Product Rule

This lesson covers the product rule for differentiating the product of two functions, as required by the Edexcel A-Level Mathematics specification (9MA0). You need to know when to use the product rule and how to apply it correctly.

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When to Use the Product Rule

Use the product rule when you need to differentiate y = u × v, where u and v are both functions of x, and it is not practical to expand the expression first.

Examples where the product rule is needed:

• y = x² sin x (polynomial × trig)
• y = xeˣ (polynomial × exponential)
• y = (2x + 1) ln x (linear × logarithmic)
• y = eˣ cos x (exponential × trig)

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The Product Rule Formula

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

dy/dx = u(dv/dx) + v(du/dx)

Or in shorter notation:

dy/dx = uv' + vu'

In words: "(first × derivative of second) + (second × derivative of first)."

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Step-by-Step Method

1. Identify u and v (the two factors).
2. Find du/dx and dv/dx.
3. Apply the formula: dy/dx = u(dv/dx) + v(du/dx).
4. Simplify the result (often by factorising).

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Worked Examples with Polynomials and Trig

Worked Example 1

Differentiate y = x² sin x.

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

dy/dx = x² cos x + sin x × 2x = x² cos x + 2x sin x

This can be factorised: x(x cos x + 2 sin x)

Worked Example 2

Differentiate y = x³ cos x.

Solution: u = x³, v = cos x du/dx = 3x², dv/dx = -sin x

dy/dx = x³(-sin x) + cos x(3x²) = -x³ sin x + 3x² cos x

Factorised: x²(3 cos x - x sin x)

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Worked Examples with Exponentials

Worked Example 3

Differentiate y = xeˣ.

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

dy/dx = xeˣ + eˣ × 1 = xeˣ + eˣ = eˣ(x + 1)

Worked Example 4

Differentiate y = x²e³ˣ.

Solution: u = x², v = e³ˣ du/dx = 2x, dv/dx = 3e³ˣ

dy/dx = x² × 3e³ˣ + e³ˣ × 2x = 3x²e³ˣ + 2xe³ˣ = xe³ˣ(3x + 2)

Worked Example 5

Differentiate y = e²ˣ sin x.

Solution: u = e²ˣ, v = sin x du/dx = 2e²ˣ, dv/dx = cos x

dy/dx = e²ˣ cos x + sin x × 2e²ˣ = e²ˣ(cos x + 2 sin x)

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Worked Examples with Logarithms

Worked Example 6

Differentiate y = x ln x.

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

dy/dx = x × (1/x) + ln x × 1 = 1 + ln x = 1 + ln x

Worked Example 7

Differentiate y = x² ln x.

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

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

Exam Tip: After applying the product rule, always look for common factors to simplify. Factorising your answer makes it cleaner and is often expected in the mark scheme.

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Finding Stationary Points Using the Product Rule

Worked Example 8

Find the coordinates of the stationary point of y = xe⁻ˣ.

Solution: 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)

Set dy/dx = 0: e⁻ˣ(1 - x) = 0

Since e⁻ˣ > 0 for all x, we need 1 - x = 0, so x = 1.

When x = 1: y = 1 × e⁻¹ = 1/e ≈ 0.368

Stationary point: (1, 1/e)

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Finding Tangents Using the Product Rule

Worked Example 9

Find the equation of the tangent to y = x sin x at x = π/2.

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

dy/dx = x cos x + sin x

At x = π/2: y = (π/2) sin(π/2) = (π/2)(1) = π/2 dy/dx = (π/2) cos(π/2) + sin(π/2) = (π/2)(0) + 1 = 1

Tangent: y - π/2 = 1(x - π/2) y = x

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Product of Three Functions

If y = uvw (three functions), the product rule extends:

dy/dx = u'vw + uv'w + uvw'

Worked Example 10

Differentiate y = x² eˣ sin x.

Solution: u = x², v = eˣ, w = sin x u' = 2x, v' = eˣ, w' = cos x

dy/dx = 2x × eˣ × sin x + x² × eˣ × sin x + x² × eˣ × cos x = x eˣ(2 sin x + x sin x + x cos x) = xeˣ(2 sin x + x sin x + x cos x)

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Common Mistakes

Mistake	Correction
dy/dx = (du/dx)(dv/dx)	WRONG — you do not multiply the derivatives. The product rule is uv' + vu'
Forgetting to factorise the answer	Factorising is not essential for correctness, but is expected for full marks on "show that" questions
Using the product rule when expanding would be simpler	For y = (x + 1)(x + 2), just expand to x² + 3x + 2 and differentiate
Choosing u and v incorrectly	Any choice of u and v gives the correct answer — but choosing wisely makes simplification easier

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Summary

• The product rule: if y = uv, then dy/dx = uv' + vu'.
• Use it when differentiating the product of two (or more) functions that cannot be easily expanded.
• Always factorise your answer where possible.
• The product rule combines naturally with the chain rule (e.g., differentiating x² e³ˣ requires both).

