Product rule — proof: Worked examples — Maths deep dive

First — see the rule before you use it

Everything on this page rests on one picture: the area of a rectangle whose sides are both growing. Look at it before any algebra.

Cell	Case class	What makes it special	Covered by
A	Two plain "nice" factors	Baseline: just slot into $u^{'} v + u v^{'}$	Ex 1
B	A factor with sign changes (trig)	Answer's sign depends on where you evaluate	Ex 2
C	A factor equals zero at the point	Terms drop out — check nothing divides by 0	Ex 3
D	Degenerate: one factor constant	Rule collapses to $(c v)^{'} = c v^{'}$	Ex 4
E	Both factors the same power of $x$	Product rule must agree with Power rule	Ex 5
F	Three factors	Apply the rule twice / all-but-one pattern	Ex 6
G	Real-world word problem (units)	Rate of a product = e.g. area growth rate	Ex 7
H	Exam twist: disguised / needs simplify	You must spot the product first	Ex 8

Warm-up: the one recipe used everywhere

Cell A — two plain factors

Worked example Example 1 (Cell A) —

$f (x) = (3 x + 1) (x^{2} - 4)$ Forecast: before reading on, guess: is $f^{'}$ a quadratic, and what's its leading coefficient? (Hint: $f$ expands to a cubic, so $f^{'}$ is a quadratic.)

Step 1 — Name the two factors. Let $u = 3 x + 1$ and $v = x^{2} - 4$ . Why this step? The rule only speaks in terms of $u, v, u^{'}, v^{'}$ ; naming them prevents mixing pieces.

Step 2 — Differentiate each factor alone. $u^{'} = 3$ , and $v^{'} = 2 x$ . Why this step? $u^{'}$ and $v^{'}$ are the ingredients; each is a one-line power-rule derivative.

Step 3 — Assemble $u^{'} v + u v^{'}$ . $f^{'} = u^{'} 3 v (x^{2} - 4) + u (3 x + 1) v^{'} (2 x)$ Why this step? This is the whole rule; keep the pieces boxed so you don't drop one.

Step 4 — Tidy up. $3 x^{2} - 12 + 6 x^{2} + 2 x = 9 x^{2} + 2 x - 12$ .

Verify: Expand first: $f = 3 x^{3} - 12 x + x^{2} - 4$ , so $f^{'} = 9 x^{2} + 2 x - 12$ . ✓ Same answer — exactly what should happen. At $x = 1$ both routes give $9 + 2 - 12 = - 1$ .

Cell B — a sign-changing factor (trig)

Trig factors are the classic "watch the sign" case: $cos$ and $sin$ flip sign across quadrants, so the value of $f^{'}$ changes character depending on where you stand.

Worked example Example 2 (Cell B) —

$f (x) = x cos x$ , a full sign table Forecast: is $f^{'}$ ever equal to $cos x$ alone? At which $x$ does the " $- x sin x$ " piece push $f^{'}$ down versus up?

Step 1 — Name and differentiate factors. $u = x \Rightarrow u^{'} = 1$ ; $v = cos x \Rightarrow v^{'} = - sin x$ . Why this step? $cos$ differentiates to $- sin$ — the minus sign is where sign-mistakes live.

Step 2 — Assemble. $f^{'} = (1) (cos x) + (x) (- sin x) = cos x - x sin x$ Why this step? Straight application; note the $-$ came from $v^{'}$ , not from the rule.

Step 3 — Build a complete sign table across all four quadrants. Here "quadrant" means which quarter-turn of the angle $x$ we sit in: QI $= (0, \frac{π}{2})$ , QII $= (\frac{π}{2}, π)$ , QIII $= (π, \frac{3 π}{2})$ , QIV $= (\frac{3 π}{2}, 2 π)$ .

$x$	quadrant	$cos x$	$sin x$	$f^{'} = cos x - x sin x$	sign of $f^{'}$
$\frac{π}{4} \approx 0.785$	QI	$+$	$+$	$0.707 - 0.785 (0.707) \approx 0.152$	$+$
$\frac{3 π}{4} \approx 2.356$	QII	$-$	$+$	$- 0.707 - 2.356 (0.707) \approx - 2.373$	$-$
$\frac{5 π}{4} \approx 3.927$	QIII	$-$	$-$	$- 0.707 - 3.927 (- 0.707) \approx 2.069$	$+$
$\frac{7 π}{4} \approx 5.498$	QIV	$+$	$-$	$0.707 - 5.498 (- 0.707) \approx 4.594$	$+$

Why this step? One point per quadrant guarantees the reader sees every sign combination of $cos$ and $sin$ , so no scenario is skipped. Notice QIII flips $f^{'}$ back to positive because two minus signs multiply positive in $- x sin x$ .

