The chain, product and quotient rules - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on The chain, product and quotient rules for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
2–11 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Given that \(y = (1 + \tan x)^3\), where \(x\) is measured in radians, find the exact value of \(\dfrac{dy}{dx}\) when \(x = \dfrac{\pi}{4}\).

[Formulae booklet: \(\dfrac{d}{dx}(\tan kx) = k\sec^2 kx\)]

Model answer

\(\dfrac{dy}{dx} = 3(1 + \tan x)^2 \times \sec^2 x\)
At \(x = \dfrac{\pi}{4}\): \(\tan\dfrac{\pi}{4} = 1\) and \(\sec^2\dfrac{\pi}{4} = \dfrac{1}{\left(\frac{1}{\sqrt{2}}\right)^2} = 2\)
\(\dfrac{dy}{dx} = 3 \times 2^2 \times 2 = 24\)

Explanation

By the chain rule, the derivative of \((1 + \tan x)^3\) is \(3(1 + \tan x)^2\) multiplied by the derivative of the inside, \(\sec^2 x\). At \(x = \dfrac{\pi}{4}\), \(\tan x = 1\) and \(\sec^2 x = 2\), giving 24. Common errors are leaving out the factor \(\sec^2 x\) and using \(\sec^2\dfrac{\pi}{4} = \sqrt{2}\).

Mark scheme

1 mark
M1 Uses the chain rule: \(\dfrac{dy}{dx} = 3(1 + \tan x)^2 \times k\sec^2 x\); or a correct final answer of 24 with no working
An answer of 24 with no working scores 3 marks. A common error is \(3(1 + \tan x)^2\), leaving out the derivative of \(\tan x\): M0.
1 mark
A1 \(\dfrac{dy}{dx} = 3(1 + \tan x)^2\sec^2 x\)
1 mark
A1 24
From \(3 \times 2^2 \times 2\), using \(\tan\dfrac{\pi}{4} = 1\) and \(\sec^2\dfrac{\pi}{4} = 2\). cao

Question 2

5 marks

The curve \(C\) has equation
\[y = \ln(\cos 2x), \quad -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}\]

Show that \(\dfrac{dy}{dx} = -2\tan 2x\).
Hence find the equation of the tangent to \(C\) at the point where \(x = \dfrac{\pi}{8}\), writing it as \(y = mx + c\) with \(m\) and \(c\) exact.

Model answer

\(\dfrac{dy}{dx} = \dfrac{1}{\cos 2x} \times (-2\sin 2x) = \dfrac{-2\sin 2x}{\cos 2x} = -2\tan 2x\), as required.
At \(x = \dfrac{\pi}{8}\): \(y = \ln\left(\cos\dfrac{\pi}{4}\right) = \ln\dfrac{1}{\sqrt{2}} = -\dfrac{1}{2}\ln 2\), and \(\dfrac{dy}{dx} = -2\tan\dfrac{\pi}{4} = -2\)
\(y + \dfrac{1}{2}\ln 2 = -2\left(x - \dfrac{\pi}{8}\right)\)
\(y = -2x + \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\)

Explanation

The derivative of \(\ln(u)\) is \(\dfrac{1}{u}\dfrac{du}{dx}\), and the derivative of \(\cos 2x\) is \(-2\sin 2x\), so \(\dfrac{dy}{dx} = -2\tan 2x\). At \(x = \dfrac{\pi}{8}\) the exact values \(\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\) and \(\tan\dfrac{\pi}{4} = 1\) give the point and the gradient. Common errors are leaving out the chain rule factor \(-2\) and giving decimals instead of exact values.