A-Level Deep Dive: Product Rule

Spec mapping

Edexcel 9MA0 specification section 9 — Differentiation, sub-strand 9.1 (product rule) covers using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions (refer to the official specification document for exact wording). The product rule $\dfrac{d}{dx}(uv) = u'v + uv'$dxd​(uv)=u′v+uv′ is foundational and is examined across 9MA0-01 (Pure Mathematics 1) and 9MA0-02 (Pure Mathematics 2), frequently in synoptic combination with section 6 (exponentials and logarithms — differentiating $x e^x$xex, $x \ln x$xlnx), section 5 (trigonometry — differentiating $x \sin x$xsinx, $x^2 \cos x$x2cosx) and section 8 (integration — integration by parts as the reverse direction). Edexcel's formula booklet does list the product, quotient and chain rules, so candidates need not memorise the formula — but they must apply it without arithmetic slips under exam pressure.

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Worked example with full mark scheme

Question (8 marks): The curve $C$C has equation $y = x^2 \ln(2x + 1)$y=x2ln(2x+1) for $x > -\tfrac{1}{2}$x>−21​.

(a) Find $\dfrac{dy}{dx}$dxdy​, simplifying your answer where possible. (5)

(b) Hence find the $x$x-coordinate of the stationary point of $C$C for $x \geq 0$x≥0, justifying your answer. (3)

Solution with mark scheme:

(a) Step 1 — identify $u$u and $v$v.

Let $u = x^2$u=x2 and $v = \ln(2x + 1)$v=ln(2x+1). Then by the product rule $\dfrac{dy}{dx} = u'v + uv'$dxdy​=u′v+uv′.

M1 — clear identification of $u$u and $v$v and the product rule structure. Examiners reward written labelling here; candidates who launch into differentiation without naming the factors often slip on which derivative belongs to which factor.

Step 2 — differentiate each factor.

$u' = 2x$u′=2x by the power rule.

$v' = \dfrac{2}{2x + 1}$v′=2x+12​ by the chain rule (derivative of $\ln(2x + 1)$ln(2x+1) is $\dfrac{1}{2x + 1}$2x+11​ multiplied by the inner derivative $2$2).

M1 — correct chain-rule application to $\ln(2x + 1)$ln(2x+1). Common error: writing $v' = \dfrac{1}{2x + 1}$v′=2x+11​ and forgetting the factor of $2$2 from the inner derivative. That loses M1 and propagates into the final A1.

A1 — both derivatives $u' = 2x$u′=2x and $v' = \dfrac{2}{2x + 1}$v′=2x+12​ correct.

Step 3 — assemble the product rule.

$\dfrac{dy}{dx} = 2x \cdot \ln(2x + 1) + x^2 \cdot \dfrac{2}{2x + 1}$dxdy​=2x⋅ln(2x+1)+x2⋅2x+12​

M1 — substitution into $u'v + uv'$u′v+uv′ in the correct order.

Step 4 — simplify.

$\dfrac{dy}{dx} = 2x \ln(2x + 1) + \dfrac{2x^2}{2x + 1}$dxdy​=2xln(2x+1)+2x+12x2​

A common neat form factors out $2x$2x:

$\dfrac{dy}{dx} = 2x\left[\ln(2x + 1) + \dfrac{x}{2x + 1}\right]$dxdy​=2x[ln(2x+1)+2x+1x​]

A1 — correct simplified expression. Either factored or unfactored form is acceptable as final answer.

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

From the factored form, $\dfrac{dy}{dx} = 0$dxdy​=0 requires $2x = 0$2x=0 or $\ln(2x + 1) + \dfrac{x}{2x + 1} = 0$ln(2x+1)+2x+1x​=0.

M1 — recognising the factorisation route and isolating the bracketed equation.

Step 2 — analyse the second factor.

For $x > 0$x>0: $\ln(2x + 1) > 0$ln(2x+1)>0 (since $2x + 1 > 1$2x+1>1) and $\dfrac{x}{2x + 1} > 0$2x+1x​>0. So the bracket is strictly positive on $x > 0$x>0.

M1 — sign analysis showing no positive root.

A1 — concluding that the unique stationary point on $x \geq 0$x≥0 is at $x = 0$x=0.

Total: 8 marks (M5 A3, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Given that $y = (2x + 1)(x^2 - 3)^4$y=(2x+1)(x2−3)4:

(a) Use the product rule and the chain rule to show that $\dfrac{dy}{dx} = (x^2 - 3)^3 \left[ 2(x^2 - 3) + 8x(2x + 1) \right]$dxdy​=(x2−3)3[2(x2−3)+8x(2x+1)]. (4)

(b) Hence write $\dfrac{dy}{dx}$dxdy​ as a fully simplified polynomial expression in $x$x multiplied by $(x^2 - 3)^3$(x2−3)3. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — identifying $u = 2x + 1$u=2x+1, $v = (x^2 - 3)^4$v=(x2−3)4, with $u' = 2$u′=2 and $v' = 4(x^2 - 3)^3 \cdot 2x = 8x(x^2 - 3)^3$v′=4(x2−3)3⋅2x=8x(x2−3)3 via the chain rule.
• M1 (AO1.1b) — substituting into $u'v + uv'$u′v+uv′: $2(x^2 - 3)^4 + (2x + 1) \cdot 8x(x^2 - 3)^3$2(x2−3)4+(2x+1)⋅8x(x2−3)3.
• M1 (AO2.1) — extracting the common factor $(x^2 - 3)^3$(x2−3)3.
• A1 (AO1.1b) — printed answer $(x^2 - 3)^3 \left[ 2(x^2 - 3) + 8x(2x + 1) \right]$(x2−3)3[2(x2−3)+8x(2x+1)].