Verify: at $x = \frac{3 π}{4}$ , $f^{'} = cos \frac{3 π}{4} - \frac{3 π}{4} sin \frac{3 π}{4} = - \frac{2}{2} - \frac{3 π}{4} \cdot \frac{2}{2} \approx - 2.373$ (negative, as tabled). ✓

Cell C — a factor that is zero at the point

Worked example Example 3 (Cell C) —

$f (x) = (x^{2} - 9) e^{x}$ , find $f^{'} (3)$ Forecast: since $x^{2} - 9 = 0$ at $x = 3$ , guess which of the two product-rule terms survives.

Step 1 — Name factors. $u = x^{2} - 9, v = e^{x}$ . Why this step? $e^{x}$ is its own derivative, so $v^{'} = e^{x}$ — a clean second factor.

Step 2 — Differentiate. $u^{'} = 2 x, v^{'} = e^{x}$ .

Step 3 — Assemble. $f^{'} = 2 x e^{x} + (x^{2} - 9) e^{x}$ Why this step? Nothing special yet — the zero only bites when we plug in a value.

Step 4 — Evaluate at $x = 3$ . $f^{'} (3) = 2 (3) e^{3} + = 0 (9 - 9) e^{3} = 6 e^{3}$ Why this step? The second term dies because $u (3) = 0$ , leaving only the " $u^{'} v$ " piece. This is a genuine feature, not a shortcut: at a root of $u$ , only $u^{'} v$ can contribute.

Verify: $6 e^{3} \approx 6 (20.0855) \approx 120.51$ . A finite-difference slope of $f$ near $x = 3$ matches. ✓ (Note: nothing here divided by zero — the rule is safe even when a factor is 0.)

Cell D — degenerate case: one factor constant

Cell E — must agree with the Power rule

Cell F — three factors (derived here, from scratch)

We do not borrow the three-factor rule — we build it from the two-factor rule so this page stands alone.

Cell G — real-world word problem (units!)

Worked example Example 7 (Cell G) — expanding photo frame

A rectangular photo has width $w (t) = 4 + 0.5 t$ cm and height $h (t) = 3 + 0.2 t$ cm, where $t$ is in seconds. How fast is the area growing at $t = 10$ s?

Forecast: will the answer be bigger or smaller than "just width's growth × current height"? (The height is also growing, so there's a second contribution.)

Step 1 — Write area as a product. $A (t) = w (t) h (t)$ . Why this step? Area is a product, so its rate of change is a textbook product-rule problem — this is literally the rectangle picture from the top of the page come to life.

Step 2 — Rates of each side. $w^{'} (t) = 0.5$ cm/s, $h^{'} (t) = 0.2$ cm/s. Why this step? These are the "growth speeds" of each side — the $u^{'}$ and $v^{'}$ ingredients.

Step 3 — Apply the rule. $A^{'} (t) = w^{'} h + w h^{'} = 0.5 h (t) + 0.2 w (t)$ Why this step? Rate of area = (widening speed × current height) + (current width × heightening speed).

Step 4 — Plug in $t = 10$ . $w (10) = 4 + 5 = 9$ cm, $h (10) = 3 + 2 = 5$ cm. $A^{'} (10) = 0.5 (5) + 0.2 (9) = 2.5 + 1.8 = 4.3 cm^{2} / s$ Why this step? Both strips contribute; ignoring the second would undercount.

Verify (units + expansion): Units: (cm/s)(cm) + (cm)(cm/s) $=$ cm²/s ✓ — an area rate. Direct check: $A (t) = (4 + 0.5 t) (3 + 0.2 t) = 12 + 0.8 t + 1.5 t + 0.1 t^{2} = 12 + 2.3 t + 0.1 t^{2}$ , so $A^{'} (t) = 2.3 + 0.2 t$ , giving $A^{'} (10) = 2.3 + 2 = 4.3$ . ✓

Cell H — exam twist (spot the hidden product)

Worked example Example 8 (Cell H) —

$f (x) = x (x - 2)$ The twist: at first glance $x (x - 2)$ looks like a single messy expression, but it is secretly a product of two factors — and the square root is a disguised power. Spotting both is the exam skill being tested.

Domain first. $x$ is only defined for $x \geq 0$ , and its derivative $\frac{1}{2} x^{- 1/2}$ needs $x > 0$ (division by $x$ ). So everything below is valid on $x > 0$ only — we will study the edge $x \to 0^{+}$ as a limit, never plug $x = 0$ in directly.

