What is 7.4: Product, quotient and chain rules?

Use (uv)'=u'v+uv' for products and (u/v)'=u'v-uv'/v^2 for quotients; brackets help preserve the order in the quotient numerator. For a composite function, differentiate the outer function and multiply by the inner derivative.

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.

Explanation

Use $(uv)'=u'v+uv'$ for products and $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ for quotients; brackets help preserve the order in the quotient numerator. For a composite function, differentiate the outer function and multiply by the inner derivative.
Also $\frac{\mathrm d}{\mathrm dx}(\sec x)=\sec x\tan x$, $\frac{\mathrm d}{\mathrm dx}(\cosec x)=-\cosec x\cot x$ and $\frac{\mathrm d}{\mathrm dx}(\cot x)=-\cosec^2x$ .
Connected rates use a shared variable, for example $\frac{\mathrm dV}{\mathrm dt}=\frac{\mathrm dV}{\mathrm dr}\frac{\mathrm dr}{\mathrm dt}$ .
For an inverse $g=f^{-1}$ , $g'(y)=1/f'(x)$ when $y=f(x)$ and $f'(x)\ne0$ .
A common error is to substitute numerical values before differentiating, which can erase the changing relationship between the variables.

Worked example

(a) Differentiate $y=\frac{\sin(2x)}{(x^2+1)^3}$ , giving one fraction. (b) Differentiate $z=\sec(3x)-2\cot x+\cosec(2x)$ .

1.(a) Take $u=\sin(2x)$ and $v=(x^2+1)^3$ .
2.Then $u'=2\cos(2x)$ and $v'=6x(x^2+1)^2$ .
3.The quotient rule and cancellation of $(x^2+1)^2$ give the stated fraction. (b) Apply the three standard derivatives and the chain rule: the terms give $3\sec(3x)\tan(3x)$ , $2\cosec^2x$ and $-2\cosec(2x)\cot(2x)$ respectively.

Answer: (a) $\frac{\mathrm dy}{\mathrm dx}=\frac{2(x^2+1)\cos(2x)-6x\sin(2x)}{(x^2+1)^4}$ ; (b) $\frac{\mathrm dz}{\mathrm dx}=3\sec(3x)\tan(3x)+2\cosec^2x-2\cosec(2x)\cot(2x)$

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.

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)

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