Quotient Rule

This lesson covers the quotient rule for differentiating one function divided by another, as required by the Edexcel A-Level Mathematics specification (9MA0). You also need to understand the connection between the quotient rule and the product rule.

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When to Use the Quotient Rule

Use the quotient rule when you need to differentiate y = u/v, where u and v are both functions of x and you cannot simplify the fraction into separate terms.

Examples where the quotient rule is needed:

• y = sin x / x
• y = eˣ / (x + 1)
• y = x² / (2x - 3)
• y = ln x / x

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The Quotient Rule Formula

If y = u/v, where u = u(x) and v = v(x), then:

dy/dx = (v(du/dx) - u(dv/dx)) / v²

Or in shorter notation:

dy/dx = (vu' - uv') / v²

In words: "(bottom × derivative of top - top × derivative of bottom) / bottom squared."

Exam Tip: The quotient rule is given in the formula booklet at A-Level. However, you should practise it enough that you can apply it quickly and accurately.

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Step-by-Step Method

1. Identify u (the numerator) and v (the denominator).
2. Find du/dx and dv/dx.
3. Apply the formula: dy/dx = (vu' - uv') / v².
4. Simplify the numerator (the denominator v² usually stays as is).

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Worked Examples

Worked Example 1

Differentiate y = x² / (x + 1).

Solution: u = x², v = x + 1 du/dx = 2x, dv/dx = 1

dy/dx = ((x + 1)(2x) - x²(1)) / (x + 1)² = (2x² + 2x - x²) / (x + 1)² = (x² + 2x) / (x + 1)² = x(x + 2) / (x + 1)²

Worked Example 2

Differentiate y = sin x / x.

Solution: u = sin x, v = x du/dx = cos x, dv/dx = 1

dy/dx = (x cos x - sin x × 1) / x² = (x cos x - sin x) / x²

Worked Example 3

Differentiate y = eˣ / (2x + 3).

Solution: u = eˣ, v = 2x + 3 du/dx = eˣ, dv/dx = 2

dy/dx = ((2x + 3)eˣ - eˣ(2)) / (2x + 3)² = eˣ(2x + 3 - 2) / (2x + 3)² = eˣ(2x + 1) / (2x + 3)²

Worked Example 4

Differentiate y = ln x / x².

Solution: u = ln x, v = x² du/dx = 1/x, dv/dx = 2x

dy/dx = (x²(1/x) - ln x (2x)) / (x²)² = (x - 2x ln x) / x⁴ = x(1 - 2 ln x) / x⁴ = (1 - 2 ln x) / x³

Exam Tip: After applying the quotient rule, simplify the numerator by expanding and collecting like terms. The denominator is almost always left as v². Factorise the numerator if possible.

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The Quotient Rule via the Product Rule

You can always rewrite u/v as u × v⁻¹ and use the product rule instead:

y = u × v⁻¹

dy/dx = u × d/dx(v⁻¹) + v⁻¹ × du/dx = u × (-v⁻²)(dv/dx) + (du/dx)/v = -u(dv/dx)/v² + (du/dx)/v = (v(du/dx) - u(dv/dx)) / v²

This gives the same result as the quotient rule. Some students prefer this approach as it avoids memorising a separate formula.

Worked Example 5

Differentiate y = x / eˣ using the product rule.

Solution: Rewrite as y = x × e⁻ˣ.

u = x, v = e⁻ˣ du/dx = 1, dv/dx = -e⁻ˣ

dy/dx = x(-e⁻ˣ) + e⁻ˣ(1) = -xe⁻ˣ + e⁻ˣ = e⁻ˣ(1 - x)

This is equivalent to: (eˣ × 1 - x × eˣ) / (eˣ)² = eˣ(1 - x) / e²ˣ = (1 - x)/eˣ = e⁻ˣ(1 - x) ✓

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Differentiating tan x

An important application of the quotient rule:

d/dx(tan x) = sec²x

Proof: tan x = sin x / cos x.

u = sin x, v = cos x du/dx = cos x, dv/dx = -sin x

dy/dx = (cos x × cos x - sin x × (-sin x)) / cos²x = (cos²x + sin²x) / cos²x = 1 / cos²x = sec²x ∎

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Finding Stationary Points

Worked Example 6

Find the stationary points of y = x² / eˣ.

