First principles and standard derivatives - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on First principles and standard derivatives 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.

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2–10 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
\[y = 3\mathrm{e}^{-2x} - 4\cos\left(\dfrac{x}{2}\right)\]
where \(x\) is measured in radians, find \(\dfrac{d^2y}{dx^2}\).

Model answer

\(\dfrac{dy}{dx} = 3 \times (-2)\mathrm{e}^{-2x} - 4 \times \left(-\dfrac{1}{2}\right)\sin\left(\dfrac{x}{2}\right) = -6\mathrm{e}^{-2x} + 2\sin\left(\dfrac{x}{2}\right)\)
\(\dfrac{d^2y}{dx^2} = -6 \times (-2)\mathrm{e}^{-2x} + 2 \times \dfrac{1}{2}\cos\left(\dfrac{x}{2}\right)\)
\(\dfrac{d^2y}{dx^2} = 12\mathrm{e}^{-2x} + \cos\left(\dfrac{x}{2}\right)\)

Explanation

Differentiate twice, using \(\mathrm{e}^{kx} \to k\mathrm{e}^{kx}\), \(\cos kx \to -k\sin kx\) and \(\sin kx \to k\cos kx\) with \(k = -2\) and \(k = \dfrac{1}{2}\). The method mark is for differentiating once correctly in form, and the accuracy marks are for each derivative. Common errors are sign slips with the cosine term and multiplying by 2 instead of by \(\dfrac{1}{2}\).

Mark scheme

1 mark
M1 Differentiates once: \(\mathrm{e}^{-2x} \to \pm k\mathrm{e}^{-2x}\) or \(\cos\left(\dfrac{x}{2}\right) \to \pm k\sin\left(\dfrac{x}{2}\right)\); or a correct final answer of \(12\mathrm{e}^{-2x} + \cos\left(\dfrac{x}{2}\right)\) with no working
An answer of \(12\mathrm{e}^{-2x} + \cos\left(\dfrac{x}{2}\right)\) with no working scores 3 marks.
1 mark
A1 \(\dfrac{dy}{dx} = -6\mathrm{e}^{-2x} + 2\sin\left(\dfrac{x}{2}\right)\)
Common errors: \(-4\cos\left(\dfrac{x}{2}\right) \to -2\sin\left(\dfrac{x}{2}\right)\) (sign error) and \(-4\cos\left(\dfrac{x}{2}\right) \to 8\sin\left(\dfrac{x}{2}\right)\) (multiplying by 2 instead of by \(\dfrac{1}{2}\)).
1 mark
A1 \(\dfrac{d^2y}{dx^2} = 12\mathrm{e}^{-2x} + \cos\left(\dfrac{x}{2}\right)\)
Depends on the M mark. cao. Accept \(12\mathrm{e}^{-2x} + \cos(0.5x)\) and equivalent typed forms. Withhold this mark for a final answer containing an unsimplified coefficient such as \(2 \times \dfrac{1}{2}\cos\left(\dfrac{x}{2}\right)\).

Question 2

5 marks

A student is asked to prove, from first principles, that the derivative of \(f(x) = 5x - 2x^2\) is \(f'(x) = 5 - 4x\).
The student's working is shown below.

Line 1: \(f(x + h) = 5x + 5h - 2x^2 - 2h^2\)
Line 2: \(\dfrac{f(x + h) - f(x)}{h} = \dfrac{5h - 2h^2}{h} = 5 - 2h\)
Line 3: Let \(h = 0\). Then \(f'(x) = 5\)

Identify the two errors in the student's working.
Prove correctly, from first principles, that \(f'(x) = 5 - 4x\).
[First principles: \(f'(x) = \displaystyle\lim_{h \to 0}\left(\dfrac{f(x + h) - f(x)}{h}\right)\)]

