Chain rule — proof, composite function derivatives: Worked examples

Everywhere below, "the machine" language is: outer function $f$ , inner function $g$ , composite $h (x) = f (g (x))$ , and the rule is $h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)$ — derivative of the outside (inside left frozen) times derivative of the inside.

The scenario matrix

Before working anything, let's list every class of case a chain-rule problem can be. Each worked example below is tagged with the cell it fills.

Cell	Case class	What's special / what can go wrong	Example
A	Power-of-polynomial (positive inner)	Basic peel; sign of answer follows $x$	Ex 1
B	Trig inside polynomial	Must keep inside intact ( $cos (x^{2}) \neq = cos 2 x$ )	Ex 2
C	Triple / multi-nesting	One factor per layer	Ex 3
D	Inside evaluates to zero at a point	Is $h^{'}$ still defined? Check the outer at $0$	Ex 4
E	Degenerate: a gear's rate $= 0$ at the point	Whole product collapses to $0$	Ex 5
F	Limiting / blow-up (denominator $\to 0$ )	Derivative $\to \pm \infty$ ; which sign?	Ex 6
G	Negative / all-sign inner (quadrant sweep)	Answer changes sign across regions	Ex 7
H	Word problem — related rates	Chain rule in time; carry units	Ex 8
I	Exam twist — abstract $f$ with a table	No formula for $f$ ; use given values	Ex 9
J	Trig-inside-trig (non-power nesting)	Both layers are transcendental	Ex 10

The rest of the page fills each cell.

Cell A — power of a polynomial

Worked example Ex 1 (Cell A):

$h (x) = (2 x^{3} - 5)^{4}$ Forecast: guess the shape of the answer before reading. Will there be a factor of $x^{2}$ ? Will the power drop to $3$ ? Jot it down.

Name the layers. Outer $f (u) = u^{4}$ , inner $g (x) = 2 x^{3} - 5$ , so $u = g (x)$ . Why this step? You cannot peel until you know which part is the "shell" and which is the "filling."
Differentiate the outer, inside frozen. $f^{'} (u) = 4 u^{3} \Rightarrow f^{'} (g (x)) = 4 (2 x^{3} - 5)^{3}$ . Why this step? Power rule on the shell; we treat the whole blob $2 x^{3} - 5$ as a single lump $u$ .
Differentiate the inner. $g^{'} (x) = 6 x^{2}$ . Why this step? This is the "gear ratio" of the inside — how fast the filling moves.
Multiply (DOTI). $h^{'} (x) = 4 (2 x^{3} - 5)^{3} \cdot 6 x^{2} = 24 x^{2} (2 x^{3} - 5)^{3} .$ Why this step? DOTI: multiply the outer rate (step 2) by the inner rate (step 3) — the two amplifications compound.

Verify: at $x = 1$ , $g (1) = 2 - 5 = - 3$ , so $h^{'} (1) = 24 \cdot 1 \cdot (- 3)^{3} = 24 \cdot (- 27) = - 648$ . A numerical slope $\frac{h ( 1.001 ) - h ( 0.999 )}{0.002} \approx - 648$ confirms it. ✓ Factor of $x^{2}$ : yes; power dropped to $3$ : yes — matches the forecast.

Cell B — trig inside a polynomial (keep the inside intact)

Worked example Ex 2 (Cell B):

$h (x) = cos (x^{2} + 1)$ Forecast: will the answer contain $sin (2 x)$ or $sin (x^{2} + 1)$ ? This is the classic trap — decide now.

Layers. Outer $f (u) = cos u$ , inner $g (x) = x^{2} + 1$ . Why this step? We must tag the outer $cos$ and the inner $x^{2} + 1$ separately so we never accidentally differentiate them together.
Outer derivative, inside frozen. $f^{'} (u) = - sin u \Rightarrow f^{'} (g (x)) = - sin (x^{2} + 1)$ . Why this step? The derivative of $cos$ is $- sin$ ; we do not touch the argument yet.
Inner derivative. $g^{'} (x) = 2 x$ . Why this step? This is the inner gear ratio; it will come out as a separate multiplied factor, not inside the $sin$ .
Multiply (DOTI). $h^{'} (x) = - sin (x^{2} + 1) \cdot 2 x = - 2 x sin (x^{2} + 1) .$ Why this step? DOTI: outer rate (step 2) times inner rate (step 3); the inner $2 x$ lands outside the $sin$ .

