First principles and standard derivatives - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on First principles and standard derivatives 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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3–9 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

A factory makes a cleaning fluid. The daily cost, £\(C\), of making \(x\) litres of the fluid is modelled by
\[C = 250 + 3x - 4\sqrt{x}, \quad 100 \leqslant x \leqslant 1500\]
Find the value of \(\dfrac{dC}{dx}\) when \(x = 400\), and explain what this value means in the context of the model.

Model answer

\(C = 250 + 3x - 4x^{\frac{1}{2}}\)
\(\dfrac{dC}{dx} = 3 - 2x^{-\frac{1}{2}} = 3 - \dfrac{2}{\sqrt{x}}\)
At \(x = 400\): \(\dfrac{dC}{dx} = 3 - \dfrac{2}{20} = 2.9\)
When 400 litres a day are made, the daily cost is increasing at a rate of £2.90 per litre, so one more litre costs about £2.90 to make.

Explanation

\(\dfrac{dC}{dx}\) is the rate of change of the cost with respect to the number of litres, so its value is in pounds per litre. The M and A marks are for differentiating \(4\sqrt{x}\) correctly and evaluating at \(x = 400\); the B mark needs both the rate and the context. A common error is to describe 2.9 as the cost itself rather than the extra cost of one more litre.

Mark scheme

1 mark
M1 Differentiates: \(\sqrt{x} = x^{\frac{1}{2}} \to x^{-\frac{1}{2}}\) or \(3x \to 3\); or a correct final answer of 2.9 with no working
An answer of 2.9 with no working scores 2 marks; the B1 for the interpretation is independent.
1 mark
A1 \(\dfrac{dC}{dx} = 3 - 2x^{-\frac{1}{2}}\), giving \(2.9\) when \(x = 400\)
Accept 2.90 or £2.90 per litre.
1 mark
B1 Interpretation in context: when 400 litres a day are being made, the daily cost is increasing at £2.90 per litre (making one more litre costs about £2.90)
Needs the idea of a rate (per extra litre, or per litre increase in output) and the context of cost. Do not accept 'the cost is £2.90', 'the gradient is 2.9' or 'the cost goes up by 2.9' with no units or no reference to litres. Follow through their value of \(\dfrac{dC}{dx}\) if positive.

Question 2

4 marks

\[f(x) = 3x^4 - 2x\]
Prove, from first principles, that \(f'(x) = 12x^3 - 2\).
[You may use \(f'(x) = \displaystyle\lim_{h \to 0}\left(\dfrac{f(x + h) - f(x)}{h}\right)\).]

Model answer

\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{3(x + h)^4 - 2(x + h) - \left(3x^4 - 2x\right)}{h}\)
\((x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4\)
\(= \dfrac{3x^4 + 12x^3h + 18x^2h^2 + 12xh^3 + 3h^4 - 2x - 2h - 3x^4 + 2x}{h}\)
\(= \dfrac{12x^3h + 18x^2h^2 + 12xh^3 + 3h^4 - 2h}{h} = 12x^3 + 18x^2h + 12xh^2 + 3h^3 - 2\)
As \(h \to 0\), \(18x^2h + 12xh^2 + 3h^3 \to 0\), so
\(f'(x) = \displaystyle\lim_{h \to 0}\left(12x^3 + 18x^2h + 12xh^2 + 3h^3 - 2\right) = 12x^3 - 2\)

Explanation

Write the gradient of the chord joining the points with \(x\)-coordinates \(x\) and \(x + h\), expand \((x + h)^4\) using the coefficients 1, 4, 6, 4, 1, cancel, divide by \(h\) and then let \(h \to 0\). The final mark needs the limit (\(h \to 0\), not just \(h = 0\)) and the conclusion \(f'(x) = 12x^3 - 2\). Common errors are wrong coefficients in the expansion of \((x + h)^4\), writing \(-2(x + h)\) as \(-2x + h\), and not dividing out the \(h\) before taking the limit.

