What is 7.4: Product, quotient and chain rules?

Chain rule: differentiate the outer function, keep the inside unchanged, multiply by the derivative of the inside, so e^4x^2-1 gives 8xe^4x^2-1 and (2x+1)^5 gives 10(2x+1)^4; writing u=4x^2-1 as an intermediate step is what the reports recommend. Product rule: (uv)'=u'v+uv'.

How many marks are 7.4 questions worth?

7.4 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 7.4 appeared in?

We have not yet indexed a verified past-paper appearance for 7.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.4

Differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions.

Draft — not yet indexed

Product, quotient and chain rules

Worked answers and methods for 7.4 on Edexcel A-level Maths 9MA0.

Definition

Specification 7.4: differentiate using the product rule, the quotient rule and the chain rule, including problems involving connected rates of change and inverse functions.

Explanation

Chain rule: differentiate the outer function, keep the inside unchanged, multiply by the derivative of the inside, so $e^{4x^2-1}$ gives $8xe^{4x^2-1}$ and $(2x+1)^5$ gives $10(2x+1)^4$ ; writing $u=4x^2-1$ as an intermediate step is what the reports recommend. Product rule: $(uv)'=u'v+uv'$ .
Quotient rule: $\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}$ , with the order of the numerator fixed and brackets kept round each product. After differentiating, factorise out the common factor, $(2x+1)^3(10x+1)$ , because the factorised form is usually what the next part needs.
Connected rates use $\dfrac{dV}{dt}=\dfrac{dV}{dh}\times\dfrac{dh}{dt}$ : differentiate the formula first, then substitute the numbers.
For an inverse function or $x$ given in terms of $y$ , $\dfrac{dy}{dx}=1\Big/\dfrac{dx}{dy}$ .
The derivatives of $\sec$ , $\cosec$ and $\cot$ are in the formulae booklet.

Worked example

Given that

y=\dfrac{\sin2x}{(x^2+1)^3}

, show that

\dfrac{dy}{dx}=\dfrac{2(x^2+1)\cos2x-6x\sin2x}{(x^2+1)^4}

(4)

(Total for Question 1 is 4 marks)

Working

1.Quotient rule with $u=\sin2x$ , $v=(x^2+1)^3$ : $u'=2\cos2x$ by the chain rule and $v'=3(x^2+1)^2\times2x=6x(x^2+1)^2$ .
2. $\dfrac{dy}{dx}=\dfrac{2\cos2x\,(x^2+1)^3-\sin2x\times6x(x^2+1)^2}{(x^2+1)^6}$ .
3.Cancel the common factor $(x^2+1)^2$ : $\dfrac{2(x^2+1)\cos2x-6x\sin2x}{(x^2+1)^4}$ , as required.

Answer: $\dfrac{dy}{dx}=\dfrac{2(x^2+1)\cos2x-6x\sin2x}{(x^2+1)^4}$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Attempts the quotient rule (or product rule) with the correct overall structure, $\dfrac{u'v-uv'}{v^2}$	M1
	$\dfrac{d}{dx}(\sin2x)=2\cos2x$ and $\dfrac{d}{dx}(x^2+1)^3=6x(x^2+1)^2$	B1
	Fully correct unsimplified derivative	A1
	Cancels $(x^2+1)^2$ to reach the printed answer with no errors	A1
Notes
B1: Chain rule on both
A1: cso

(4 marks)

Common mistakes

Don't differentiate a product $uv$ as $u'v'$ and omit both product-rule terms.
Don't apply product, quotient and chain rules as separate fragments and lose factors in the combined derivative.

Mark-scheme wording to use

“M1: Attempts the quotient rule to obtain an expression of the form (P(x² + 4) − Qx(2x − 3))/(x² + 4)² P, Q > 0 … A1: Fully correct differentiation in any form with correct bracketing which may be implied by subsequent working. … A1: f′(x) = (−2x² + 6x + 8)/(x² + 4)² or simplified equivalent, e.g. numerator terms in a different order.” — Pearson 9MA0, June 2024 Paper 1, Q5(a)
“M1 for attempting the product and chain rules. A1 for dy/dx = (2x + 1)^4 + 8x(2x + 1)^3. M1 for taking out the common factor (2x + 1)^3. A1 for (2x + 1)^3(10x + 1), hence n = 3, A = 10 and B = 1” — Pearson 9MA0, Sample Paper 2, Q3
“M1 for differentiating V with respect to h. A1 for dV/dh = 50*pi*h - pi*h^2. M1 for using dV/dt = (dV/dh)(dh/dt). dM1 for substituting h = 10 and dV/dt = 160*pi into a complete equation. A1 for 0.4 cm s^(-1)” — Pearson 9MA0, Specimen Paper 2, Q8(a)

What the examiner wants

An intermediate step for the chain rule, then factorise rather than expand. The June 2024 Paper 2 report says candidates struggled with the exponential $e^{4x^2-1}$ and the most successful wrote a substitution step before applying the chain rule, and that expanding brackets instead of using the chain rule and factorisation made a later question longer and less accurate.