(b)

• M1 (AO1.1b) — expanding the bracket: $2(x^2 - 3) + 8x(2x + 1) = 2x^2 - 6 + 16x^2 + 8x = 18x^2 + 8x - 6$2(x2−3)+8x(2x+1)=2x2−6+16x2+8x=18x2+8x−6.
• A1 (AO2.5) — fully simplified answer $\dfrac{dy}{dx} = 2(x^2 - 3)^3 (9x^2 + 4x - 3)$dxdy​=2(x2−3)3(9x2+4x−3).

Total: 6 marks split AO1 = 4, AO2 = 2. The "show that" structure is the Edexcel signature for product-rule questions: candidates are guided to factor out the common chain-rule term, which sets up part (b) cleanly.

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Synoptic links

Connects to:

1. Section 9 — Chain rule (combined applications): the product rule almost never appears alone at A-Level. $x^2 e^x$x2ex is a pure product; $x^2 e^{3x}$x2e3x requires the product rule with the chain rule applied to $e^{3x}$e3x. Recognising which rule applies to which factor is the AO2 reasoning examiners reward.

2. Section 9 — Quotient rule: for $y = u/v$y=u/v, $\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2}$dxdy​=v2u′v−uv′​. This is structurally a product rule applied to $u \cdot v^{-1}$u⋅v−1 followed by simplification — many A* candidates derive it on the fly rather than memorising it.

3. Section 8 — Integration by parts: the reverse direction of the product rule. From $\dfrac{d}{dx}(uv) = u'v + uv'$dxd​(uv)=u′v+uv′, integrating both sides gives $\int u'v\,dx + \int uv'\,dx = uv$∫u′vdx+∫uv′dx=uv, which rearranges to $\int u\,dv = uv - \int v\,du$∫udv=uv−∫vdu. Every integration-by-parts question is a disguised product-rule question.

4. Modelling — power as the product of force and velocity: in Mechanics applications, $P(t) = F(t) \cdot v(t)$P(t)=F(t)⋅v(t), and rates of change of power require the product rule. Similarly, kinetic energy $\tfrac{1}{2}m v^2$21​mv2 with mass varying in time (rocket-equation contexts) needs the product rule on $m(t) \cdot v(t)^2$m(t)⋅v(t)2.

5. Section 9 — Second derivatives and stationary points: finding $\dfrac{d^2y}{dx^2}$dx2d2y​ for a function defined as a product means applying the product rule twice — once for $\dfrac{dy}{dx}$dxdy​, and again to differentiate the resulting product. Stationary-point classification questions at A-Level routinely require this iterated application.

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Mark-scheme literacy

Product-rule questions on 9MA0 distribute AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Identifying $u$u and $v$v, differentiating each correctly (often with chain rule), substituting into $u'v + uv'$u′v+uv′
AO2 (reasoning / interpretation)	20–30%	Spotting common factors for elegant simplification, justifying which factor to differentiate first, recognising synoptic chain-rule applications
AO3 (problem-solving)	5–15%	Stationary-point analysis, modelling-context interpretation, connected rates of change

Examiner-rewarded phrasing: "Let $u = \ldots$u=… and $v = \ldots$v=…, so by the product rule $\dfrac{dy}{dx} = u'v + uv'$dxdy​=u′v+uv′"; "applying the chain rule to the second factor"; "factorising out $(x^2 - 3)^3$(x2−3)3 as the lowest power present". Phrases that lose marks: writing $\dfrac{d}{dx}(uv) = u'v'$dxd​(uv)=u′v′ (the most catastrophic single error in calculus); leaving the answer in unsimplified form when the question demands a polynomial; mishandling sign when one factor is a difference.

A specific Edexcel pattern: in printed-answer ("show that") questions, the printed form is deliberately factored to guide candidates. If your working produces an unfactored expression that doesn't visibly match, do not just write the printed line — show the factoring step explicitly. Examiners look for the working that bridges your line to the printed line.

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Grade-band model answers

3-mark question

Question: Differentiate $y = x^2 e^x$y=x2ex with respect to $x$x.

Grade C response (~210 words):

Using the product rule with $u = x^2$u=x2 and $v = e^x$v=ex:

$u' = 2x$u′=2x and $v' = e^x$v′=ex.

So $\dfrac{dy}{dx} = 2x \cdot e^x + x^2 \cdot e^x = 2x e^x + x^2 e^x$dxdy​=2x⋅ex+x2⋅ex=2xex+x2ex.

This can be written as $e^x(2x + x^2)$ex(2x+x2) or $x e^x (2 + x)$xex(2+x).