AS & A Level · Practice questions · The product and quotient rules

u, v and order.

Six original Cambridge-style questions on the product and quotient rules, at Paper 2 and Paper 3 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Name u and v, differentiate each, combine.

Product: \((uv)' = uv' + vu'\).
Quotient: \(\left(\dfrac{u}{v}\right)' = \dfrac{vu' - uv'}{v^2}\).
Chain rule inside \(u\) or \(v\) (for example \(e^{3x}\), \(\sin 2x\)).
Factorise the answer, especially common exponentials.
Stationary points: a product is zero when a factor is; a fraction is zero when its numerator is.

Recall

[2 marks]

Differentiate \(y = x^3e^{2x}\), simplifying your answer.

\(\dfrac{dy}{dx} = 3x^2e^{2x} + x^3 \times 2e^{2x}\). ✓
\(= x^2e^{2x}(3 + 2x)\). ✓

Calculation

[3 marks]

Differentiate \(y = \dfrac{3x - 1}{x^2 + 2}\), simplifying the numerator.

\(\dfrac{dy}{dx} = \dfrac{(x^2 + 2)(3) - (3x - 1)(2x)}{(x^2 + 2)^2}\). ✓
Numerator: \(3x^2 + 6 - 6x^2 + 2x = -3x^2 + 2x + 6\). ✓
\(\dfrac{dy}{dx} = \dfrac{6 + 2x - 3x^2}{(x^2 + 2)^2}\). ✓

Calculation

[5 marks]

Find the \(x\)-coordinates of the stationary points of \(y = e^{-x}\sin 2x\) for \(0 < x < \pi\). Give your answers correct to 3 significant figures.

\(\dfrac{dy}{dx} = e^{-x} \times 2\cos 2x + \sin 2x \times (-e^{-x}) = e^{-x}(2\cos 2x - \sin 2x)\). ✓
\(e^{-x} \ne 0\), so \(2\cos 2x = \sin 2x\), giving \(\tan 2x = 2\). ✓
For \(0 < 2x < 2\pi\): \(2x = 1.1071\) or \(1.1071 + \pi = 4.2487\). ✓
\(x = 0.554,\ 2.12\). ✓

Proof

[4 marks]

Use the quotient rule to show that (a) \(\dfrac{d}{dx}\sec x = \sec x\tan x\) (b) \(\dfrac{d}{dx}\cot x = -\operatorname{cosec}^2 x\).

(a) \(\sec x = \dfrac{1}{\cos x}\): \(\dfrac{\cos x \times 0 - 1 \times (-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x}\). ✓
\(= \dfrac{1}{\cos x} \times \dfrac{\sin x}{\cos x} = \sec x\tan x\). ✓
(b) \(\cot x = \dfrac{\cos x}{\sin x}\): \(\dfrac{\sin x(-\sin x) - \cos x\cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}\). ✓
\(= -\dfrac{1}{\sin^2 x} = -\operatorname{cosec}^2 x\). ✓

Analysis

[4 marks]

Find the equation of the tangent to the curve \(y = x\ln x\) at the point where \(x = e\), and find where this tangent crosses the \(x\)-axis.

At \(x = e\): \(y = e\ln e = e\). ✓
\(\dfrac{dy}{dx} = x \times \dfrac{1}{x} + \ln x = 1 + \ln x = 2\) at \(x = e\). ✓
Tangent: \(y - e = 2(x - e)\), so \(y = 2x - e\). ✓
Crosses the \(x\)-axis when \(y = 0\): \(x = \dfrac{e}{2}\). ✓

Analysis

[6 marks]

The concentration, \(C\) mg per litre, of a drug in a patient's blood \(t\) hours after an injection is modelled by \(C = 12te^{-0.5t}\).

(a) Find \(\dfrac{dC}{dt}\).

(b) Find the maximum concentration and the time at which it occurs.

(a) \(\dfrac{dC}{dt} = 12e^{-0.5t} + 12t(-0.5)e^{-0.5t} = 6e^{-0.5t}(2 - t)\). ✓
(b) \(\dfrac{dC}{dt} = 0\) when \(t = 2\) (the exponential is never zero). \(\dfrac{dC}{dt} > 0\) just before and \(< 0\) just after, so this is a maximum. ✓
\(C = 24e^{-1} = 8.83\) mg per litre, after 2 hours. ✓
(c) At \(t = 4\): \(6e^{-2}(2 - 4) = -12e^{-2} = -1.62\) mg per litre per hour. ✓
Negative: the concentration is falling at that time. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.