Mark scheme

1 mark
B1 Gives the gradient of a chord in a correct form, e.g. \(\dfrac{f(x + h) - f(x)}{h} = \dfrac{3(x + h)^4 - 2(x + h) - \left(3x^4 - 2x\right)}{h}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof, so stating \(f'(x) = 12x^3 - 2\), or differentiating by rule, earns nothing. Allow \(\delta x\), or any other letter, for \(h\) if used consistently. Condone missing brackets if they are recovered. Also allow \(\dfrac{f(x) - f(x - h)}{h}\) or \(\dfrac{f(x + h) - f(x - h)}{2h}\), used correctly.
1 mark
M1 Expands \((x + h)^4\) with the correct structure \(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4\) (condone one slip in a coefficient) and the \(3x^4\) terms cancel
The expansion may come from Pascal's triangle, the binomial expansion or repeated multiplication. Failing to multiply every term of the expansion by 3 is condoned for this mark but loses the A1.
1 mark
A1 Reaches \(\dfrac{12x^3h + 18x^2h^2 + 12xh^3 + 3h^4 - 2h}{h} = 12x^3 + 18x^2h + 12xh^2 + 3h^3 - 2\) with no algebraic errors
The \(-2h\) from \(-2(x + h) + 2x\) must be correct. Dividing through by \(h\) must be seen before the limit is taken.
1 mark
A1* Completes the limiting argument: \(f'(x) = \displaystyle\lim_{h \to 0}\left(12x^3 + 18x^2h + 12xh^2 + 3h^3 - 2\right) = 12x^3 - 2\), cso
\(h \to 0\) must be referred to at least once (it need not be on every line). Do not allow \(h = 0\) if there is never a reference to \(h \to 0\). They must reach \(12x^3 + 18x^2h + 12xh^2 + 3h^3 - 2\) (the \(h\) divided out) before taking the limit, and must conclude \(f'(x) = 12x^3 - 2\) (or \(\dfrac{dy}{dx} = \), or 'gradient ='), not stop at the limit expression.

Question 3

5 marks

The curve \(C\) has equation \(y = x^3 - 5x + 2\).
The points \(A\) and \(B\) lie on \(C\) and have \(x\)-coordinates \(2 - h\) and \(2 + h\) respectively, where \(h > 0\).

Show that the gradient of the chord \(AB\) is \(7 + h^2\).

Given that the chord \(AB\) is parallel to the line with equation \(y = 11x\), find the coordinates of \(A\) and the coordinates of \(B\).

By considering what happens to the chord \(AB\) as \(h \to 0\), deduce the gradient of the tangent to \(C\) at the point \(P(2, 0)\), explaining your reasoning.

Model answer

\(y_A = (2 - h)^3 - 5(2 - h) + 2\) and \(y_B = (2 + h)^3 - 5(2 + h) + 2\)
\(y_B - y_A = \left(8 + 12h + 6h^2 + h^3\right) - \left(8 - 12h + 6h^2 - h^3\right) - 5\left[(2 + h) - (2 - h)\right] = 24h + 2h^3 - 10h = 14h + 2h^3\)
Gradient of \(AB = \dfrac{14h + 2h^3}{(2 + h) - (2 - h)} = \dfrac{14h + 2h^3}{2h} = 7 + h^2\)
Parallel to \(y = 11x\): \(7 + h^2 = 11 \Rightarrow h^2 = 4 \Rightarrow h = 2\) (as \(h > 0\))
\(A\): \(x = 0\), \(y = 0 - 0 + 2 = 2\), so \(A(0, 2)\); \(B\): \(x = 4\), \(y = 64 - 20 + 2 = 46\), so \(B(4, 46)\)
As \(h \to 0\), \(A\) and \(B\) both move along \(C\) towards \(P(2, 0)\), and the chord \(AB\) approaches the tangent to \(C\) at \(P\).
So the gradient of the tangent at \(P\) is \(\displaystyle\lim_{h \to 0}\left(7 + h^2\right) = 7\)

Explanation

The chord joins points on either side of \(P\), so its horizontal change is \(2h\); after expanding, the even powers of \(h\) cancel and dividing by \(2h\) gives \(7 + h^2\). Setting this equal to 11 gives \(h = 2\), and as \(h \to 0\) the chord becomes the tangent at \(P\), whose gradient is the limit 7. Common errors are dividing by \(h\) instead of \(2h\) and describing the limit as 'putting \(h = 0\)'.