Mark scheme

1 mark
M1 Uses the chain rule on \(\ln(\cos 2x)\): \(\dfrac{k\sin 2x}{\cos 2x}\). An answer with no working scores 0 marks for the show-that.
An answer with no working scores 0 marks for the show-that (the first two marks); an answer of \(y = -2x + \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\) with no working scores 3 marks (the last three marks).
1 mark
A1* \(\dfrac{dy}{dx} = \dfrac{-2\sin 2x}{\cos 2x} = -2\tan 2x\), cso
The line \(\dfrac{-2\sin 2x}{\cos 2x}\) must be seen before the given answer. A common error is \(\dfrac{1}{\cos 2x}\) with no chain rule.
1 mark
B1 \(y = \ln\left(\cos\dfrac{\pi}{4}\right) = \ln\dfrac{1}{\sqrt{2}} = -\dfrac{1}{2}\ln 2\) at the point
Accept \(\ln\dfrac{\sqrt{2}}{2}\) or \(-\ln\sqrt{2}\).
1 mark
M1 Uses the gradient \(-2\tan\dfrac{\pi}{4} = -2\) and their point in a straight-line equation
May use the given derivative even if the show-that is not complete.
1 mark
A1 \(y = -2x + \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\)
Accept exact equivalents, e.g. \(c = \dfrac{\pi}{4} - \ln\sqrt{2}\), or the exact values of \(m\) and \(c\) clearly stated. Decimals (\(c = 0.439\)) score A0.

Question 3

7 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

By writing \(\operatorname{cosec} x = (\sin x)^{-1}\), use the chain rule to show that
\[\dfrac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x\cot x\]

The curve \(C\) has equation
\[y = \operatorname{cosec} x + 2\sin x, \quad 0 < x < \pi\]
Find the exact coordinates of the three stationary points of \(C\).
Determine, with a reason, the nature of the stationary point where \(x = \dfrac{\pi}{2}\).

Model answer

\(\operatorname{cosec} x = (\sin x)^{-1}\), so \(\dfrac{d}{dx}(\operatorname{cosec} x) = -(\sin x)^{-2} \times \cos x = -\dfrac{1}{\sin x} \times \dfrac{\cos x}{\sin x} = -\operatorname{cosec} x\cot x\), as required.
\(\dfrac{dy}{dx} = -\operatorname{cosec} x\cot x + 2\cos x = \cos x\left(2 - \dfrac{1}{\sin^2 x}\right)\)
\(\dfrac{dy}{dx} = 0\): \(\cos x = 0 \Rightarrow x = \dfrac{\pi}{2}\), or \(\sin^2 x = \dfrac{1}{2} \Rightarrow \sin x = \dfrac{1}{\sqrt{2}}\) (as \(\sin x > 0\)) \(\Rightarrow x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}\)
\(x = \dfrac{\pi}{4}\): \(y = \sqrt{2} + 2 \times \dfrac{1}{\sqrt{2}} = 2\sqrt{2}\); \(x = \dfrac{\pi}{2}\): \(y = 1 + 2 = 3\); \(x = \dfrac{3\pi}{4}\): \(y = 2\sqrt{2}\)
Points: \(\left(\dfrac{\pi}{4}, 2\sqrt{2}\right)\), \(\left(\dfrac{\pi}{2}, 3\right)\), \(\left(\dfrac{3\pi}{4}, 2\sqrt{2}\right)\)
Near \(x = \dfrac{\pi}{2}\), \(2 - \dfrac{1}{\sin^2 x} > 0\), and \(\cos x\) changes from positive to negative, so \(\dfrac{dy}{dx}\) changes from positive to negative: \(\left(\dfrac{\pi}{2}, 3\right)\) is a maximum.

Explanation

Writing \(\operatorname{cosec} x\) as \((\sin x)^{-1}\) and using the chain rule gives \(-\dfrac{\cos x}{\sin^2 x}\), which splits into \(-\operatorname{cosec} x\cot x\). Taking out the factor \(\cos x\) from the derivative of the curve gives one stationary point from \(\cos x = 0\) and two from \(\sin^2 x = \dfrac{1}{2}\); the sign of the gradient either side of \(\dfrac{\pi}{2}\) shows a maximum. A common error is to divide by \(\cos x\) and lose the stationary point at \(x = \dfrac{\pi}{2}\).