Forecast: what power of $x$ is $x$ , and what's its derivative? Guess whether $f^{'}$ blows up as $x \to 0^{+}$ .

Step 1 — Rewrite the disguised factor as a power. $x = x^{1/2}$ . Set $u = x^{1/2}, v = x - 2$ . Why this step? Exams hide products behind radicals; converting to $x^{1/2}$ lets the Power rule handle it. $u^{'} = \frac{1}{2} x^{- 1/2}$ (valid for $x > 0$ ).

Step 2 — Differentiate the other factor. $v^{'} = 1$ .

Step 3 — Assemble. $f^{'} = \frac{1}{2} x^{- 1/2} (x - 2) + x^{1/2} (1) = \frac{x - 2}{2 x} + x$ Why this step? Direct product rule; keep the negative exponent as $1/ (2 x)$ for clarity.

Step 4 — Combine over one denominator (exam-neat form). $f^{'} = \frac{x - 2}{2 x} + \frac{2 x}{2 x} = \frac{3 x - 2}{2 x} (x > 0)$ Why this step? Examiners want a single simplified fraction; common denominator $2 x$ .

Step 5 — Limiting behaviour as $x \to 0^{+}$ . Approaching the left edge of the domain: numerator $\to - 2$ , denominator $\to 0^{+}$ , so $f^{'} \to - \infty$ : the slope becomes vertical and steeply negative near the origin. Why this step? The matrix asks for limiting behaviour — the derivative can be unbounded even though $f$ itself is finite ( $f (0) = 0$ ). We take the limit from the right because the domain is $x > 0$ .

Verify: at $x = 1$ : $f^{'} = \frac{3 - 2}{2} = 0.5$ . Direct: $f = x^{3/2} - 2 x^{1/2}$ , so $f^{'} = \frac{3}{2} x^{1/2} - x^{- 1/2}$ ; at $x = 1$ : $1.5 - 1 = 0.5$ . ✓ At $x = 4$ : $\frac{10}{4} = 2.5$ and $\frac{3}{2} (2) - \frac{1}{2} = 3 - 0.5 = 2.5$ . ✓

Matrix completeness check

Recall Did we really cover every cell? (Each line below is "prompt ::: answer" — tap to check yourself)

A → Ex 1 (plain factors) ::: ✓ B → Ex 2 (trig, sign per quadrant, full table) ::: ✓ C → Ex 3 (a factor equals zero) ::: ✓ D → Ex 4 (constant factor, degenerate) ::: ✓ E → Ex 5 (agrees with power rule) ::: ✓ F → Ex 6 (three factors, derived here) ::: ✓ G → Ex 7 (word problem with units) ::: ✓ H → Ex 8 (disguised product + limiting slope) ::: ✓

Rapid recall

Recall How to read these lines

Each line is written as prompt ::: answer. Cover the part after the :::, try to answer, then reveal. This is the vault's standard self-quiz format.

Differentiate

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

9 x^{2} + 2 x - 12

Differentiate

x cos x

cos x - x sin x

For

f = (x^{2} - 9) e^{x}

, why does one term vanish at

x = 3

Because

u (3) = x^{2} - 9 = 0

kills the

u v^{'}

term, leaving

u^{'} v = 6 e^{3}

When one factor is a constant

c

, the product rule becomes…

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

(the other term is

0 \cdot v

For expanding rectangle area,

A^{'} =

w^{'} h + w h^{'}

(widening speed × height + width × heightening speed).

\frac{d}{d x} [x (x - 2)] =

\frac{3 x - 2}{2 x}

(valid for

x > 0

)

Connections

Product rule — proof — the parent this page drills.
Power rule — Examples 5 and 8 lean on it.
Chain rule — needed once a factor is itself a composition.
Quotient rule — the sibling for divisions instead of products.
Differentiability implies continuity — the hidden lemma behind the original proof.
Leibniz rule (nth derivative of a product) — where the three-factor pattern generalises.
Limit definition of the derivative — the engine underneath every step.

Worked examples — Product rule — proof

First — see the rule before you use it

The scenario matrix

Warm-up: the one recipe used everywhere

Cell A — two plain factors

Cell B — a sign-changing factor (trig)

Cell C — a factor that is zero at the point

Cell D — degenerate case: one factor constant

Cell E — must agree with the Power rule

Cell F — three factors (derived here, from scratch)

Cell G — real-world word problem (units!)

Cell H — exam twist (spot the hidden product)

Matrix completeness check

Rapid recall

Connections