Verify: at $x = 0$ , $h^{'} (0) = - 2 \cdot 0 \cdot sin (1) = 0$ . Makes sense: $x^{2} + 1$ has zero slope at $x = 0$ (bottom of the parabola), so the composite is momentarily flat. ✓

Cell C — triple nesting

Worked example Ex 3 (Cell C):

$h (x) = sin^{3} (4 x) = (sin (4 x))^{3}$ Forecast: how many factors will the answer have? (Hint: count the layers.)

There are three layers, peel outside-in:

Layer 1 (outermost): cube, $(\cdot)^{3}$ .
Layer 2: $sin (\cdot)$ .
Layer 3 (innermost): $4 x$ .

Cube's derivative: $3 (sin 4 x)^{2}$ . Why? Power rule, inside ( $sin 4 x$ ) frozen.
$sin$ 's derivative: $cos (4 x)$ . Why? Peel the next shell, its inside ( $4 x$ ) frozen.
Inner's derivative: $\frac{d}{d x} (4 x) = 4$ . Why this step? The innermost gear ratio: $4 x$ changes $4$ times as fast as $x$ ; this becomes the last multiplied factor.
Multiply all three factors (DOTI, thrice): $h^{'} (x) = 3 (sin 4 x)^{2} \cdot cos (4 x) \cdot 4 = 12 sin^{2} (4 x) cos (4 x) .$ Why this step? DOTI applied at each layer boundary: one factor per shell, all multiplied together.

Verify: three layers → three factors → matches forecast. At $x = \frac{π}{8}$ : $4 x = \frac{π}{2}$ , $sin = 1, cos = 0$ , so $h^{'} = 12 \cdot 1 \cdot 0 = 0$ . Correct — $sin^{3} (4 x)$ peaks at $4 x = π /2$ , so slope is $0$ there. ✓

Cell D — the inside hits zero

This is the case people fear from the proof (the " $Δ u = 0$ " worry). Let's see it concretely.

Worked example Ex 4 (Cell D):

$h (x) = e^{x^{2} - 4}$ , evaluate $h^{'}$ where the inside is $0$ . Forecast: the inside $g (x) = x^{2} - 4$ is zero at $x = \pm 2$ . Is $h^{'}$ still defined there? Guess yes/no.

Layers. Outer $f (u) = e^{u}$ , inner $g (x) = x^{2} - 4$ . Why this step? Identify the shell ( $e^{u}$ ) and filling ( $x^{2} - 4$ ) before peeling — the exponential is the outer machine.
Outer derivative: $f^{'} (u) = e^{u}$ (its own derivative) $\Rightarrow f^{'} (g (x)) = e^{x^{2} - 4}$ . Why this step? $e^{u}$ is the unique function equal to its own derivative; we evaluate it at the inner value $u = x^{2} - 4$ .
Inner derivative: $g^{'} (x) = 2 x$ . Why this step? Power rule on $x^{2} - 4$ gives the inner gear ratio $2 x$ .
Multiply (DOTI): $h^{'} (x) = e^{x^{2} - 4} \cdot 2 x = 2 x e^{x^{2} - 4} .$ Why this step? DOTI: compound the outer rate (step 2) with the inner rate (step 3).
Plug in the zero-inside point $x = 2$ : $g (2) = 0$ , so $h^{'} (2) = 2 \cdot 2 \cdot e^{0} = 4 \cdot 1 = 4.$ Why this step? This is the whole point of the cell — testing whether "inside $= 0$ " breaks anything. It doesn't.

Verify: numerically $\frac{h ( 2.001 ) - h ( 1.999 )}{0.002} \approx 4.00$ . ✓ And $h^{'} (2) > 0$ : at $x = 2$ the inside is increasing ( $2 x = 4 > 0$ ), so the composite rises. Sign is right.

Cell E — degenerate: a gear's rate is zero

Worked example Ex 5 (Cell E):

$h (x) = cos (x^{3})$ , evaluate $h^{'}$ at $x = 0$ . Forecast: at $x = 0$ the inner rate $g^{'} (x) = 3 x^{2}$ vanishes — and it happens the outer rate does too. Predict $h^{'} (0)$ .

Layers. Outer $f (u) = cos u$ , inner $g (x) = x^{3}$ . Why this step? Tag shell ( $cos$ ) and filling ( $x^{3}$ ) so we can track which gear stalls.
Outer derivative: $- sin (x^{3})$ . Why this step? Derivative of $cos$ is $- sin$ , evaluated at the inner value $x^{3}$ (inside frozen).
Inner derivative: $3 x^{2}$ . Why this step? Power rule on $x^{3}$ gives the inner gear ratio — this is the factor that will vanish at $x = 0$ .
Multiply (DOTI): $h^{'} (x) = - 3 x^{2} sin (x^{3})$ . Why this step? DOTI: outer rate times inner rate, both packaged into one product.
At $x = 0$ : $h^{'} (0) = - 3 \cdot 0^{2} \cdot sin (0) = 0$ . Here the inner gear $3 x^{2}$ is zero and the outer $- sin (x^{3}) = - sin 0 = 0$ — both factors happen to vanish, so the product is emphatically $0$ .

