) \to E (0) = 0

. Without it the proof stalls.

TF7. "For a constant inner function

g (x) = c

, the naïve

Δ u

-cancellation proof works fine."

False — here

Δ u = 0

for every

Δ x

, so dividing by

Δ u

is always undefined; only the error-function proof survives, giving

h^{'} (x) = f^{'} (c) \cdot 0 = 0

correctly (assuming

f

is differentiable at

c

TF8. "If

h = f \circ g

is differentiable at

x

, then both

f

and

g

must be differentiable there."

False — a composite can be smooth even when a piece isn't; e.g.

g (x) = ∣ x ∣

(not differentiable at

0

) fed into

f (u) = u^{2}

gives

h (x) = x^{2}

, differentiable at

0

. The chain rule is sufficient, not necessary.

Spot the error

SE1. "

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

Missing the inner rate. The full chain gives

3 (x^{2})^{2} \cdot 2 x = 6 x^{5}

, matching

\frac{d}{d x} x^{6} = 6 x^{5}

. They forgot to multiply by

g^{'} (x) = 2 x

SE2. "

\frac{d}{d x} cos 4 x = \frac{1}{2} (cos 4 x)^{- 1/2}

Two inner rates dropped. Peeling the onion needs

\times (- sin 4 x)

(rate of

cos

)

\times 4

(rate of

4 x

), giving

\frac{- 2 sin 4 x}{cos 4 x}

SE3. "Since

E (s) \to 0

, we can just replace

E (Δ u)

0

before taking the limit and simplify early."

Illegal —

E (Δ u)

is generally nonzero for finite

Δ x

; you may only send it to

0

inside the limit as

Δ x \to 0

. Substituting early changes an exact identity into an approximation.

SE4. "

\frac{d}{d x} e^{s i n x} = e^{c o s x}

The outer

e^{u}

has derivative

e^{u}

evaluated at the unchanged inside

u = sin x

, times the inner rate

cos x

e^{s i n x} cos x

. They mistakenly differentiated inside the exponent.

SE5. "To differentiate

f (g (x))

I differentiate

g

first, get

g^{'}

, then feed

g^{'}

into

f^{'}

f^{'} (g^{'} (x))

Wrong composition. You feed the original inside

g (x)

into

f^{'}

and then multiply by

g^{'} (x)

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

, not

f^{'} (g^{'} (x))

SE6. "The naïve split

\frac{f ( u + Δ u ) - f ( u )}{Δ u} \cdot \frac{Δ u}{Δ x}

is a valid proof because it gives the right answer."

A correct answer from an invalid step is luck, not proof. The step divides by

Δ u

, which can be

0

for infinitely many

Δ x \neq = 0

; a proof must be valid for all admissible

Δ x

Why questions

WH1. Why do the two rates multiply rather than add?

Because the wiggles act in sequence like gears:

g

scales

Δ x

into

Δ u \approx g^{'} Δ x

, then

f

scales that into

Δ y \approx f^{'} Δ u

, so

Δ y \approx f^{'} g^{'} Δ x

— compounding, hence a product.

WH2. Why does the proof start from the limit definition of the derivative rather than the Leibniz form?

The Leibniz "cancellation" is only a mnemonic; a genuine derivation must reduce

h^{'}

to a difference-quotient limit, which is the actual definition of a derivative.

WH3. Why is the error function

E

made continuous at

0

(by setting

E (0) = 0

So the single identity

f (u + s) - f (u) = (f^{'} (u) + E (s)) s

holds for all

s

including

s = 0

, and so

E (Δ u) \to E (0) = 0

once

Δ u \to 0

— removing every division-by-zero worry.

WH4. Why evaluate

f^{'}

g (x)

and not at

x

The outer machine only ever sees the number

u = g (x)

, never

x

itself. Its slope must be read at the value actually entering it, which is

g (x)

WH5. Why does "

g

differentiable

\Rightarrow

g

continuous" matter more than "

g

differentiable"?

The product formula needs

g^{'} (x)

(differentiability) and

Δ u \to 0

(continuity) to kill the error term. Continuity is the bridge that makes the limit of

E (Δ u)

collapse to

0

WH6. Why does implicit differentiation "just work" without a new rule?

Because it is the chain rule: treating

y

y (x)

, differentiating any expression in

y

carries a factor

\frac{d y}{d x}

exactly as the chain rule demands (see Implicit differentiation).

WH7. Why can the same trick prove the Inverse function derivative?

Chain-rule the identity

f (f^{- 1} (x)) = x

: you get

f^{'} (f^{- 1} (x)) \cdot (f^{- 1})^{'} (x) = 1

, so

(f^{- 1})^{'} (x) = 1/ f^{'} (f^{- 1} (x))

— the inverse-derivative formula falls straight out.

Edge cases

EC1. What is

h^{'} (x)

when the inner function is constant,

g (x) = c

(with

f

differentiable at

c

h (x) = f (c)

is itself constant, so

h^{'} (x) = 0

; the formula agrees since

g^{'} (x) = 0

gives

f^{'} (c) \cdot 0 = 0

. (If

f

were not differentiable at

c

, the formula simply wouldn't apply — but

h

is still constant, so

h^{'} = 0

directly.)

EC2. What happens at a point where

g^{'} (x) = 0

but

f^{'} (g (x)) \neq = 0

The composite has zero slope there:

h^{'} (x) = f^{'} (g (x)) \cdot 0 = 0

. A stationary inner wiggle produces no output wiggle, no matter how steep

f

is.

EC3. Can

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

while

g^{'} (x) = 0

Not via the chain rule — whenever its hypotheses hold, the formula forces

h^{'} = f^{'} \cdot 0 = 0

. If

h^{'} \neq = 0

is observed, some hypothesis (differentiability of

f

g (x)

) must have failed, so the formula never applied.

EC4. Does the chain rule apply to

g (x) = x^{2} sin (1/ x)

(with

g (0) = 0

) at

x = 0

Yes if

g

is differentiable at

0

and

f

g (0)

— and this is precisely the pathological case where

Δ u = 0

for infinitely many

Δ x

, showing exactly why the error-function proof (not naïve cancellation) is required.

EC5. For a triple nest

f (g (p (x)))

, how many factors and where is each derivative evaluated?

Three factors:

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

— each outer derivative read at the value actually fed into it, all multiplied together (peel outside-in). (We name the innermost function $p$ so it isn't mistaken for the error-function argument.)

EC6. If

g

has a corner (not differentiable) at

x_{0}

but

f

is smooth, is

f \circ g

necessarily non-differentiable at

x_{0}

No — smoothing can occur. With

f (u) = u^{2}

and

g (u) = ∣ u ∣

, the composite

x^{2}

is perfectly differentiable at

0

; the chain rule simply doesn't apply, but the composite may still be differentiable by direct check.

Connections

Chain rule — proof, composite function derivatives — parent; every trap here targets a step of that proof.
Derivative — limit definition — WH2 and SE3 hinge on the difference-quotient definition.
Continuity implies differentiability fails converse — WH5, EC6 use "differentiable ⇒ continuous."
Implicit differentiation — WH6 recasts it as the chain rule.
Inverse function derivative — WH7 derives it by chain-ruling $f (f^{- 1} (x)) = x$ .
Product rule, Quotient rule — sibling rules often combined with the chain rule.