The chain, product and quotient rules - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on The chain, product and quotient rules for Edexcel A-Level Maths (Paper 1). 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.

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2–10 marks each
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19 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

By writing \(\cot x = \dfrac{\cos x}{\sin x}\), use the quotient rule to prove that
\[\dfrac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x\]
[You may use the formula \(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x) - f(x)g'(x)}{\left(g(x)\right)^2}\).]

Model answer

\(u = \cos x\), \(v = \sin x\), \(u' = -\sin x\), \(v' = \cos x\)
\(\dfrac{d}{dx}(\cot x) = \dfrac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2 x} = \dfrac{-\left(\sin^2 x + \cos^2 x\right)}{\sin^2 x}\)
\(= -\dfrac{1}{\sin^2 x} = -\operatorname{cosec}^2 x\), as required.

Explanation

The quotient rule gives \(-\left(\sin^2 x + \cos^2 x\right)\) in the numerator, which the Pythagorean identity reduces to \(-1\). The final mark is for using the identity clearly and reaching \(-\operatorname{cosec}^2 x\). Common errors are the order of the numerator, which changes the sign, and forgetting to square the denominator.

Mark scheme

1 mark
M1 Applies the quotient rule with \(u = \cos x\), \(v = \sin x\): \(\dfrac{\sin x \times (\pm\sin x) - \cos x \times (\pm\cos x)}{\sin^2 x}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof and the result is in the formulae booklet. A numerator the wrong way round is M0.
1 mark
A1 \(\dfrac{-\sin^2 x - \cos^2 x}{\sin^2 x}\)
1 mark
A1* Uses \(\sin^2 x + \cos^2 x = 1\) to reach \(-\dfrac{1}{\sin^2 x} = -\operatorname{cosec}^2 x\), cso
The identity must be used explicitly (e.g. \(-(\sin^2 x + \cos^2 x)\) seen, then \(-1\)). Going straight from \(\dfrac{-\sin^2 x - \cos^2 x}{\sin^2 x}\) to \(-\operatorname{cosec}^2 x\) is A0.

Question 2

4 marks

\[y = (x + 3)\sqrt{2x - 1}, \quad x > \dfrac{1}{2}\]
Show that \(\dfrac{dy}{dx}\) can be written in the form \(\dfrac{ax + b}{\sqrt{2x - 1}}\), where \(a\) and \(b\) are integers to be found.

Model answer

\(u = x + 3\), \(v = (2x - 1)^{\frac{1}{2}}\), \(u' = 1\), \(v' = \dfrac{1}{2}(2x - 1)^{-\frac{1}{2}} \times 2 = (2x - 1)^{-\frac{1}{2}}\)
\(\dfrac{dy}{dx} = (2x - 1)^{\frac{1}{2}} + (x + 3)(2x - 1)^{-\frac{1}{2}}\)
\(= (2x - 1)^{-\frac{1}{2}}\left[(2x - 1) + (x + 3)\right] = \dfrac{3x + 2}{\sqrt{2x - 1}}\)
So \(a = 3\) and \(b = 2\).

Explanation

Use the product rule, with the chain rule for \(\sqrt{2x - 1}\), then take out the factor \((2x - 1)^{-\frac{1}{2}}\) (or use a common denominator) to reach a single fraction. The chain rule factor 2 cancels with the \(\dfrac{1}{2}\) from the power. Common errors are leaving out the inner derivative 2 and making sign or bracket slips when combining the two terms.

Mark scheme

1 mark
M1 Uses the product rule: \(\sqrt{2x - 1} + (x + 3) \times \dfrac{d}{dx}\left(\sqrt{2x - 1}\right)\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the form is given, so the working must be seen. Multiplying the two derivatives is M0.
1 mark
M1 Uses the chain rule on \((2x - 1)^{\frac{1}{2}}\) to reach \(k(2x - 1)^{-\frac{1}{2}}\)
1 mark
A1 \(\dfrac{dy}{dx} = (2x - 1)^{\frac{1}{2}} + (x + 3)(2x - 1)^{-\frac{1}{2}}\)
1 mark
A1 Combines over the common denominator: \(\dfrac{2x - 1 + x + 3}{\sqrt{2x - 1}} = \dfrac{3x + 2}{\sqrt{2x - 1}}\), so \(a = 3\), \(b = 2\)
cso. The step multiplying \(\sqrt{2x - 1}\) by \(\dfrac{\sqrt{2x - 1}}{\sqrt{2x - 1}}\) (or taking out \((2x - 1)^{-\frac{1}{2}}\)) must be seen.

