Quotient rule — proof: Worked examples — Maths deep dive

This page is a drill through every kind of quotient you can meet. We do not just plug into the formula — we build each case from the picture, forecast the answer, work it step by step, and verify it. If you have not yet seen the proof, read Quotient rule — proof first; here we use the result

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

The scenario matrix

Every quotient problem falls into one of these cells. Our examples (E1–E8) are labelled with the cell they cover, so together they touch every box.

Cell	What makes it special	Covered by
A. Polynomial / polynomial	plain algebra, real roots of slope	E1
B. Trig quotient	$u^{'}, v^{'}$ are trig; identities simplify	E2
C. Reciprocal ( $u = 1$ )	degenerate numerator, $u^{'} = 0$	E3
D. Constant on top ( $u = c$ )	$u^{'} = 0$ but $u \neq = 1$	E3 (note)
E. Sign of the slope at a point	which quadrant of behaviour: increasing vs decreasing	E4
F. Exponential over polynomial	new $u^{'}$ type, limiting behaviour $x \to \infty$	E5
G. Point where the answer is zero	numerator of $f^{'}$ vanishes (flat tangent)	E1, E4
H. Degenerate: $v (x) = 0$	rule does NOT apply — must flag it	E6
I. Real-world word problem	concentration / rate of change	E7
J. Exam twist: nested / with chain rule	$v$ itself a composite	E8

E1 — Cell A + G: polynomial over polynomial

Worked example Differentiate

$f (x) = \frac{x ^{2}}{x + 1}$ and find where the slope is zero. Forecast: Guess — will the slope ever be zero? A parabola on top, a line on bottom... expect two flat points.

Step 1. Name the parts: $u = x^{2}, v = x + 1$ , so $u^{'} = 2 x, v^{'} = 1$ . Why this step? The rule needs four ingredients; label them so you cannot mix them up.

Step 2. Plug into "Lo·dHi − Hi·dLo over Lo²": $f^{'} (x) = \frac{( 2 x ) ( x + 1 ) - ( x ^{2} ) ( 1 )}{( x + 1 ) ^{2}} .$ Why this step? Direct substitution — Lo ( $v$ ) comes first in each product, protecting the sign.

Step 3. Simplify the numerator: $= \frac{2 x ^{2} + 2 x - x ^{2}}{( x + 1 ) ^{2}} = \frac{x ^{2} + 2 x}{( x + 1 ) ^{2}} = \frac{x ( x + 2 )}{( x + 1 ) ^{2}} .$ Why this step? Factoring the top exposes the zeros of the slope: $x = 0$ and $x = - 2$ (cell G).

Step 4. At $x = 1$ : $f^{'} (1) = \frac{1 \cdot 3}{2 ^{2}} = \frac{3}{4} > 0$ . Why this step? Sign check — positive means the function is climbing there.

Verify: $f = \frac{x ^{2}}{x + 1}$ . Rewrite $x^{2} = (x + 1) (x - 1) + 1$ , so $f = x - 1 + \frac{1}{x + 1}$ , giving $f^{'} = 1 - \frac{1}{( x + 1 ) ^{2}}$ . At $x = 1$ : $1 - \frac{1}{4} = \frac{3}{4}$ . ✓ Matches.

E2 — Cell B: trig quotient (derive $sec^{2} x$ and $csc^{2} x$ )

Worked example Show

$\frac{d}{d x} tan x = sec^{2} x$ and $\frac{d}{d x} cot x = - csc^{2} x$ . Forecast: Both should be "1 over a squared trig", but the signs will differ — why? (Hint: which one is decreasing?)

Step 1. For $tan x = \frac{sin x}{cos x}$ : $u = sin x, v = cos x$ , so $u^{'} = cos x, v^{'} = - sin x$ . Why this step? We need the derivatives of the trig pieces — see Derivatives of trig functions.

Step 2. Apply the rule: $\frac{d}{d x} tan x = \frac{c o s x c o s x - s i n x ( - s i n x )}{c o s ^{2} x} = \frac{c o s ^{2} x + s i n ^{2} x}{c o s ^{2} x} .$ Why this step? The minus times minus flips $- sin x$ into $+ sin^{2} x$ — this is where the identity will kick in.

Step 3. Use $cos^{2} x + sin^{2} x = 1$ : $= \frac{1}{cos ^{2} x} = sec^{2} x$ . Always $\geq 1$ , always positive — $tan$ never decreases.

Step 4. For $cot x = \frac{cos x}{sin x}$ : $u = cos x, v = sin x$ , $u^{'} = - sin x, v^{'} = cos x$ : $\frac{- s i n x s i n x - c o s x c o s x}{s i n ^{2} x} = \frac{- ( s i n ^{2} x + c o s ^{2} x )}{s i n ^{2} x} = - \frac{1}{s i n ^{2} x} = - csc^{2} x .$ Why the minus? Here both terms in the numerator are negative — $cot$ is a decreasing function everywhere it is defined.