Exam tip

Name the outermost rule first, differentiate nested functions carefully, then simplify only after every factor is present.

Worked practice

Tier 1 · Easy

Differentiate

y=x^2e^{3x}

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\frac{\mathrm dy}{\mathrm dx}=e^{3x}(2x+3x^2)$	3
Notes
Apply the product rule: $\frac{\mathrm dy}{\mathrm dx}=(2x)e^{3x}+x^2(3e^{3x})=e^{3x}(2x+3x^2)$ .

(3 marks)

Tier 2 · Standard

The volume of a sphere is increasing at

12\pi\,\text{cm}^3\text{s}^{-1}

. Find the rate at which its radius is increasing when the radius is

3\,\text{cm}

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac13\,\text{cm s}^{-1}$	4
Notes
For $V=\tfrac43\pi r^3$ , the chain rule gives $dV/dt=4\pi r^2\,dr/dt$ . Substitute $dV/dt=12\pi$ and $r=3$ : $12\pi=4\pi(3^2)\,dr/dt=36\pi\,dr/dt$ . Therefore $dr/dt=1/3\,\text{cm s}^{-1}$ .

(4 marks)

Tier 3 · Hard

(a) A sphere has radius

r

cm and volume

V=\frac43\pi r^3

. At an instant when

r=3

, its volume is increasing at

24\pi\,\text{cm}^3\text{s}^{-1}

. Find

\frac{\mathrm dr}{\mathrm dt}

. (b) The function

f(x)=x^3+x+1

has inverse

g

. Given that

g(3)=1

, find

g'(3)

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	(a) $\frac{\mathrm dr}{\mathrm dt}=\frac23\,\text{cm s}^{-1}$ (b) $g'(3)=\frac14$	7
Notes
(a) Differentiate with respect to time: $\frac{\mathrm dV}{\mathrm dt}=4\pi r^2\frac{\mathrm dr}{\mathrm dt}$ . At $r=3$ , $24\pi=36\pi\frac{\mathrm dr}{\mathrm dt}$ , so $\frac{\mathrm dr}{\mathrm dt}=\frac23\,\text{cm s}^{-1}$ . (b) Since $g=f^{-1}$ , $g'(3)=1/f'(g(3))$ . Now $f'(x)=3x^2+1$ and $g(3)=1$ , so $g'(3)=1/f'(1)=\frac14$ .

(7 marks)

Tier 1 · Easy

Differentiate

y=(3x^2-1)^4

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{\mathrm dy}{\mathrm dx}=24x(3x^2-1)^3$	2
Notes
Apply the chain rule: multiply $4(3x^2-1)^3$ by the inner derivative $6x$ . This gives $24x(3x^2-1)^3$ .

(2 marks)

Tier 2 · Standard

Differentiate

y=\dfrac{\ln(2x+1)}{x^2+4}

, giving your answer as a single fraction.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2(x^2+4)-2x(2x+1)\ln(2x+1)}{(2x+1)(x^2+4)^2}$	4
Notes
Let $u=\ln(2x+1)$ and $v=x^2+4$ . Then $u'=2/(2x+1)$ and $v'=2x$ . The quotient rule gives $\dfrac{u'v-uv'}{v^2}$ . Multiplying the numerator and denominator by $2x+1$ gives $\dfrac{2(x^2+4)-2x(2x+1)\ln(2x+1)}{(2x+1)(x^2+4)^2}$ .

(4 marks)

Tier 3 · Hard

A conical pile has radius

r

cm and height

h

cm, with

h=2r

at all times. Its volume is

V=\dfrac13\pi r^2h

. Material is added at

18\pi\,\text{cm}^3\text{s}^{-1}

. Find

\dfrac{\mathrm dr}{\mathrm dt}

and

\dfrac{\mathrm dh}{\mathrm dt}

at the instant when

h=4

cm.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\dfrac{\mathrm dr}{\mathrm dt}=\dfrac94\,\text{cm s}^{-1}$ $\dfrac{\mathrm dh}{\mathrm dt}=\dfrac92\,\text{cm s}^{-1}$	6
Notes
Using $h=2r$ , $V=\frac23\pi r^3$ , so $\dfrac{\mathrm dV}{\mathrm dt}=2\pi r^2\dfrac{\mathrm dr}{\mathrm dt}$ . When $h=4$ , $r=2$ . Hence $18\pi=2\pi(2^2)\dfrac{\mathrm dr}{\mathrm dt}$ , giving $\dfrac{\mathrm dr}{\mathrm dt}=9/4$ . Since $h=2r$ , differentiating with respect to time gives $\dfrac{\mathrm dh}{\mathrm dt}=2\dfrac{\mathrm dr}{\mathrm dt}=9/2$ .