Question 3

6 marks

The curve \(C\) has equation
\[y = \dfrac{\ln x}{x^2}, \quad x > 0\]
[You may use the formula \(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x) - f(x)g'(x)}{\left(g(x)\right)^2}\).]

Show that \(\dfrac{dy}{dx} = \dfrac{1 - 2\ln x}{x^3}\).
Hence find the exact coordinates of the stationary point of \(C\).

Model answer

\(u = \ln x\), \(v = x^2\), \(u' = \dfrac{1}{x}\), \(v' = 2x\)
\(\dfrac{dy}{dx} = \dfrac{x^2 \times \dfrac{1}{x} - \ln x \times 2x}{\left(x^2\right)^2} = \dfrac{x - 2x\ln x}{x^4} = \dfrac{1 - 2\ln x}{x^3}\), as required.
Stationary point: \(1 - 2\ln x = 0 \Rightarrow \ln x = \dfrac{1}{2} \Rightarrow x = \mathrm{e}^{\frac{1}{2}}\)
\(y = \dfrac{\frac{1}{2}}{\left(\mathrm{e}^{\frac{1}{2}}\right)^2} = \dfrac{1}{2\mathrm{e}}\)
The stationary point is \(\left(\sqrt{\mathrm{e}}, \dfrac{1}{2\mathrm{e}}\right)\).

Explanation

Use the quotient rule with \(u = \ln x\) and \(v = x^2\), then cancel the common factor \(x\) to reach the printed form. The stationary point is where the numerator is zero, \(\ln x = \dfrac{1}{2}\), giving exact coordinates in terms of \(\mathrm{e}\). Common errors are forgetting to square the denominator, writing the numerator the wrong way round, and giving decimal coordinates.

Mark scheme

1 mark
M1 Applies the quotient rule with \(u = \ln x\), \(v = x^2\): \(\dfrac{x^2 \times \frac{1}{x} - 2x\ln x}{\left(x^2\right)^2}\) (condone slips in \(u'\) or \(v'\) but not the order of the numerator). An answer with no working scores 0 marks for the show-that; exact coordinates \(\left(\sqrt{\mathrm{e}}, \dfrac{1}{2\mathrm{e}}\right)\) with no working score 3 marks (SC).
An answer with no working scores 0 marks for the show-that (the first three marks): the derivative is given, so it must be shown. Correct exact coordinates \(\left(\sqrt{\mathrm{e}}, \dfrac{1}{2\mathrm{e}}\right)\) with no working score 3 marks (SC: the last three marks), as they follow directly from the given derivative. Alternative: product rule on \(x^{-2}\ln x\), giving \(x^{-2} \times \dfrac{1}{x} - 2x^{-3}\ln x\). A numerator the wrong way round (\(uv' - vu'\) instead of \(vu' - uv'\)) is M0. The last three marks may be scored from the given derivative even if the show-that is not complete.
1 mark
A1 \(\dfrac{dy}{dx} = \dfrac{x - 2x\ln x}{x^4}\) oe
1 mark
A1* \(\dfrac{dy}{dx} = \dfrac{1 - 2\ln x}{x^3}\), cso, with the factor \(x\) cancelled
A correct intermediate line before the given answer must be seen. The denominator must be squared (\(x^4\)); a common error is \(x^2\) in the denominator.
1 mark
M1 Sets \(1 - 2\ln x = 0\) and solves: \(\ln x = \dfrac{1}{2}\)
May use the given derivative. Setting the denominator to zero is M0.
1 mark
A1 \(x = \mathrm{e}^{\frac{1}{2}}\) (or \(\sqrt{\mathrm{e}}\))
1.65 alone is A0 (exact coordinates asked for).
1 mark
A1 \(y = \dfrac{1}{2\mathrm{e}}\), so the stationary point is \(\left(\sqrt{\mathrm{e}}, \dfrac{1}{2\mathrm{e}}\right)\)
Accept \(\dfrac{1}{2}\mathrm{e}^{-1}\). Both coordinates exact.

Question 4

8 marks

A motorboat travels in a straight line across a lake. Its speed, \(v\ \mathrm{m\,s^{-1}}\), \(t\) seconds after it passes a marker buoy, is modelled by
\[v = 8 + \dfrac{40t}{t^2 + 16}, \quad 0 \leqslant t \leqslant 20\]
[You may use the formula \(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x) - f(x)g'(x)}{\left(g(x)\right)^2}\).]