Verify: at $x = \frac{π}{4}$ , $sec^{2} \frac{π}{4} = (2)^{2} = 2$ and $- csc^{2} \frac{π}{4} = - 2$ . ✓

E3 — Cell C + D: reciprocal and constant-on-top

E4 — Cell E + G: reading the sign of the slope in every region

Worked example For

$g (x) = \frac{x}{x ^{2} + 1}$ , find $g^{'}$ , and classify the sign of the slope in each region of $x$ . Forecast: This function rises then falls (it peaks). Predict: slope positive for small $∣ x ∣$ , negative for large $∣ x ∣$ , zero somewhere.

Step 1. $u = x, v = x^{2} + 1$ , so $u^{'} = 1, v^{'} = 2 x$ : $g^{'} (x) = \frac{1 \cdot ( x ^{2} + 1 ) - x \cdot 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{x ^{2} + 1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} .$ Why this step? After simplifying, the denominator is always $> 0$ , so the sign of $g^{'}$ is decided entirely by the top $1 - x^{2}$ .

Step 2. Solve $1 - x^{2} = 0 \Rightarrow x = \pm 1$ (cell G, flat tangents). Why this step? These split the number line into the three regions we forecast.

Step 3. Test each region (cell E — every sign):

Region	sample $x$	$1 - x^{2}$	slope
$x < - 1$	$- 2$	$- 3$	negative (falling)
$- 1 < x < 1$	$0$	$+ 1$	positive (rising)
$x > 1$	$2$	$- 3$	negative (falling)

Why this step? This is the whole "quadrant coverage" idea: we do not stop at one point, we cover every sign the slope can take.

Verify: $g^{'} (0) = \frac{1}{1} = 1 > 0$ ✓; $g^{'} (2) = \frac{1 - 4}{5 ^{2}} = \frac{- 3}{25} < 0$ ✓; $g^{'} (1) = 0$ ✓ (peak at $x = 1$ ).

E5 — Cell F: exponential over polynomial + limiting behaviour

Worked example Differentiate

$h (x) = \frac{e ^{x}}{x}$ (for $x \neq = 0$ ) and describe $h^{'}$ as $x \to + \infty$ . Forecast: $e^{x}$ grows insanely fast, so eventually the top wins everything. Predict $h^{'}$ becomes large and positive.

Step 1. $u = e^{x}, v = x$ , so $u^{'} = e^{x}, v^{'} = 1$ : $h^{'} (x) = \frac{e ^{x} \cdot x - e ^{x} \cdot 1}{x ^{2}} = \frac{e ^{x} ( x - 1 )}{x ^{2}} .$ Why this step? Factoring $e^{x}$ out of the numerator exposes the sign-controlling factor $(x - 1)$ .

Step 2. Sign map (another all-signs sweep):

$0 < x < 1$ : $x - 1 < 0 \Rightarrow h^{'} < 0$ (falling),
$x > 1$ : $h^{'} > 0$ (rising, minimum at $x = 1$ ),
$x < 0$ : $x^{2} > 0$ , $e^{x} > 0$ , $x - 1 < 0 \Rightarrow h^{'} < 0$ (falling).

Why this step? Covers the negative- $x$ branch too — do not forget the region left of the missing point.

Step 3. Limit $x \to + \infty$ : $h^{'} = e^{x} \cdot \frac{x - 1}{x ^{2}} \to + \infty$ because $e^{x}$ dominates any power of $x$ . Why this step? This is the "limiting behaviour" cell — the exponential outruns the polynomial.

Verify: at $x = 1$ , $h^{'} (1) = \frac{e ^{1} ( 0 )}{1} = 0$ ✓ (the minimum). At $x = 2$ , $h^{'} (2) = \frac{e ^{2} ( 1 )}{4} = \frac{e ^{2}}{4} \approx 1.847 > 0$ ✓.

E6 — Cell H: the degenerate case $v (x) = 0$

Worked example What is

$\frac{d}{d x} \frac{x - 3}{x - 2}$ at $x = 2$ ? Forecast: Careful — the quotient rule requires $v (x) \neq = 0$ . At $x = 2$ the bottom is zero. Predict: the derivative does not exist there because the function itself does not exist there.

Step 1. $u = x - 3, v = x - 2$ , so $u^{'} = 1, v^{'} = 1$ . For $x \neq = 2$ : $\frac{d}{d x} \frac{x - 3}{x - 2} = \frac{1 \cdot ( x - 2 ) - ( x - 3 ) \cdot 1}{( x - 2 ) ^{2}} = \frac{( x - 2 ) - ( x - 3 )}{( x - 2 ) ^{2}} = \frac{1}{( x - 2 ) ^{2}} .$ Why this step? Valid everywhere except $x = 2$ .

Step 2. At $x = 2$ : $v (2) = 0$ , so the hypothesis of the rule fails. The function has a vertical asymptote there; there is no tangent line, hence no derivative. Why this step? This is the honesty cell — a rule with a precondition. Never blindly plug $x = 2$ into $\frac{1}{( x - 2 ) ^{2}}$ and call it "infinity"; the correct statement is undefined / does not exist.

