Quotient rule — proof: Exercises — Maths deep dive

Intuition How to use this page

Each problem sits inside a collapsible Solution callout — try it first with the page covered, then reveal. The difficulty climbs in five steps:

L1 Recognition — can you spot the pieces $u, v, u^{'}, v^{'}$ ?
L2 Application — plug into the rule and simplify cleanly.
L3 Analysis — what does the derivative tell you (signs, zeros, behaviour)?
L4 Synthesis — combine with the Chain rule, Product rule — proof, and Derivatives of trig functions.
L5 Mastery — proof-level and edge-case reasoning.

The one rule we lean on throughout, from the parent proof: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}, v \neq = 0.$ In words: "Lo dee-Hi minus Hi dee-Lo, over Lo-Lo." Here $u$ is the top ("Hi"), $v$ is the bottom ("Lo"), and a prime $^{'}$ means "the derivative of", i.e. the instantaneous rate of change. The final "Lo-Lo" is shorthand for $v^{2}$ — the bottom multiplied by itself, not two separate factors.

Level 1 — Recognition

These ask only: identify the four ingredients and place them.

Exercise 1.1

For $f (x) = \frac{3 x + 2}{x - 5}$ , name $u, v, u^{'}, v^{'}$ . Do not simplify a final answer yet.

Recall Solution

The top is $u = 3 x + 2$ ; the bottom is $v = x - 5$ .

$u^{'} = 3$ (the slope of the straight line $3 x + 2$ ).
$v^{'} = 1$ (the slope of $x - 5$ ). Placed into the rule: $f^{'} (x) = \frac{( 3 ) ( x - 5 ) - ( 3 x + 2 ) ( 1 )}{( x - 5 ) ^{2}} .$ WHAT we did: matched each piece to a slot. WHY: every quotient-rule mistake starts with mis-labelling — get this right and the algebra is mechanical. Domain: valid only for $x \neq = 5$ , since there the bottom $v = x - 5$ is zero and $f$ is undefined.

Exercise 1.2

For $g (x) = \frac{cos x}{x ^{2}}$ , write down $u, v, u^{'}, v^{'}$ .

Recall Solution

$u = cos x, v = x^{2}$ . Then $u^{'} = - sin x$ (from Derivatives of trig functions) and $v^{'} = 2 x$ . $g^{'} (x) = \frac{( - s i n x ) ( x ^{2} ) - ( c o s x ) ( 2 x )}{( x ^{2} ) ^{2}} .$ WHAT we did: slotted a trig top over a polynomial bottom. WHY it matters: recognising $u, v$ works identically whether the pieces are trig, polynomial, or exponential — the type of function never changes which slot it goes in, only how you differentiate it. Domain: valid only for $x \neq = 0$ , since there the bottom $v = x^{2}$ is zero and $g$ is undefined.

Level 2 — Application

Now: apply the rule and simplify fully.

Exercise 2.1

Differentiate $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ and simplify.

Recall Solution

$u = x^{2} - 1, v = x^{2} + 1$ ; $u^{'} = 2 x, v^{'} = 2 x$ . $f^{'} (x) = \frac{( 2 x ) ( x ^{2} + 1 ) - ( x ^{2} - 1 ) ( 2 x )}{( x ^{2} + 1 ) ^{2}} .$ WHAT next: expand the numerator carefully. $= \frac{2 x ^{3} + 2 x - 2 x ^{3} + 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{4 x}{( x ^{2} + 1 ) ^{2}} .$ WHY the cancellation is nice: the $2 x^{3}$ terms kill each other, leaving a clean $4 x$ . So $f^{'} = 0$ only at $x = 0$ . Domain: all real $x$ — here $v = x^{2} + 1 \geq 1$ is never zero, so no points are excluded.

Exercise 2.2

Differentiate $h (x) = \frac{e ^{x}}{x}$ (take $\frac{d}{d x} e^{x} = e^{x}$ as known).