Show that \(\dfrac{dv}{dt} = \dfrac{40\left(16 - t^2\right)}{\left(t^2 + 16\right)^2}\).
Using the model, find the maximum speed of the motorboat, justifying that it is a maximum.
Find the acceleration of the motorboat when \(t = 10\), and state what this value tells you about its motion. Give the acceleration correct to 3 significant figures.

Model answer

\(\dfrac{dv}{dt} = \dfrac{40\left(t^2 + 16\right) - 40t(2t)}{\left(t^2 + 16\right)^2} = \dfrac{40t^2 + 640 - 80t^2}{\left(t^2 + 16\right)^2} = \dfrac{40\left(16 - t^2\right)}{\left(t^2 + 16\right)^2}\), as required.
\(\dfrac{dv}{dt} = 0 \Rightarrow t^2 = 16 \Rightarrow t = 4\) (\(t \geqslant 0\))
\(v = 8 + \dfrac{40 \times 4}{16 + 16} = 8 + 5 = 13\ \mathrm{m\,s^{-1}}\)
For \(t < 4\), \(16 - t^2 > 0\) so \(\dfrac{dv}{dt} > 0\); for \(t > 4\), \(\dfrac{dv}{dt} < 0\). So \(v\) has a maximum of \(13\ \mathrm{m\,s^{-1}}\) at \(t = 4\).
At \(t = 10\): \(\dfrac{dv}{dt} = \dfrac{40(16 - 100)}{116^2} = -\dfrac{3360}{13456} = -0.250\ \mathrm{m\,s^{-2}}\) (3 s.f.)
The acceleration is negative, so it is slowing down at about \(0.25\ \mathrm{m\,s^{-2}}\).

Explanation

Apply the quotient rule (from the formulae booklet) to the fraction; the constant 8 disappears. The speed is greatest where \(\dfrac{dv}{dt}\) changes from positive to negative, at \(t = 4\), and a negative value of \(\dfrac{dv}{dt}\) means the motorboat is decelerating. Common errors are writing the quotient-rule numerator the wrong way round, which changes the sign, and giving \(t = 4\) instead of the maximum speed.

Mark scheme

1 mark
M1 Applies the quotient rule to \(\dfrac{40t}{t^2 + 16}\): \(\dfrac{40\left(t^2 + 16\right) - 40t \times 2t}{\left(t^2 + 16\right)^2}\) (condone one slip). An answer with no working scores 0 marks for the show-that; answers of 13 and \(-0.250\) with no working score 2 marks each (SC).
An answer with no working scores 0 marks for the show-that (the first two marks). The later answers use the given derivative: a maximum speed of 13 with no working scores 2 marks (SC: M1A1; the justification B1 needs a reason), an acceleration of awrt \(-0.250\) with no working scores 2 marks (SC: M1A1), and the interpretation B1 is for a correct statement. The constant 8 must differentiate to 0.
1 mark
A1* \(\dfrac{dv}{dt} = \dfrac{40\left(16 - t^2\right)}{\left(t^2 + 16\right)^2}\), cso
A correct unsimplified numerator, e.g. \(640 + 40t^2 - 80t^2\), must be seen before the given answer.
1 mark
M1 Sets \(\dfrac{dv}{dt} = 0\): \(16 - t^2 = 0\), giving \(t = 4\) (rejecting \(t = -4\))
May use the given derivative.
1 mark
A1 Maximum speed \(13\ \mathrm{m\,s^{-1}}\)
From \(v = 8 + \dfrac{160}{32}\). Units not required.
1 mark
B1 Justification: \(\dfrac{dv}{dt} > 0\) for \(0 \leqslant t < 4\) and \(< 0\) for \(4 < t \leqslant 20\) (since \(16 - t^2\) changes sign and the denominator is positive), so maximum
Or a second derivative shown negative at \(t = 4\), or values of \(v\) at \(t = 0\) (8), \(t = 4\) (13) and \(t = 20\) (9.92) compared. A bare statement 'it is a maximum' is B0.
1 mark
M1 Substitutes \(t = 10\) into \(\dfrac{dv}{dt}\)
1 mark
A1 Acceleration \(=\) awrt \(-0.250\ \mathrm{m\,s^{-2}}\), i.e. in [−0.2505, −0.2495)
From \(-\dfrac{3360}{13456} = -0.2497\ldots\). Accept \(-0.25\) only if \(-0.2497\ldots\) or awrt \(-0.250\) is seen first.
1 mark
B1 Interpretation: the acceleration is negative, so it is slowing down (decelerating) at about \(0.25\ \mathrm{m\,s^{-2}}\) at \(t = 10\)
Needs 'slowing down' or 'decelerating' (oe); accept 'its speed is decreasing'. Follow through the sign of their acceleration.

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