Verify: for $x \neq = 2$ , $\frac{1}{( x - 2 ) ^{2}} > 0$ always — the function is increasing on both branches. At $x = 3$ , $h^{'} (3) = \frac{1}{1} = 1 > 0$ ✓. At $x = 2$ : undefined (both $f$ and $f^{'}$ ). ✓

E7 — Cell I: real-world word problem (concentration)

Worked example A tank starts with

$2$ kg of salt in $10$ L of water. Fresh salt is added at $t$ kg total and water at $t$ L total, so the concentration is $C (t) = \frac{2 + t}{10 + t} (kg per L) .$ How fast is concentration changing at $t = 0$ ? Forecast: Salt fraction $\frac{2}{10} = 0.2$ is below the fraction being added ( $\frac{1}{1}$ ), so concentration should be rising — expect $C^{'} > 0$ .

Step 1. $u = 2 + t, v = 10 + t$ , so $u^{'} = 1, v^{'} = 1$ : $C^{'} (t) = \frac{1 \cdot ( 10 + t ) - ( 2 + t ) \cdot 1}{( 10 + t ) ^{2}} = \frac{( 10 + t ) - ( 2 + t )}{( 10 + t ) ^{2}} = \frac{8}{( 10 + t ) ^{2}} .$ Why this step? The $t$ 's cancel in the top, leaving a constant $8$ — the "gap" between what's inside and what's being added.

Step 2. At $t = 0$ : $C^{'} (0) = \frac{8}{1 0 ^{2}} = \frac{8}{100} = 0.08$ kg/L per unit time. Why this step? Positive, as forecast — concentration climbs because we add saltier stuff than the current mix.

Step 3. As $t \to \infty$ , $C^{'} \to 0$ and $C \to 1$ : the concentration flattens toward the incoming ratio. Why this step? Limiting behaviour — the tank "forgets" its start.

Verify: $C (0) = 0.2$ , $C (1) = \frac{3}{11} \approx 0.2727$ ; average rate $\approx 0.0727$ , close to instantaneous $0.08$ at $t = 0$ (rate is dropping). Units: $\frac{kg/L}{time}$ ✓. $C^{'} (0) = 0.08$ ✓.

E8 — Cell J: exam twist (quotient + chain rule)

Worked example Differentiate

$k (x) = \frac{x}{x ^{2} + 1}$ . Forecast: The bottom is a composite, so we need Chain rule to differentiate $v$ . Predict a clean result since this is a well-known "sigmoid-shaped" function.

Step 1. $u = x \Rightarrow u^{'} = 1$ . $v = (x^{2} + 1)^{1/2}$ ; by the chain rule $v^{'} = \frac{1}{2} (x^{2} + 1)^{- 1/2} \cdot 2 x = \frac{x}{x ^{2} + 1}$ . Why this step? $v$ is not elementary — differentiate the outside power, then multiply by the inside's derivative $2 x$ .

Step 2. Quotient rule: $k^{'} (x) = \frac{1 \cdot x ^{2} + 1 - x \cdot \frac{x}{x ^{2} + 1}}{( x ^{2} + 1 ) ^{2}} = \frac{x ^{2} + 1 - \frac{x ^{2}}{x ^{2} + 1}}{x ^{2} + 1} .$ Why this step? Denominator squared removes the root: $(x^{2} + 1)^{2} = x^{2} + 1$ .

Step 3. Combine the top over $x^{2} + 1$ : $numerator = \frac{( x ^{2} + 1 ) - x ^{2}}{x ^{2} + 1} = \frac{1}{x ^{2} + 1} .$ So $k^{'} (x) = \frac{1}{x ^{2} + 1} \cdot \frac{1}{x ^{2} + 1} = \frac{1}{( x ^{2} + 1 ) ^{3/2}} .$ Why this step? Always positive — $k$ is increasing everywhere, matching its S-curve shape.

Verify: at $x = 0$ , $k^{'} (0) = \frac{1}{1 ^{3/2}} = 1$ ✓; at $x = 1$ , $k^{'} (1) = \frac{1}{2 ^{3/2}} = \frac{1}{2 2} \approx 0.3536$ ✓.

Recall Self-test: which cell is each problem?

Match to the matrix without looking. $\frac{e ^{x}}{x}$ ::: Cell F (exponential over polynomial, limiting behaviour). $\frac{1}{x ^{2} + 1}$ ::: Cell C (reciprocal, $u^{'} = 0$ ). $\frac{x - 3}{x - 2}$ at $x = 2$ ::: Cell H (degenerate, $v = 0$ , undefined). $\frac{x}{x ^{2} + 1}$ ::: Cell J (needs the chain rule inside).

Connections

Quotient rule — proof — the parent result we apply throughout.
Derivatives of trig functions — supplies $u^{'}, v^{'}$ in E2.
Chain rule — needed for the composite denominator in E8.
Limit definition of derivative — where the rule itself comes from.
Product rule — proof — the twin rule; recall $u / v = u \cdot v^{- 1}$ .
Continuity — why the degenerate $v = 0$ case (E6) breaks the tangent.