Verify: $\frac{h ( 0.001 ) - h ( - 0.001 )}{0.002} \approx 0$ . ✓ (By symmetry $cos (x^{3})$ is even, so its slope at $0$ must be $0$ — sanity confirmed.)

Cell F — limiting / blow-up (derivative $\to \pm \infty$ )

Worked example Ex 6 (Cell F):

$h (x) = 1 - x^{2}$ (upper unit semicircle), behaviour near $x \to 1^{-}$ . Forecast: as $x$ approaches $1$ from below, the semicircle plunges toward the axis. Does the slope go to $0$ , a finite number, or $\pm \infty$ ? Which sign?

Layers. Outer $f (u) = u^{1/2} = u$ , inner $g (x) = 1 - x^{2}$ .
Outer derivative: $f^{'} (u) = \frac{1}{2} u^{- 1/2} = \frac{1}{2 u}$ . Why this step? Power rule with exponent $\frac{1}{2}$ ; the $u^{- 1/2}$ is why a blow-up is possible — as $u \to 0$ this factor explodes.
Inner derivative: $g^{'} (x) = - 2 x$ . Why this step? Power rule on $1 - x^{2}$ ; the minus sign here is what makes the final slope negative (curve falling).
Multiply (DOTI): $h^{'} (x) = \frac{1}{2 1 - x ^{2}} \cdot (- 2 x) = \frac{- x}{1 - x ^{2}} .$ Why this step? DOTI: outer rate (step 2) times inner rate (step 3), then simplify.
Limit $x \to 1^{-}$ : numerator $\to - 1$ ; denominator $1 - x^{2} \to 0^{+}$ (a shrinking positive number). So $h^{'} (x) \to \frac{- 1}{0 ^{+}} = - \infty.$

Figure below (s01): the upper semicircle in violet, with three dashed tangent lines at $x = 0, 0.6, 0.9$ that get steeper and steeper, tilting toward vertical as we approach $x = 1$ . The minus sign says each tangent tilts downward (curve falling); the magnitude $\to \infty$ says it becomes vertical.

Verify: at $x = 0.99$ , $h^{'} = \frac{- 0.99}{1 - 0.9801} = \frac{- 0.99}{0.0199} \approx \frac{- 0.99}{0.1411} \approx - 7.02$ — large and negative, heading to $- \infty$ . ✓

Cell G — all-sign inner (a quadrant/region sweep)

Worked example Ex 7 (Cell G):

$h (x) = (x^{2} - 1)^{3}$ ; describe the sign of $h^{'}$ across all regions of $x$ . Forecast: where is the composite increasing vs decreasing? Sketch a guess with three break-points.

Layers. Outer $f (u) = u^{3}$ , inner $g (x) = x^{2} - 1$ .
Outer derivative: $3 u^{2} = 3 (x^{2} - 1)^{2}$ . Why this step? Power rule on the cube; crucially this is a square, so this factor is $\geq 0$ always — it can touch zero but never flip the sign.
Inner derivative: $2 x$ . Why this step? Power rule on $x^{2} - 1$ ; this is the only factor that changes sign, so it controls where $h$ rises or falls.
Multiply (DOTI): $h^{'} (x) = 3 (x^{2} - 1)^{2} \cdot 2 x = 6 x (x^{2} - 1)^{2} .$ Why this step? DOTI: outer rate times inner rate; the sign-carrying factor is the inner $2 x$ .
Sign analysis — cover every region. The factor $(x^{2} - 1)^{2} \geq 0$ never flips sign; it is $0$ at $x = \pm 1$ . So the sign of $h^{'}$ is the sign of $6 x$ :

Region	$6 x$	$(x^{2} - 1)^{2}$	$h^{'} (x)$	Behaviour
$x < - 1$	$-$	$+$	$-$	decreasing
$x = - 1$	$-$	$0$	$0$	flat (touch)
$- 1 < x < 0$	$-$	$+$	$-$	decreasing
$x = 0$	$0$	$+$	$0$	flat (min)
$0 < x < 1$	$+$	$+$	$+$	increasing
$x = 1$	$+$	$0$	$0$	flat (touch)
$x > 1$	$+$	$+$	$+$	increasing

Figure below (s02): the curve $(x^{2} - 1)^{3}$ in violet with the three horizontal tangents marked at $x = - 1, 0, 1$ (orange dots, magenta level segments); labels flag $x = 0$ as the true minimum and $x = \pm 1$ as saddle-flats.