Solution: u = x², v = eˣ du/dx = 2x, dv/dx = eˣ

dy/dx = (eˣ(2x) - x²(eˣ)) / (eˣ)² = eˣ(2x - x²) / e²ˣ = (2x - x²) / eˣ = x(2 - x) / eˣ

Set dy/dx = 0: x(2 - x)/eˣ = 0

Since eˣ > 0 for all x: x(2 - x) = 0 x = 0 or x = 2

When x = 0: y = 0/e⁰ = 0. Point: (0, 0) When x = 2: y = 4/e² ≈ 0.541. Point: (2, 4/e²)

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Finding Tangents

Worked Example 7

Find the equation of the tangent to y = (x + 1)/(x - 1) at x = 2.

Solution: u = x + 1, v = x - 1 du/dx = 1, dv/dx = 1

dy/dx = ((x - 1)(1) - (x + 1)(1)) / (x - 1)² = (x - 1 - x - 1) / (x - 1)² = -2 / (x - 1)²

At x = 2: dy/dx = -2/(1)² = -2 y = (2 + 1)/(2 - 1) = 3

Tangent: y - 3 = -2(x - 2) → y = -2x + 7

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Worked Example 8

Differentiate y = (3x + 2) / √(x + 1).

Solution: u = 3x + 2, v = (x + 1)^(1/2) du/dx = 3, dv/dx = (1/2)(x + 1)^(-1/2)

dy/dx = ((x + 1)^(1/2) × 3 - (3x + 2) × (1/2)(x + 1)^(-1/2)) / (x + 1)

Multiply numerator and denominator by (x + 1)^(1/2):

= (3(x + 1) - (3x + 2)/2) / (x + 1)^(3/2) = (3x + 3 - 3x/2 - 1) / (x + 1)^(3/2) = ((6x + 6 - 3x - 2) / 2) / (x + 1)^(3/2) = (3x + 4) / (2(x + 1)^(3/2))

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Common Mistakes

Mistake	Correction
Getting the numerator the wrong way round	It is vu' - uv', NOT uv' - vu'. The "v" (denominator) goes first
Forgetting to square the denominator	The denominator is always v²
Using the quotient rule when simplification would be easier	For y = x³/x, just simplify to y = x² first
Not factorising the numerator	Simplify after applying the rule to get cleaner results

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Summary

• The quotient rule: if y = u/v, then dy/dx = (vu' - uv') / v².
• Remember the order: "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."
• Alternatively, rewrite u/v as u × v⁻¹ and use the product rule.
• d/dx(tan x) = sec²x is derived from the quotient rule applied to sin x / cos x.
• After applying the quotient rule, always simplify the numerator.

A-Level Deep Dive: Quotient Rule

Spec mapping

Edexcel 9MA0 specification section 9 — Differentiation, sub-strand 9.6 (quotient rule) covers using the chain, product and quotient rules, including problems involving connected rates of change and inverse functions (refer to the official specification document for exact wording). The quotient rule $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$dxd​(vu​)=v2u′v−uv′​ is examined explicitly on Paper 1 — Pure Mathematics and is synoptic with section 5 (Trigonometry, where $\tan x = \sin x / \cos x$tanx=sinx/cosx allows derivation of $\dfrac{d}{dx}\tan x = \sec^2 x$dxd​tanx=sec2x), section 6 (Exponentials, with quotients such as $e^x/(x+1)$ex/(x+1)), section 2 (rational functions and their asymptotes) and section 9.7 (stationary points of rational functions). The quotient rule is derivable from the product rule combined with the chain rule applied to $v^{-1}$v−1, and Edexcel will occasionally test that derivation directly. The rule appears in the formula booklet, but candidates lose marks for misquoting the numerator order $u'v - uv'$u′v−uv′ (not $uv' - u'v$uv′−u′v).

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Worked example with full mark scheme

Question (8 marks):

The curve $C$C has equation $y = \dfrac{x^2 + 1}{2x - 3}$y=2x−3x2+1​ for $x \neq \tfrac{3}{2}$x=23​.

