A-Level Maths 9709 / Topic 3.4 / AS & A Level

When two functions meet.

Papers 2 and 3. The derivative of a product is not the product of the derivatives. The product rule and the quotient rule handle functions such as \(x^2\ln x\) and \(\dfrac{2x - 4}{3x + 2}\), and lead straight to tangents and stationary points.

Topic 3.4All differentiation lessons →

The key idea

Product rule: if \(y = uv\), then \(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\). Quotient rule: if \(y = \dfrac{u}{v}\), then \(\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\). In the quotient rule the order matters: bottom times derivative of top comes first.

Fig. 1: y = xe⁻ˣ drawn to scale. The product rule gives dy/dx = e⁻ˣ(1 − x), which is zero only at x = 1, so the curve has a horizontal tangent at its maximum point (1, 1/e) ≈ (1, 0.368)

Section 01

The product rule.

Name the two factors \(u\) and \(v\), differentiate each (chain rule if needed), then combine: first times derivative of second, plus second times derivative of first. Factorising the answer makes later steps much easier.

Worked example Power times log

Differentiate \(y = x^2\ln x\).

\(u = x^2\), \(\dfrac{du}{dx} = 2x\); \(v = \ln x\), \(\dfrac{dv}{dx} = \dfrac{1}{x}\).
\(\dfrac{dy}{dx} = x^2 \times \dfrac{1}{x} + \ln x \times 2x\).
\(= x + 2x\ln x = x(1 + 2\ln x)\).

Answer: \(\dfrac{dy}{dx} = x(1 + 2\ln x)\)

Worked example Product with a chain inside

Differentiate \(y = xe^{1 - x^2}\).

\(u = x\), \(\dfrac{du}{dx} = 1\); \(v = e^{1 - x^2}\), \(\dfrac{dv}{dx} = -2xe^{1 - x^2}\) (chain rule).
\(\dfrac{dy}{dx} = x \times (-2x)e^{1 - x^2} + e^{1 - x^2} \times 1\).
Factorise: \(= e^{1 - x^2}(1 - 2x^2)\).

Answer: \(\dfrac{dy}{dx} = (1 - 2x^2)e^{1 - x^2}\)

Worked example Exponential times trig

Differentiate \(y = e^{2x}\sin 3x\).

\(u = e^{2x}\), \(\dfrac{du}{dx} = 2e^{2x}\); \(v = \sin 3x\), \(\dfrac{dv}{dx} = 3\cos 3x\).
\(\dfrac{dy}{dx} = e^{2x} \times 3\cos 3x + \sin 3x \times 2e^{2x}\).

Answer: \(\dfrac{dy}{dx} = e^{2x}(2\sin 3x + 3\cos 3x)\)

Your turn

Differentiate \(y = x^3\cos 2x\).

Examiner trap

\(\dfrac{d}{dx}(uv) \ne \dfrac{du}{dx} \times \dfrac{dv}{dx}\). For \(y = x^2e^{3x}\), multiplying derivatives gives \(6xe^{3x}\), which is wrong; the product rule gives \(xe^{3x}(2 + 3x)\).

Section 02

The quotient rule.

For a fraction, \(u\) is the top and \(v\) is the bottom. The numerator of the answer has a minus sign, so the order matters: 'bottom times d(top), minus top times d(bottom), all over bottom squared'.

Worked example A linear fraction

Differentiate \(y = \dfrac{2x - 4}{3x + 2}\).

\(u = 2x - 4\), \(u' = 2\); \(v = 3x + 2\), \(v' = 3\).
\(\dfrac{dy}{dx} = \dfrac{(3x + 2)(2) - (2x - 4)(3)}{(3x + 2)^2}\).
Numerator: \(6x + 4 - 6x + 12 = 16\).

Answer: \(\dfrac{dy}{dx} = \dfrac{16}{(3x + 2)^2}\)

Worked example The derivative of tan x

Use the quotient rule to show that \(\dfrac{d}{dx}\tan x = \sec^2 x\).