Mark scheme

1 mark
M1 Uses the chain rule: \(\dfrac{d}{dx}(\sin x)^{-1} = -(\sin x)^{-2} \times \pm\cos x\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working.
1 mark
A1 \(-\dfrac{\cos x}{\sin^2 x}\)
1 mark
A1* \(-\dfrac{1}{\sin x} \times \dfrac{\cos x}{\sin x} = -\operatorname{cosec} x\cot x\), cso
Depends on the M mark (cso). The split into the two factors must be seen.
1 mark
M1 Sets \(\dfrac{dy}{dx} = -\operatorname{cosec} x\cot x + 2\cos x = 0\) and factorises, e.g. \(\cos x\left(2 - \operatorname{cosec}^2 x\right) = 0\) or \(\cos x\left(2\sin^2 x - 1\right) = 0\)
May use the given derivative. Dividing by \(\cos x\) loses \(x = \dfrac{\pi}{2}\); this can still score M1 but not the next A1.
1 mark
A1 \(x = \dfrac{\pi}{4}\), \(x = \dfrac{\pi}{2}\) and \(x = \dfrac{3\pi}{4}\)
All three, exact, and no others in \(0 < x < \pi\). Degrees or decimals score A0.
1 mark
A1 \(\left(\dfrac{\pi}{4}, 2\sqrt{2}\right)\), \(\left(\dfrac{\pi}{2}, 3\right)\) and \(\left(\dfrac{3\pi}{4}, 2\sqrt{2}\right)\)
Depends on the second M mark. All three points, with exact \(y\)-coordinates (\(\sqrt{2} + \sqrt{2} = 2\sqrt{2}\)).
1 mark
B1 A maximum at \(x = \dfrac{\pi}{2}\), with a valid reason, e.g. \(\dfrac{dy}{dx} = \cos x\left(2 - \operatorname{cosec}^2 x\right)\) changes from positive to negative there (\(2 - \operatorname{cosec}^2 x > 0\) near \(\dfrac{\pi}{2}\) while \(\cos x\) changes sign), or \(\dfrac{d^2y}{dx^2} = -1 < 0\)
Accept a correct comparison: \(y = 3\) at \(x = \dfrac{\pi}{2}\) is greater than \(y = 2\sqrt{2}\) at the stationary points on either side, and there are no other stationary points between them, so it is a maximum. 'Maximum' with no reason is B0.

Question 4

8 marks

The curve \(C\) has equation
\[y = \dfrac{\sin x}{3 + \cos x}, \quad 0 \leqslant x < 2\pi\]
where \(x\) is measured in radians.

Show that \(\dfrac{dy}{dx} = \dfrac{3\cos x + 1}{(3 + \cos x)^2}\).
Hence find the \(x\)-coordinates of the stationary points of \(C\), giving your answers to 3 significant figures.
Find the exact greatest value of \(y\), and state the range of values that \(y\) takes.
[Quotient rule: \(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x) - f(x)g'(x)}{\left(g(x)\right)^2}\)]

Model answer

\(\dfrac{dy}{dx} = \dfrac{\cos x(3 + \cos x) - \sin x(-\sin x)}{(3 + \cos x)^2} = \dfrac{3\cos x + \cos^2 x + \sin^2 x}{(3 + \cos x)^2} = \dfrac{3\cos x + 1}{(3 + \cos x)^2}\), as required.
\(\dfrac{dy}{dx} = 0 \Rightarrow \cos x = -\dfrac{1}{3} \Rightarrow x = 1.91\) or \(x = 2\pi - 1.91 = 4.37\) (3 s.f.)
With \(\cos x = -\dfrac{1}{3}\), \(\sin x = \pm\sqrt{1 - \dfrac{1}{9}} = \pm\dfrac{2\sqrt{2}}{3}\) (positive at \(x = 1.91\), negative at \(x = 4.37\)) and \(3 + \cos x = \dfrac{8}{3}\)
So \(y = \pm\dfrac{2\sqrt{2}}{3} \times \dfrac{3}{8} = \pm\dfrac{\sqrt{2}}{4}\)
The greatest value of \(y\) is \(\dfrac{\sqrt{2}}{4}\) and the least is \(-\dfrac{\sqrt{2}}{4}\); as \(C\) is continuous, \(-\dfrac{\sqrt{2}}{4} \leqslant y \leqslant \dfrac{\sqrt{2}}{4}\).