(a) Find $\dfrac{dy}{dx}$dxdy​, giving your answer as a single simplified fraction. (4)

(b) Hence find the $x$x-coordinates of the stationary points of $C$C, giving your answers in exact form. (4)

Solution with mark scheme:

(a) Step 1 — identify $u$u and $v$v.

Let $u = x^2 + 1$u=x2+1 so $u' = 2x$u′=2x. Let $v = 2x - 3$v=2x−3 so $v' = 2$v′=2.

M1 — correct identification of $u$u, $v$v and their derivatives. Common error: students differentiate the numerator and denominator separately and divide, producing $2x/2 = x$2x/2=x, which is wrong.

Step 2 — apply the quotient rule.

$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{2x(2x - 3) - (x^2 + 1)(2)}{(2x - 3)^2}$dxdy​=v2u′v−uv′​=(2x−3)22x(2x−3)−(x2+1)(2)​

M1 — correct quoted formula with correct substitution; the numerator must read $u'v - uv'$u′v−uv′ (not $uv' - u'v$uv′−u′v).

Step 3 — expand and simplify the numerator.

$2x(2x - 3) = 4x^2 - 6x; \qquad (x^2 + 1)(2) = 2x^2 + 2$2x(2x−3)=4x2−6x;(x2+1)(2)=2x2+2

So the numerator is $4x^2 - 6x - 2x^2 - 2 = 2x^2 - 6x - 2$4x2−6x−2x2−2=2x2−6x−2.

A1 — simplified numerator $2x^2 - 6x - 2$2x2−6x−2 (or equivalently $2(x^2 - 3x - 1)$2(x2−3x−1)).

Step 4 — present the simplified derivative.

$\dfrac{dy}{dx} = \dfrac{2(x^2 - 3x - 1)}{(2x - 3)^2}$dxdy​=(2x−3)22(x2−3x−1)​

A1 — final form with denominator left as $(2x - 3)^2$(2x−3)2 (do not expand the denominator; examiners specifically reward leaving $v^2$v2 in factored form because part (b) needs the numerator zero).

(b) Step 1 — set the numerator to zero.

Stationary points occur where $\dfrac{dy}{dx} = 0$dxdy​=0. Since $(2x - 3)^2 > 0$(2x−3)2>0 for $x \neq \tfrac{3}{2}$x=23​, this happens iff the numerator is zero:

$2(x^2 - 3x - 1) = 0 \implies x^2 - 3x - 1 = 0$2(x2−3x−1)=0⟹x2−3x−1=0

M1 — recognising that only the numerator need vanish, with the implicit comment that $(2x - 3)^2 \neq 0$(2x−3)2=0 on the domain.

Step 2 — solve the quadratic exactly.

The discriminant is $9 + 4 = 13$9+4=13, giving:

$x = \dfrac{3 \pm \sqrt{13}}{2}$x=23±13​​

M1 — correct use of the quadratic formula (or completion of the square).

A1 — both roots stated.

A1 (AO2.5) — answers given in exact form as the question demanded; a decimal answer here would lose this mark.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Given $f(x) = \dfrac{e^x}{x + 1}$f(x)=x+1ex​ for $x \neq -1$x=−1:

(a) Find $f'(x)$f′(x), simplifying your answer as far as possible. (3)

(b) Show that $f(x)$f(x) has a stationary point at $x = 0$x=0 and determine its nature using the second derivative or otherwise. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — quoting the quotient rule with $u = e^x$u=ex, $v = x + 1$v=x+1, $u' = e^x$u′=ex, $v' = 1$v′=1.
• M1 (AO1.1b) — substituting to obtain $\dfrac{e^x(x + 1) - e^x \cdot 1}{(x + 1)^2}$(x+1)2ex(x+1)−ex⋅1​.
• A1 (AO2.5) — factorising $e^x$ex from the numerator: $f'(x) = \dfrac{e^x \cdot x}{(x + 1)^2} = \dfrac{x e^x}{(x + 1)^2}$f′(x)=(x+1)2ex⋅x​=(x+1)2xex​.