Model answer

Error in Line 1: \(-2(x + h)^2 = -2x^2 - 4xh - 2h^2\), not \(-2x^2 - 2h^2\); the term \(-4xh\) is missing.
Error in Line 3: \(h\) cannot be set equal to 0, because the working divides by \(h\); the derivative is the limit of the chord gradient as \(h \to 0\).
Correct proof:
\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{5(x + h) - 2(x + h)^2 - \left(5x - 2x^2\right)}{h}\)
\(= \dfrac{5x + 5h - 2x^2 - 4xh - 2h^2 - 5x + 2x^2}{h} = \dfrac{5h - 4xh - 2h^2}{h} = 5 - 4x - 2h\)
As \(h \to 0\), \(-2h \to 0\), so \(f'(x) = \displaystyle\lim_{h \to 0}(5 - 4x - 2h) = 5 - 4x\).

Explanation

The student squared \(x + h\) as \(x^2 + h^2\), which loses the \(-4xh\) term and so the \(-4x\) in the derivative, and then put \(h = 0\) instead of taking the limit as \(h \to 0\). A correct proof expands \((x + h)^2\) fully, cancels, divides by \(h\) and then lets \(h \to 0\). Both are common errors in first-principles proofs, and either one loses the final mark.

Mark scheme

1 mark
B1 Identifies the error in Line 1: \(-2(x + h)^2\) has been expanded as \(-2x^2 - 2h^2\), missing the term \(-4xh\); it should be \(-2x^2 - 4xh - 2h^2\). An answer with no working scores 0 marks for the proof.
An answer with no working scores 0 marks for the proof (the last three marks); the first two marks are for the errors, which may be described in words. Accept '\((x + h)^2\) is not \(x^2 + h^2\)' or 'the \(2xh\) term is missing from \((x + h)^2\)', oe. Do not accept 'Line 1 is wrong' with no reason.
1 mark
B1 Identifies the error in Line 3: \(h\) cannot be made equal to 0 (the expression came from dividing by \(h\)); the derivative is the limit of the chord gradient as \(h \to 0\)
Accept 'should take the limit as \(h\) tends to 0, not substitute \(h = 0\)', oe. Do not accept 'the answer should be \(5 - 4x\)' on its own, or 'the \(-2h\) should not disappear'.
1 mark
M1 Writes a correct chord gradient and expands correctly: \(\dfrac{5(x + h) - 2(x + h)^2 - \left(5x - 2x^2\right)}{h}\) with \((x + h)^2 = x^2 + 2xh + h^2\)
The \(-\left(5x - 2x^2\right)\) must be present, with its bracket or recovered.
1 mark
A1 \(\dfrac{5h - 4xh - 2h^2}{h} = 5 - 4x - 2h\), with no errors
1 mark
A1* \(f'(x) = \displaystyle\lim_{h \to 0}(5 - 4x - 2h) = 5 - 4x\), cso
Depends on the M mark (cso). \(h \to 0\) must be referred to at least once; \(h = 0\) with no reference to a limit is A0. The conclusion \(f'(x) = 5 - 4x\) must be seen.

Question 3

6 marks

Light from the surface of a lake is absorbed as it passes down through the water. In a model, the intensity of the light, \(I\) lux, at a depth of \(h\) metres below the surface of the lake is given by
\[I = 1200 \times 0.7^h, \quad 0 \leqslant h \leqslant 15\]

Find the value of \(\dfrac{dI}{dh}\) when \(h = 2\), and interpret your answer in the context of the model.
Find, according to the model, the depth at which the intensity of the light is decreasing at a rate of 50 lux per metre. Give your answer to 3 significant figures.

Model answer

\(\dfrac{dI}{dh} = 1200 \times 0.7^h\ln 0.7\)
When \(h = 2\): \(\dfrac{dI}{dh} = 1200 \times 0.49 \times \ln 0.7 = -209.7\ldots \approx -210\)
At a depth of 2 m the intensity of the light is decreasing at about 210 lux per metre of depth.
\(1200 \times 0.7^h\ln 0.7 = -50\)
\(0.7^h = \dfrac{50}{1200 \times (-\ln 0.7)} = 0.11682\ldots\)
\(h = \dfrac{\ln 0.11682\ldots}{\ln 0.7} = 6.02\) metres (3 s.f.)