Verify: $h^{'} (0.5) = 6 (0.5) (0.25 - 1)^{2} = 3 \cdot 0.5625 = 1.6875 > 0$ ✓ (increasing). $h^{'} (- 0.5) = 6 (- 0.5) (0.5625) = - 1.6875 < 0$ ✓ (decreasing). $h^{'} (\pm 1) = 0$ ✓.

Worked example Ex 8 (Cell H): A spherical balloon is inflated so its

radius grows at $\frac{d r}{d t} = 2 cm/s$ . How fast is the volume growing when $r = 5 cm$ ? Forecast: volume is 3D, radius is 1D — do you expect the volume rate to be bigger or smaller than $2$ ? Guess a ballpark number.

Write the relationship. $V = \frac{4}{3} π r^{3}$ , and $r$ depends on time $t$ . Why this step? We're asked for $\frac{d V}{d t}$ but only know $\frac{d r}{d t}$ ; we first need a formula that ties $V$ to the shared variable $r$ , otherwise there is nothing to differentiate.
Chain rule in time. With $V = f (r)$ and $r = r (t)$ , $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t} .$ Why this step? This is exactly $\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ from the parent note, with $u = r$ , $x = t$ — the topic Related rates.
Outer rate. $\frac{d V}{d r} = \frac{4}{3} π \cdot 3 r^{2} = 4 π r^{2}$ . Why this step? Power rule on $r^{3}$ ; the $\frac{4}{3} π \cdot 3$ collapses to $4 π r^{2}$ — which is exactly the sphere's surface area, a handy sanity anchor.
Assemble with values (DOTI). At $r = 5$ : $\frac{d V}{d t} = d V / d r 4 π (5)^{2} \cdot d r / d t 2 = 4 π \cdot 25 \cdot 2 = 200 π \approx 628.3.$ Why this step? DOTI in time: multiply the outer rate $\frac{d V}{d r}$ (step 3) by the inner rate $\frac{d r}{d t}$ (given).
Attach units. $\frac{d V}{d r}$ has units $cm^{2}$ (area), $\frac{d r}{d t}$ has $cm/s$ , so the product has $cm^{2} \cdot cm/s = cm^{3} / s$ . Therefore $\frac{d V}{d t} \approx 628.3 cm^{3} / s .$ Why this step? Units are part of the answer in a word problem — a rate of volume change must come out in volume-per-time.

Verify: units multiply correctly ( $cm^{2} \cdot cm/s = cm^{3} / s$ ✓). Revisiting the forecast: yes, $628.3$ is far bigger than $2$ — exactly as predicted, because a large sphere gains a lot of volume for each extra cm of radius. Numeric: $200 π = 628.318 \dots$ ✓ The takeaway: related rates is just the chain rule with time as the driving variable.

Cell I — exam twist (abstract $f$ from a table, no formula)

Worked example Ex 9 (Cell I): You are told

$h (x) = f (g (x))$ and given only this table. Find $h^{'} (1)$ .

$x$	$g (x)$	$g^{'} (x)$	$f (x)$	$f^{'} (x)$
$1$	$3$	$4$	$6$	$- 2$
$3$	$2$	$5$	$7$	$10$

Forecast: which two numbers from this table do you actually need? Circle them before computing.

Write the rule at the point. $h^{'} (1) = f^{'} (g (1)) \cdot g^{'} (1)$ . Why this step? This is the whole chain rule instantiated at $x = 1$ ; the twist is only that $f, g$ have no formula, so every quantity must be read off the table.
Find the inner value. $g (1) = 3$ (row $x = 1$ ). Why this step? The outer derivative is evaluated at $u = g (1)$ , so we must first learn where to look. The answer is $3$ , not $1$ — this is the #1 exam trap.
Read $f^{'}$ at that inner value. $f^{'} (3) = 10$ (row $x = 3$ ). Why this step? We need $f^{'}$ at $u = 3$ (from step 2), which lives in the $x = 3$ row — not the $x = 1$ row.
Read the inner rate. $g^{'} (1) = 4$ (row $x = 1$ ). Why this step? The second chain-rule factor $g^{'} (1)$ is read directly at the original point $x = 1$ .
Multiply (DOTI): $h^{'} (1) = f^{'} (g (1)) \cdot g^{'} (1) = f^{'} (3) \cdot g^{'} (1) = 10 \cdot 4 = 40.$ Why this step? DOTI: outer rate $f^{'} (3)$ times inner rate $g^{'} (1)$ — the only two table entries that actually matter.