(6 marks)

Tier 2 · Standard

A point

P(x,y)

moves along the curve

x^2+4y^2=100

in the first quadrant. At the instant when

P=(6,4)

, the

x

-coordinate is increasing at

2

units per second. Find the rate of change of the

y

-coordinate and state whether

P

is moving upwards or downwards.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac{\mathrm dy}{\mathrm dt}=-\dfrac34$ units per second $P$ is moving downwards	4
Notes
Differentiate $x^2+4y^2=100$ with respect to time: $2x\dfrac{\mathrm dx}{\mathrm dt}+8y\dfrac{\mathrm dy}{\mathrm dt}=0$ . At $P=(6,4)$ with $\dfrac{\mathrm dx}{\mathrm dt}=2$ , this gives $2(6)(2)+8(4)\dfrac{\mathrm dy}{\mathrm dt}=0$ . Hence $\dfrac{\mathrm dy}{\mathrm dt}=-24/32=-3/4$ units per second. The negative sign shows that $P$ is moving downwards.

(4 marks)

Tier 3 · Hard

A rectangle has width

w

cm and length

l

cm, where

l=2w+3

at all times. Its area is increasing at

95\,\text{cm}^2\text{s}^{-1}

. Find the rates of change of

w

l

and the perimeter when

w=4

cm.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\dfrac{\mathrm dw}{\mathrm dt}=5\,\text{cm s}^{-1}$ $\dfrac{\mathrm dl}{\mathrm dt}=10\,\text{cm s}^{-1}$ $\dfrac{\mathrm dP}{\mathrm dt}=30\,\text{cm s}^{-1}$	6
Notes
Using $l=2w+3$ , the area is $A=wl=2w^2+3w$ . Differentiating with respect to time gives $\dfrac{\mathrm dA}{\mathrm dt}=(4w+3)\dfrac{\mathrm dw}{\mathrm dt}$ . At $w=4$ , $95=19\,\dfrac{\mathrm dw}{\mathrm dt}$ , so $\dfrac{\mathrm dw}{\mathrm dt}=5$ . Differentiating $l=2w+3$ gives $\dfrac{\mathrm dl}{\mathrm dt}=10$ . Since $P=2l+2w$ , $\dfrac{\mathrm dP}{\mathrm dt}=2(10)+2(5)=30\,\text{cm s}^{-1}$ .

(6 marks)

Tier 3 · Hard

A straight ladder of length

10

m rests with its foot on horizontal ground and its top against a vertical wall. The foot moves away from the wall at

0.6

m s

^{-1}

. Let

\theta

be the angle between the ladder and the ground. Find

\dfrac{\mathrm d\theta}{\mathrm dt}

and the vertical velocity of the top when the foot is

6

m from the wall.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{3}{40}$ rad s $^{-1}$ The top moves downwards at $\dfrac9{20}$ m s $^{-1}$	6
Notes
If the foot is $x$ metres from the wall, $x=10\cos\theta$ . At $x=6$ , the height is $8$ and hence $\sin\theta=4/5$ , $\cos\theta=3/5$ . Differentiating gives $\mathrm dx/\mathrm dt=-10\sin\theta\,\mathrm d\theta/\mathrm dt$ . Thus $0.6=-8\,\mathrm d\theta/\mathrm dt$ , so $\mathrm d\theta/\mathrm dt=-3/40$ . With height $y=10\sin\theta$ , $\mathrm dy/\mathrm dt=10\cos\theta\,\mathrm d\theta/\mathrm dt=6(-3/40)=-9/20$ , so the top moves downwards.

(6 marks)

Q10

Tier 3 · Hard

10.

For

x>0

, the curve has equation

y=(x^2+1)^3e^{-x}

. Differentiate the function in factorised form. Hence find the exact

x

-coordinates of all stationary points and classify each one.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\dfrac{\mathrm dy}{\mathrm dx}=e^{-x}(x^2+1)^2(-x^2+6x-1)$ $x=3-2\sqrt2$ is a local minimum $x=3+2\sqrt2$ is a local maximum	6
Notes
Using the product and chain rules, $y'=6x(x^2+1)^2e^{-x}-(x^2+1)^3e^{-x}=e^{-x}(x^2+1)^2(-x^2+6x-1)$ . The positive factors never vanish, so stationary points satisfy $x^2-6x+1=0$ , giving $x=3\pm2\sqrt2$ . The quadratic factor in $y'$ is negative before the smaller root, positive between the roots and negative after the larger root. Therefore the smaller root is a local minimum and the larger root a local maximum.

(6 marks)

Verified exam appearances

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