Explanation

Use \(\dfrac{d}{dh}\left(a^h\right) = a^h\ln a\); because \(0.7 < 1\), \(\ln 0.7 < 0\) and the derivative is negative, showing that the intensity decreases with depth. A rate of decrease of 50 lux per metre means \(\dfrac{dI}{dh} = -50\), which is solved with logarithms. Common errors are treating \(0.7^h\) as a power of \(h\), setting \(I = 50\) instead of \(\dfrac{dI}{dh} = -50\), and describing \(-210\) as the intensity rather than a rate per metre.

Mark scheme

1 mark
M1 Differentiates \(0.7^h\) to \(0.7^h\ln 0.7\): \(\dfrac{dI}{dh} = k \times 0.7^h\ln 0.7\); or correct final answers of awrt \(-210\) and awrt 6.02 with no working
Correct answers of awrt \(-210\) and awrt 6.02 (metres) with no working score 4 marks (M1A0A1 and M1A1); the B1 needs the interpretation. Writing \(0.7^h = \mathrm{e}^{h\ln 0.7}\) and differentiating is acceptable. \(0.7^h \to h \times 0.7^{h - 1}\) scores M0.
1 mark
A1 \(\dfrac{dI}{dh} = 1200 \times 0.7^h\ln 0.7\)
Accept \(1200\ln(0.7) \times 0.7^h\), or a decimal coefficient of awrt −428, i.e. in [−428.5, −427.5), times \(0.7^h\).
1 mark
A1 \(\dfrac{dI}{dh} =\) awrt −210, i.e. in [−210.5, −209.5)
Depends on the M mark. The value is \(-209.72\ldots\); accept \(-209.7\) or \(-209.72\). Units are not required here.
1 mark
B1 Interprets: at a depth of 2 m, the intensity of the light is decreasing at a rate of about 210 lux per metre (as the depth increases)
Needs 'decreasing' (or 'falling', or a correct explanation of the minus sign) and 'per metre' of depth. Follow through their negative value. Do not accept 'the intensity is 210 lux' or 'the light decreases by 210'.
1 mark
M1 Sets \(1200 \times 0.7^h\ln 0.7 = -50\) (or the magnitude equal to 50), reaches \(0.7^h = k\) with \(k > 0\), and uses logarithms to find \(h\)
Depends on the first M mark. \(0.7^h = \dfrac{50}{1200 \times (-\ln 0.7)} = 0.1168\ldots\), so \(h = \dfrac{\ln(0.1168\ldots)}{\ln 0.7}\). Accept exact or decimal working, and solving with logarithms to any base. Setting \(I = 50\) is M0.
1 mark
A1 \(h =\) awrt 6.02 (metres), i.e. in [6.015, 6.025)
The value is 6.0198..., so accept any value in [6.015, 6.025), e.g. 6.020. Units not required.

Question 4

7 marks

A lift rises from the ground floor of a building. In a model, its height, \(h\) metres, above the ground floor \(t\) seconds after it starts is given by
\[h = 12 - 12\cos\left(\dfrac{\pi t}{12}\right), \quad 0 \leqslant t \leqslant 12\]
where the angle is in radians.

Find \(\dfrac{dh}{dt}\) and \(\dfrac{d^2h}{dt^2}\).

Find the exact speed of the lift when \(t = 2\).

Find, according to the model, the greatest speed of the lift and the value of \(t\) at which it occurs.

By considering the sign of \(\dfrac{d^2h}{dt^2}\), describe how the speed of the lift changes during the 12 seconds, and state the height at which the lift stops.