Recall Solution

$u = e^{x}, v = x$ ; $u^{'} = e^{x}, v^{'} = 1$ . $h^{'} (x) = \frac{e ^{x} \cdot x - e ^{x} \cdot 1}{x ^{2}} = \frac{e ^{x} ( x - 1 )}{x ^{2}} .$ WHY factor $e^{x}$ : it shows the sign of $h^{'}$ is controlled by $(x - 1)$ , since $e^{x} > 0$ and $x^{2} > 0$ always. So $h$ decreases for $x < 1$ , increases for $x > 1$ . Domain: valid only for $x \neq = 0$ , since there the bottom $v = x$ is zero and $h$ is undefined.

Exercise 2.3

Differentiate $r (x) = \frac{1}{x ^{2} + 4}$ .

Recall Solution

Use the reciprocal special case (Example 3 of the parent): $u = 1, u^{'} = 0$ , so $(\frac{1}{v})^{'} = - \frac{v ^{'}}{v ^{2}}$ . Here $v = x^{2} + 4, v^{'} = 2 x$ : $r^{'} (x) = - \frac{2 x}{( x ^{2} + 4 ) ^{2}} .$ WHY this shortcut is legal: setting $u = 1$ inside the full quotient rule kills the $u^{'} v$ term, leaving exactly $- v^{'} / v^{2}$ . Domain: all real $x$ — here $v = x^{2} + 4 \geq 4$ is never zero, so no points are excluded.

Level 3 — Analysis

Now use the derivative to answer a question about the function.

Exercise 3.1

For $f (x) = \frac{x}{x ^{2} + 1}$ , find all $x$ where the tangent line is horizontal.

Recall Solution

$u = x, v = x^{2} + 1$ ; $u^{'} = 1, v^{'} = 2 x$ . $f^{'} (x) = \frac{( 1 ) ( x ^{2} + 1 ) - ( x ) ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{x ^{2} + 1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} .$ A horizontal tangent means slope $= 0$ , i.e. numerator $= 0$ : $1 - x^{2} = 0 \Rightarrow x = \pm 1$ . Check the sign of $f^{'}$ : for $∣ x ∣ < 1$ numerator $> 0$ (rising), for $∣ x ∣ > 1$ numerator $< 0$ (falling). So $x = 1$ is a maximum, $x = - 1$ a minimum. Domain: all real $x$ — here $v = x^{2} + 1 \geq 1$ is never zero.

In the figure below, the solid curve is $f$ ; the pink dashed line marks the horizontal tangent at $x = 1$ and the blue dashed line the one at $x = - 1$ . Notice both dashed lines are perfectly flat (slope $0$ ), and the curve rises between them and falls outside them — exactly matching the sign analysis.

Exercise 3.2

At $x = 1$ , is $f (x) = \frac{x ^{2}}{x + 1}$ (from the parent's Example 1) increasing or decreasing, and by how much?

Recall Solution

Parent found $f^{'} (x) = \frac{x ( x + 2 )}{( x + 1 ) ^{2}}$ . Evaluate at $x = 1$ : $f^{'} (1) = \frac{1 \cdot 3}{2 ^{2}} = \frac{3}{4} .$ This is positive, so $f$ is increasing at $x = 1$ with slope $\frac{3}{4}$ . WHY it matters: the parent's "backwards subtraction" mistake would give $- \frac{3}{4}$ — same size, wrong sign. Checking the sign of a known-increasing function catches that error instantly. Domain: valid only for $x \neq = - 1$ , where the bottom $v = x + 1$ is zero.

Level 4 — Synthesis

Combine the quotient rule with other rules.

Exercise 4.1

Differentiate $y = \frac{sin x}{1 + cos x}$ and simplify using $sin^{2} x + cos^{2} x = 1$ .