Verify: the two needed numbers were $f^{'} (3) = 10$ and $g^{'} (1) = 4$ ; product $= 40$ . Consistent with the forecast if you circled those two. ✓

Cell J — trig-inside-trig (both layers transcendental)

Worked example Ex 10 (Cell J):

$h (x) = sin (cos x)$ Forecast: neither layer is a power — both are trig. Will the answer be $cos (cos x)$ times something, or $sin (sin x)$ ? And what sign will the extra factor bring? Guess first.

Layers. Outer $f (u) = sin u$ , inner $g (x) = cos x$ . Why this step? With two trig layers it is doubly important to name them — it is easy to accidentally differentiate the inner as if it were the outer.
Outer derivative, inside frozen. $f^{'} (u) = cos u \Rightarrow f^{'} (g (x)) = cos (cos x)$ . Why this step? Derivative of $sin$ is $cos$ , evaluated at the inner value $u = cos x$ — the argument $cos x$ stays intact inside the outer $cos$ .
Inner derivative. $g^{'} (x) = - sin x$ . Why this step? Derivative of $cos x$ is $- sin x$ ; the minus sign is the reason the final answer carries a leading negative.
Multiply (DOTI): $h^{'} (x) = cos (cos x) \cdot (- sin x) = - sin x cos (cos x) .$ Why this step? DOTI: outer rate (step 2) times inner rate (step 3).

Verify: at $x = 0$ : $h^{'} (0) = - sin 0 \cdot cos (cos 0) = 0 \cdot cos 1 = 0$ . Sanity: $cos x$ has zero slope at $x = 0$ (top of the cosine wave), so the composite is flat there. ✓ At $x = \frac{π}{2}$ : $h^{'} = - sin \frac{π}{2} cos (cos \frac{π}{2}) = - 1 \cdot cos 0 = - 1$ . ✓

Recall Self-test: name the cell, then solve

Cover the worked solutions. For each, first say which matrix cell it is, then differentiate.

Differentiate $(2 x^{3} - 5)^{4}$ . ::: Cell A: $24 x^{2} (2 x^{3} - 5)^{3}$ . Differentiate $cos (x^{2} + 1)$ (watch the trap). ::: Cell B: $- 2 x sin (x^{2} + 1)$ , not $- sin (2 x)$ . Differentiate $sin^{3} (4 x)$ ; how many factors? ::: Cell C: $12 sin^{2} (4 x) cos (4 x)$ ; three layers → three factors. Is $h^{'}$ defined where the inside of $e^{x^{2} - 4}$ is zero? Value at $x = 2$ ? ::: Cell D: yes; $h^{'} (2) = 4$ . Why is $h^{'} (0) = 0$ for $cos (x^{3})$ ? ::: Cell E: the inner rate $3 x^{2} = 0$ at $x = 0$ , collapsing the product. What is $lim_{x \to 1^{-}} \frac{d}{d x} 1 - x^{2}$ ? ::: Cell F: $- \infty$ (tangent goes vertical, falling). Where does $(x^{2} - 1)^{3}$ have horizontal tangents, and which is a true min? ::: Cell G: $x = - 1, 0, 1$ ; only $x = 0$ is a min (others are saddle-flats). Balloon: $\frac{d V}{d t}$ at $r = 5$ , $\frac{d r}{d t} = 2$ ? ::: Cell H: $200 π \approx 628.3 cm^{3} / s$ . From the table, $h^{'} (1)$ where $h = f (g (x))$ ? ::: Cell I: $f^{'} (3) \cdot g^{'} (1) = 10 \cdot 4 = 40$ . Differentiate $sin (cos x)$ . ::: Cell J: $- sin x cos (cos x)$ .

Connections

Chain rule — proof, composite function derivatives — the parent; this page is its applied stress-test.
Related rates — Cell H is a direct related-rates use of $\frac{d V}{d t} = \frac{d V}{d r} \frac{d r}{d t}$ .
Derivative — limit definition — Cells D–F lean on the limit meaning of a derivative (blow-up, zero rate).
Product rule and Quotient rule — often combined with the chain rule in tougher exam problems.
Implicit differentiation — the abstract- $f$ mindset of Cell I generalises here.
Inverse function derivative and Continuity implies differentiability fails converse — related follow-ups.

Worked examples — Chain rule — proof, composite function derivatives