Model answer

\(\dfrac{dh}{dt} = 12 \times \dfrac{\pi}{12}\sin\left(\dfrac{\pi t}{12}\right) = \pi\sin\left(\dfrac{\pi t}{12}\right)\)
\(\dfrac{d^2h}{dt^2} = \dfrac{\pi^2}{12}\cos\left(\dfrac{\pi t}{12}\right)\)
\(t = 2\): speed \(= \pi\sin\dfrac{\pi}{6} = \dfrac{\pi}{2}\) m s\(^{-1}\)
The greatest value of \(\sin\left(\dfrac{\pi t}{12}\right)\) is 1, when \(\dfrac{\pi t}{12} = \dfrac{\pi}{2}\), i.e. \(t = 6\): greatest speed \(\pi\) m s\(^{-1}\) (3.14 m s\(^{-1}\)).
For \(0 \leqslant t < 6\), \(\cos\left(\dfrac{\pi t}{12}\right) > 0\), so \(\dfrac{d^2h}{dt^2} > 0\) and the speed increases; for \(6 < t \leqslant 12\), \(\dfrac{d^2h}{dt^2} < 0\) and the speed decreases, reaching 0 at \(t = 12\).
The lift stops at \(h = 12 - 12\cos\pi = 24\) m.

Explanation

Differentiating \(\cos kt\) gives \(-k\sin kt\), so the velocity is \(\pi\sin\left(\dfrac{\pi t}{12}\right)\) and the second derivative, the acceleration, is \(\dfrac{\pi^2}{12}\cos\left(\dfrac{\pi t}{12}\right)\). The speed is greatest when the sine is 1, at \(t = 6\), and the acceleration is positive before this and negative after it, so the lift speeds up and then slows to a stop at 24 m. Common errors are leaving out the factor \(\dfrac{\pi}{12}\) when differentiating \(\cos\left(\dfrac{\pi t}{12}\right)\) and using a calculator in degree mode.

Mark scheme

1 mark
M1 Differentiates \(\cos\left(\dfrac{\pi t}{12}\right) \to \pm k\sin\left(\dfrac{\pi t}{12}\right)\); or correct final answers of \(\dfrac{dh}{dt} = \pi\sin\left(\dfrac{\pi t}{12}\right)\), \(\dfrac{d^2h}{dt^2} = \dfrac{\pi^2}{12}\cos\left(\dfrac{\pi t}{12}\right)\), \(\dfrac{\pi}{2}\) m s\(^{-1}\), \(\pi\) m s\(^{-1}\) at \(t = 6\) and 24 m with no working
Correct answers to each part with no working score full marks, provided the description in point 6 is given in words.
1 mark
A1 \(\dfrac{dh}{dt} = \pi\sin\left(\dfrac{\pi t}{12}\right)\) and \(\dfrac{d^2h}{dt^2} = \dfrac{\pi^2}{12}\cos\left(\dfrac{\pi t}{12}\right)\)
Both needed. Accept \(12 \times \dfrac{\pi}{12}\sin\left(\dfrac{\pi t}{12}\right)\) unsimplified.
1 mark
A1 Speed at \(t = 2\): \(\pi\sin\dfrac{\pi}{6} = \dfrac{\pi}{2}\) m s\(^{-1}\)
Depends on the M mark. Exact; 1.57 alone scores A0. A calculator in degree mode gives 0.0287: A0.
1 mark
M1 Greatest speed where \(\sin\left(\dfrac{\pi t}{12}\right) = 1\), or where \(\dfrac{d^2h}{dt^2} = 0\): \(\dfrac{\pi t}{12} = \dfrac{\pi}{2}\)
1 mark
A1 Greatest speed \(\pi\) m s\(^{-1}\) (awrt 3.14, i.e. in [3.135, 3.145)) at \(t = 6\)
Both the speed and the time needed.
1 mark
B1 \(\dfrac{d^2h}{dt^2} > 0\) for \(0 \leqslant t < 6\), so the speed increases (the lift accelerates); \(\dfrac{d^2h}{dt^2} < 0\) for \(6 < t \leqslant 12\), so the speed decreases (the lift decelerates) until it is 0 at \(t = 12\)
Both halves needed, linked to the sign of the second derivative.
1 mark
B1 The lift stops at a height of 24 m
From \(h(12) = 12 - 12\cos\pi = 24\).

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