Recall Solution

$u = sin x, v = 1 + cos x$ ; $u^{'} = cos x, v^{'} = - sin x$ . $y^{'} = \frac{c o s x ( 1 + c o s x ) - s i n x ( - s i n x )}{( 1 + c o s x ) ^{2}} = \frac{c o s x + c o s ^{2} x + s i n ^{2} x}{( 1 + c o s x ) ^{2}} .$ Use $cos^{2} x + sin^{2} x = 1$ : $= \frac{c o s x + 1}{( 1 + c o s x ) ^{2}} = \frac{1}{1 + c o s x} .$ WHY the collapse: the numerator becomes exactly one factor of the denominator, so one $(1 + cos x)$ cancels. Domain: valid wherever $1 + cos x \neq = 0$ , i.e. $x \neq = π + 2 π k$ (where $cos x = - 1$ ).

Exercise 4.2

Differentiate $F (x) = \frac{( 2 x + 1 ) ^{3}}{x}$ , using the Chain rule on the top.

Recall Solution

$u = (2 x + 1)^{3}$ . Chain rule on $u$ : outer power $3 (\cdot)^{2}$ times inner derivative $2$ , so $u^{'} = 3 (2 x + 1)^{2} \cdot 2 = 6 (2 x + 1)^{2}$ . $v = x, v^{'} = 1$ . $F^{'} (x) = \frac{6 ( 2 x + 1 ) ^{2} \cdot x - ( 2 x + 1 ) ^{3} \cdot 1}{x ^{2}} .$ WHAT next: pull out the common factor $(2 x + 1)^{2}$ from the numerator: $= \frac{( 2 x + 1 ) ^{2} [ 6 x - ( 2 x + 1 ) ]}{x ^{2}} = \frac{( 2 x + 1 ) ^{2} ( 4 x - 1 )}{x ^{2}} .$ WHY factor before expanding: it reveals slope zeros at $x = - \frac{1}{2}$ (double) and $x = \frac{1}{4}$ without brute-force algebra. Domain: valid only for $x \neq = 0$ , where the bottom $v = x$ is zero.

Exercise 4.3

Show that differentiating $\frac{u}{v}$ gives the same answer as the product rule applied to $u \cdot v^{- 1}$ .

Recall Solution

Write $f = u \cdot v^{- 1}$ . Product rule: $f^{'} = u^{'} \cdot v^{- 1} + u \cdot (v^{- 1})^{'}$ . By the chain rule, $(v^{- 1})^{'} = - v^{- 2} v^{'} = - \frac{v ^{'}}{v ^{2}}$ . So $f^{'} = \frac{u ^{'}}{v} - \frac{u v ^{'}}{v ^{2}} .$ Put over the common denominator $v^{2}$ : $= \frac{u ^{'} v - u v ^{'}}{v ^{2}} .$ This is exactly the quotient rule. WHY it's satisfying: the quotient rule isn't a new law — it's the product rule plus "one over $v$ ". Domain: valid wherever $v \neq = 0$ , since $v^{- 1}$ requires it.

Level 5 — Mastery

Proof-level and degenerate-case reasoning.

Exercise 5.1

Recall from Limit definition of derivative that the proof uses a small nudge $h$ in the input, forming $\frac{f ( x + h ) - f ( x )}{h}$ and letting $h \to 0$ . The proof's final step needs $v (x + h) \to v (x)$ as this nudge shrinks. Which property of $v$ guarantees this, and where would the argument break if $v$ lacked it? Give a one-line concrete example.

Recall Solution

What $h$ is here: $h$ is a tiny step in $x$ ; " $h \to 0$ " means we shrink that step toward nothing, so the point $x + h$ slides back onto $x$ . The property that forces $v (x + h) \to v (x)$ as $x + h \to x$ is continuity of $v$ ; differentiability implies it. It makes the denominator $v (x + h) v (x) \to v (x)^{2}$ . If $v$ were discontinuous at $x$ : say $v$ jumps, so $v (x + h) \to L \neq = v (x)$ . Then the denominator tends to $L v (x)$ , not $v (x)^{2}$ , and the "clean" formula is simply false there. Example: $v (x) = {1, 2, x < 0 x \geq 0$ is discontinuous at $0$ — no derivative exists there, so the quotient rule cannot apply.

Exercise 5.2

Use the quotient rule to derive $\frac{d}{d x} cot x$ where $cot x = \frac{cos x}{sin x}$ . Then state where the result is undefined.