Mark scheme

1 mark
M1 Writes the \(y\)-coordinates of \(A\) and \(B\) as \((2 - h)^3 - 5(2 - h) + 2\) and \((2 + h)^3 - 5(2 + h) + 2\), attempts to expand \((2 - h)^3\) and \((2 + h)^3\), and uses \(\dfrac{y_B - y_A}{(2 + h) - (2 - h)}\). An answer with no working scores 0 marks for this mark and the next.
An answer with no working scores 0 marks for the show-that (the first two marks); correct coordinates \(A(0, 2)\) and \(B(4, 46)\) with no working score 2 marks (the M1 A1 for that part), and a gradient of 7 with no reasoning scores 0 for the last mark. The denominator must be \(2h\) (not \(h\)) for M1.
1 mark
A1* \(y_B - y_A = \left(8 + 12h + 6h^2 + h^3\right) - \left(8 - 12h + 6h^2 - h^3\right) - 5(2h) = 14h + 2h^3\), so the gradient is \(\dfrac{14h + 2h^3}{2h} = 7 + h^2\), cso
At least one correct unsimplified line, such as \(\dfrac{14h + 2h^3}{2h}\) or \(\dfrac{24h + 2h^3 - 10h}{2h}\), must be seen before the given answer, with no errors. The constant terms (8 and +2) must be seen to cancel or be shown to cancel.
1 mark
M1 Sets \(7 + h^2 = 11\) (the gradient of \(y = 11x\)) and solves to find \(h = 2\)
May use the given expression even if the show-that was not completed. \(h = -2\) must be rejected (or not used) since \(h > 0\); using both \(h = \pm 2\) gives the same two points, which is condoned.
1 mark
A1 \(A(0, 2)\) and \(B(4, 46)\)
Both points are needed for the mark, as coordinates or as \(x = 0, y = 2\) and \(x = 4, y = 46\). The points must be labelled correctly: \(A\) has the smaller \(x\)-coordinate.
1 mark
B1 As \(h \to 0\), \(A\) and \(B\) both move along \(C\) towards \(P\), so the chord \(AB\) approaches the tangent at \(P\); hence the gradient of the tangent is \(\displaystyle\lim_{h \to 0}\left(7 + h^2\right) = 7\)
Needs the limiting idea (\(h \to 0\), or \(A\) and \(B\) approaching \(P\)) and the value 7. Do not accept 'put \(h = 0\)' with no idea of a limit. A value of 7 found only by differentiating (\(3x^2 - 5\) at \(x = 2\)), with no reference to the chord, scores B0.

Question 4

7 marks

\[f(x) = \sin x + \cos x\]
where \(x\) is measured in radians.

Prove, from first principles, that \(f'(x) = \cos x - \sin x\).
You may use without proof the formulae for \(\sin(A + B)\) and \(\cos(A + B)\), and you may assume that, as \(h \to 0\), \(\dfrac{\sin h}{h} \to 1\) and \(\dfrac{\cos h - 1}{h} \to 0\).
[\(\sin(A + B) = \sin A\cos B + \cos A\sin B\) and \(\cos(A + B) = \cos A\cos B - \sin A\sin B\)]

Hence find the values of \(x\), for \(0 \leqslant x < 2\pi\), at which the gradient of the curve \(y = f(x)\) is zero. Give your answers in terms of \(\pi\).

Model answer

\(\dfrac{f(x + h) - f(x)}{h} = \dfrac{\sin(x + h) + \cos(x + h) - \sin x - \cos x}{h}\)
\(= \dfrac{\sin x\cos h + \cos x\sin h + \cos x\cos h - \sin x\sin h - \sin x - \cos x}{h}\)
\(= (\sin x + \cos x)\left(\dfrac{\cos h - 1}{h}\right) + (\cos x - \sin x)\left(\dfrac{\sin h}{h}\right)\)
As \(h \to 0\), \(\dfrac{\cos h - 1}{h} \to 0\) and \(\dfrac{\sin h}{h} \to 1\), so
\(f'(x) = \displaystyle\lim_{h \to 0}\left[(\sin x + \cos x)\left(\dfrac{\cos h - 1}{h}\right) + (\cos x - \sin x)\left(\dfrac{\sin h}{h}\right)\right] = (\sin x + \cos x) \times 0 + (\cos x - \sin x) \times 1\)
\(f'(x) = \cos x - \sin x\), as required.
Gradient zero: \(\cos x - \sin x = 0 \Rightarrow \tan x = 1\)
\(x = \dfrac{\pi}{4}\) or \(x = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}\)