\(\tan x = \dfrac{\sin x}{\cos x}\): \(u = \sin x\), \(u' = \cos x\); \(v = \cos x\), \(v' = -\sin x\).
\(\dfrac{dy}{dx} = \dfrac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}\).
\(= \dfrac{1}{\cos^2 x} = \sec^2 x\).

Answer: \(\dfrac{d}{dx}\tan x = \sec^2 x\), as required.

Worked example Log over x

Differentiate \(y = \dfrac{\ln x}{x}\).

\(u = \ln x\), \(u' = \dfrac{1}{x}\); \(v = x\), \(v' = 1\).
\(\dfrac{dy}{dx} = \dfrac{x \times \frac{1}{x} - \ln x \times 1}{x^2}\).

Answer: \(\dfrac{dy}{dx} = \dfrac{1 - \ln x}{x^2}\)

Examiner trap

Swapping the terms in the numerator, \(u v' - v u'\), changes the sign of the whole answer. For \(\dfrac{2x - 4}{3x + 2}\) it would give \(-\dfrac{16}{(3x + 2)^2}\). Also, never forget to square the denominator.

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If \(y = uv\), which is \(\dfrac{dy}{dx}\)?

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Differentiate \(xe^{x}\).

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If \(y = \dfrac{u}{v}\), which is \(\dfrac{dy}{dx}\)?

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Differentiate \(\dfrac{1}{x + 2}\) using the quotient rule.

Section 03

Stationary points and tangents.

The rules give \(\dfrac{dy}{dx}\) as a product or a fraction. Set it equal to zero: a product is zero when a factor is zero, and a fraction is zero when its numerator is zero. Since \(e^{\ldots}\) is never zero, it can always be ignored when solving.

Worked example Maximum of xe⁻ˣ

Find the stationary point of \(y = xe^{-x}\) and determine its nature.

\(\dfrac{dy}{dx} = x(-e^{-x}) + e^{-x} = e^{-x}(1 - x)\).
\(e^{-x} > 0\), so \(\dfrac{dy}{dx} = 0\) only when \(x = 1\); then \(y = e^{-1} = \dfrac{1}{e}\).
\(\dfrac{d^2y}{dx^2} = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2)\), which is \(-\dfrac{1}{e} < 0\) at \(x = 1\).

Answer: Maximum at \(\left(1, \dfrac{1}{e}\right)\) (Fig. 1)

Worked example Stationary points of a fraction

Find the coordinates of the stationary points of \(y = \dfrac{x^2 + 3}{x - 1}\).

\(\dfrac{dy}{dx} = \dfrac{(x - 1)(2x) - (x^2 + 3)(1)}{(x - 1)^2} = \dfrac{x^2 - 2x - 3}{(x - 1)^2}\).
Numerator zero: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\), so \(x = 3\) or \(x = -1\).
\(x = 3\): \(y = \dfrac{12}{2} = 6\). \(x = -1\): \(y = \dfrac{4}{-2} = -2\).

Answer: \((3, 6)\) and \((-1, -2)\)

Examiner trap

When solving \(e^{-x}(1 - x) = 0\), do not 'divide by zero' or try to solve \(e^{-x} = 0\): it has no solution. Only the other factor gives stationary points.

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Differentiate \(y = x^2e^{3x}\).

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Find the gradient of \(y = \dfrac{x}{2x + 1}\) at the point where \(x = 1\).

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The curve \(y = x^2\ln x\), \(x > 0\), has one stationary point. Find its \(x\)-coordinate.

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Find the gradient of \(y = \dfrac{\sin x}{1 + \cos x}\) at \(x = \dfrac{\pi}{3}\).

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Six original Cambridge-style questions

Differentiating products and quotients, deriving the derivatives of \(\sec x\) and \(\cot x\), stationary points, a tangent and a drug-concentration model.