Recall Solution

$u = cos x, v = sin x$ ; $u^{'} = - sin x, v^{'} = cos x$ . $\frac{d}{d x} cot x = \frac{( - s i n x ) ( s i n x ) - ( c o s x ) ( c o s x )}{s i n ^{2} x} = \frac{- s i n ^{2} x - c o s ^{2} x}{s i n ^{2} x} .$ Numerator $= - (sin^{2} x + cos^{2} x) = - 1$ : $= \frac{- 1}{s i n ^{2} x} = - csc^{2} x .$ Undefined where $sin x = 0$ , i.e. $x = k π$ (there $cot x$ itself blows up).

Exercise 5.3

For $f (x) = \frac{u ( x )}{v ( x )}$ with $u (a) = 0$ and $v (a) \neq = 0$ , show the tangent slope at $x = a$ reduces to $\frac{u ^{'} ( a )}{v ( a )}$ . Interpret geometrically.

Recall Solution

General: $f^{'} (a) = \frac{u ^{'} ( a ) v ( a ) - u ( a ) v ^{'} ( a )}{v ( a ) ^{2}}$ . Since $u (a) = 0$ , the term $u (a) v^{'} (a) = 0$ vanishes: $f^{'} (a) = \frac{u ^{'} ( a ) v ( a )}{v ( a ) ^{2}} = \frac{u ^{'} ( a )}{v ( a )} .$ Geometric meaning: at a root of the top, the fraction crosses zero, and near there the denominator acts like a constant $v (a)$ . So the curve locally looks like the straight line $\frac{u ^{'} ( a )}{v ( a )} (x - a)$ — the bottom's change $v^{'}$ doesn't matter at that instant because it's multiplied by the (zero) height of the fraction.

In the figure below the solid curve is $f = (x - 1) / (x^{2} + 1)$ , which has $u (a) = 0$ at $a = 1$ (marked). Watch the pink dashed tangent touch the curve exactly where it crosses the $x$ -axis: its slope is $u^{'} (1) / v (1) = 1/2$ . Even though the denominator $x^{2} + 1$ is changing there, that change leaves no fingerprint on the slope — precisely because $u (a) = 0$ .

Exercise 5.4

Verify the quotient rule cannot be replaced by $\frac{u ^{'}}{v ^{'}}$ using $f (x) = \frac{x ^{3}}{x ^{2}}$ .

Recall Solution

Truthfully $f (x) = \frac{x ^{3}}{x ^{2}} = x$ (for $x \neq = 0$ ), so $f^{'} (x) = 1$ . The wrong "term-wise" rule: $\frac{u ^{'}}{v ^{'}} = \frac{3 x ^{2}}{2 x} = \frac{3 x}{2}$ , which is not constant — clearly wrong. Correct quotient rule: $\frac{( 3 x ^{2} ) ( x ^{2} ) - ( x ^{3} ) ( 2 x )}{( x ^{2} ) ^{2}} = \frac{3 x ^{4} - 2 x ^{4}}{x ^{4}} = \frac{x ^{4}}{x ^{4}} = 1$ . ✓

Connections

Quotient rule — proof — the parent; every solution here plugs into its formula.
Product rule — proof — Exercise 4.3 rebuilds the quotient rule from it.
Chain rule — Exercises 4.2 and 4.3 use it on the top / on $v^{- 1}$ .
Limit definition of derivative — the engine behind Exercise 5.1's reasoning.
Continuity — justifies $v (x + h) \to v (x)$ in Exercise 5.1.
Derivatives of trig functions — Exercises 1.2, 4.1, 5.2 draw on $sin^{'}, cos^{'}$ .

Exercises — Quotient rule — proof

Level 1 — Recognition

Exercise 1.1

Exercise 1.2

Level 2 — Application

Exercise 2.1

Exercise 2.2

Exercise 2.3

Level 3 — Analysis

Exercise 3.1

Exercise 3.2

Level 4 — Synthesis

Exercise 4.1

Exercise 4.2

Exercise 4.3

Level 5 — Mastery

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Connections

Concept Map