(b)

• M1 (AO1.1b) — setting $f'(x) = 0$f′(x)=0: since $e^x > 0$ex>0 for all real $x$x and $(x + 1)^2 > 0$(x+1)2>0 for $x \neq -1$x=−1, we need $x = 0$x=0.
• M1 (AO2.1) — testing the sign of $f'(x)$f′(x) on either side of $x = 0$x=0 (or computing $f''(0)$f′′(0)): for $x$x slightly negative, $f'(x) < 0$f′(x)<0; for $x$x slightly positive, $f'(x) > 0$f′(x)>0.
• A1 (AO2.4) — concluding that $x = 0$x=0 is a local minimum, with $f(0) = 1$f(0)=1.

Total: 6 marks split AO1 = 3, AO2 = 3. This question is more reasoning-heavy than (a) suggests because of the sign-analysis demand in (b).

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Synoptic links

Connects to:

1. Section 9.5 — Product and chain rules: the quotient rule is derivable from the product rule applied to $u \cdot v^{-1}$u⋅v−1. Setting $y = u v^{-1}$y=uv−1 and applying the product rule with $\dfrac{d}{dx}(v^{-1}) = -v^{-2} v'$dxd​(v−1)=−v−2v′ (chain rule) gives $y' = u' v^{-1} - u v^{-2} v' = (u' v - u v')/v^2$y′=u′v−1−uv−2v′=(u′v−uv′)/v2. Edexcel sometimes asks candidates to derive the quotient rule from the product rule — full marks require both product and chain rule citations.

2. Section 5 — Trigonometric derivatives: the standard result $\dfrac{d}{dx}\tan x = \sec^2 x$dxd​tanx=sec2x is proved by writing $\tan x = \sin x / \cos x$tanx=sinx/cosx and applying the quotient rule: $\dfrac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x$cos2xcosx⋅cosx−sinx⋅(−sinx)​=cos2xcos2x+sin2x​=cos2x1​=sec2x. The same technique gives $\dfrac{d}{dx}\cot x = -\csc^2 x$dxd​cotx=−csc2x.

3. Section 2 — Rational functions and asymptotes: rational functions $y = p(x)/q(x)$y=p(x)/q(x) have vertical asymptotes where $q(x) = 0$q(x)=0 and the quotient rule's denominator $q(x)^2$q(x)2 inherits exactly those zeros. Stationary-point analysis for rational functions therefore lives entirely inside quotient-rule calculations.

4. Section 9.7 — Second derivatives: computing $f''(x)$f′′(x) for a quotient $f(x) = u/v$f(x)=u/v requires applying the quotient rule a second time to $f'(x) = (u'v - uv')/v^2$f′(x)=(u′v−uv′)/v2. The algebra grows quickly; A* candidates often factorise $v$v from numerator and denominator before the second differentiation to keep the expression manageable.

5. Section 8 — Integration techniques: integration by parts and integration of rational functions are the integral-calculus partners of the quotient rule. $\int \dfrac{f'(x)}{f(x)} dx = \ln|f(x)| + C$∫f(x)f′(x)​dx=ln∣f(x)∣+C is recognisable precisely because differentiating $\ln|f(x)|$ln∣f(x)∣ via the chain rule produces this quotient-shaped derivative.

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Mark-scheme literacy

Quotient rule questions on 9MA0 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Correct identification of $u$u and $v$v, accurate differentiation, correct substitution into $(u'v - uv')/v^2$(u′v−uv′)/v2
AO2 (reasoning / interpretation)	30–40%	Simplifying the numerator algebraically, leaving the denominator in factored form, identifying when the numerator-only condition determines stationary points
AO3 (problem-solving)	5–15%	Modelling rates of change of rational quantities; choosing between quotient rule and rewriting as a product with a negative index

Examiner-rewarded phrasing: "applying the quotient rule with $u = \ldots$u=… and $v = \ldots$v=…"; "since $v^2 > 0$v2>0 on the domain, stationary points occur iff the numerator vanishes"; "leaving the denominator as $(2x - 3)^2$(2x−3)2 to expose the zeros of the derivative". Phrases that lose marks: writing $u v' - u' v$uv′−u′v for the numerator (sign-reversed); placing $v$v in the denominator instead of $v^2$v2; expanding the denominator unnecessarily and obscuring the structure.