Explanation

Expand \(\sin(x + h)\) and \(\cos(x + h)\), then collect the terms into multiples of \(\dfrac{\cos h - 1}{h}\) and \(\dfrac{\sin h}{h}\) so that the given limits can be applied as \(h \to 0\). The final proof mark needs the limiting language and the conclusion \(f'(x) = \cos x - \sin x\); then \(\tan x = 1\) gives two values in the interval. Common errors are writing the limits as equalities with no '\(h \to 0\)', and missing the second solution \(\dfrac{5\pi}{4}\).

Mark scheme

1 mark
B1 Gives the gradient of a chord: \(\dfrac{\sin(x + h) + \cos(x + h) - \sin x - \cos x}{h}\). An answer with no working scores 0 marks for the proof; answers of \(\dfrac{\pi}{4}\) and \(\dfrac{5\pi}{4}\) with no working score 2 marks (SC).
An answer with no working scores 0 marks for the proof (the first five marks): the derivative is given, so the proof must be seen. Answers of \(x = \dfrac{\pi}{4}\) and \(x = \dfrac{5\pi}{4}\) with no working score 2 marks (SC: the last two marks), as they follow directly from the given derivative. Allow \(\delta x\) for \(h\) if used consistently. Condone missing brackets if recovered. Also allow \(\dfrac{f(x) - f(x - h)}{h}\) or \(\dfrac{f(x + h) - f(x - h)}{2h}\) with the matching expansions.
1 mark
M1 Uses the compound angle formulae to expand both \(\sin(x + h)\) and \(\cos(x + h)\): \(\sin x\cos h \pm \cos x\sin h\) and \(\cos x\cos h \pm \sin x\sin h\)
Condone sign slips in the expansions for this mark.
1 mark
A1 \(\dfrac{\sin x\cos h + \cos x\sin h + \cos x\cos h - \sin x\sin h - \sin x - \cos x}{h}\) oe
May be implied by later work.
1 mark
M1 (dM1) Isolates \(\dfrac{\cos h - 1}{h}\) and \(\dfrac{\sin h}{h}\), e.g. \((\sin x + \cos x)\dfrac{\cos h - 1}{h} + (\cos x - \sin x)\dfrac{\sin h}{h}\), and replaces them with 0 and 1
Depends on the B mark and the first M mark (not on the A mark just before). Going straight from the unsplit fraction to \(\cos x - \sin x\) is dM0. Condone limit notation left in after the limits are applied.
1 mark
A1* Uses correct limiting language to conclude \(f'(x) = \cos x - \sin x\) with no errors, cso
\(h \to 0\) must be seen at some point, e.g. \(f'(x) = \displaystyle\lim_{h \to 0}\left[\ldots\right] = \cos x - \sin x\). Just stating \(\dfrac{\sin h}{h} = 1\) and \(\dfrac{\cos h - 1}{h} = 0\), or 'when \(h = 0\)', with no reference to \(h \to 0\) is A0. Withhold this mark for mixed variables (e.g. \(\theta\) and \(x\)) or bracketing errors that are not recovered. The final line must be \(f'(x) = \cos x - \sin x\) (or \(\dfrac{dy}{dx} =\)).
1 mark
M1 Sets \(\cos x - \sin x = 0\), reaches \(\tan x = 1\) (or \(\sin x = \cos x\)) and finds one value of \(x\)
May be scored using the given derivative even if the proof is incomplete. Squaring both sides is acceptable if any extra roots are then rejected.
1 mark
A1 \(x = \dfrac{\pi}{4}\) and \(x = \dfrac{5\pi}{4}\) only
Both values and no others in \(0 \leqslant x < 2\pi\). Degrees (\(45^\circ\), \(225^\circ\)) or decimals (0.785, 3.93) score A0. A common error is giving only \(\dfrac{\pi}{4}\).

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