Explanation

The quotient rule gives a numerator of \(3\cos x + \cos^2 x + \sin^2 x\), which simplifies to \(3\cos x + 1\) using \(\sin^2 x + \cos^2 x = 1\). The stationary points are where \(\cos x = -\dfrac{1}{3}\); finding \(\sin x\) exactly from the identity gives the greatest and least values \(\pm\dfrac{\sqrt{2}}{4}\), so these are the ends of the range. Common errors are a sign error with \(\dfrac{d}{dx}(\cos x)\), missing the second solution \(2\pi - 1.91\), and giving the greatest value as a rounded decimal.

Mark scheme

1 mark
M1 Uses the quotient rule with \(f(x) = \sin x\) and \(g(x) = 3 + \cos x\) in the correct order. An answer with no working scores 0 marks for the show-that.
An answer with no working scores 0 marks for the show-that (the first three marks); correct \(x\)-coordinates, \(\dfrac{\sqrt{2}}{4}\) and the range with no working score 3 marks (SC: the A1 for the \(x\)-coordinates, the A1 for \(\dfrac{\sqrt{2}}{4}\) and the B1 for the range), as the exact working for the greatest value must be seen for its method mark.
1 mark
A1 \(\dfrac{\cos x(3 + \cos x) - \sin x(-\sin x)}{(3 + \cos x)^2}\) oe
1 mark
A1* Uses \(\sin^2 x + \cos^2 x = 1\): \(\dfrac{3\cos x + \cos^2 x + \sin^2 x}{(3 + \cos x)^2} = \dfrac{3\cos x + 1}{(3 + \cos x)^2}\), cso
Depends on the M mark (cso).
1 mark
M1 Sets the numerator equal to 0: \(\cos x = -\dfrac{1}{3}\), and finds at least one value of \(x\)
May use the given derivative.
1 mark
A1 \(x =\) awrt 1.91 and awrt 4.37, i.e. in [1.905, 1.915) and [4.365, 4.375)
Both needed, and no others in \(0 \leqslant x < 2\pi\). The second is \(2\pi - 1.9106\ldots\). Degrees (109.5°, 250.5°) score A0.
1 mark
M1 Uses \(\cos x = -\dfrac{1}{3}\) and \(\sin x = \pm\sqrt{1 - \dfrac{1}{9}} = \pm\dfrac{2\sqrt{2}}{3}\) (exact) to find \(y\)
Depends on the second M mark. Using rounded values of \(x\) in a calculator gives \(0.354\), which does not show the exact value: M0.
1 mark
A1 Greatest value of \(y = \dfrac{\frac{2\sqrt{2}}{3}}{\frac{8}{3}} = \dfrac{\sqrt{2}}{4}\)
Accept \(\dfrac{1}{2\sqrt{2}}\) or \(\dfrac{2\sqrt{2}}{8}\).
1 mark
B1 \(-\dfrac{\sqrt{2}}{4} \leqslant y \leqslant \dfrac{\sqrt{2}}{4}\)
Follow through their exact greatest value \(k\) as \(-k \leqslant y \leqslant k\). Accept \(\left[-\dfrac{\sqrt{2}}{4}, \dfrac{\sqrt{2}}{4}\right]\) or the set in set notation. Strict inequalities score B0.

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