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Cambridge IGCSE · Topic 14.4
IGCSE Add Maths 0606 / Topic 14 / IGCSE

Two functions, one derivative.

The chain rule handles a function inside another. When two functions are multiplied or divided, you need two more rules. The product rule and the quotient rule are not printed in the exam, so learn them by heart, then practise the part that earns most marks: tidying and factorising the answer so you can use it.

Topic 14All calculus lessons →
The key idea

If \(y = uv\), then \(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\). If \(y = \dfrac{u}{v}\), then \(\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}}\). Use the chain rule for each of \(u'\) and \(v'\) when they need it. Then take out common factors: a factorised derivative shows the form the question asks for and tells you at once where \(\dfrac{dy}{dx} = 0\).

uv vΔu uΔv ΔuΔv u Δu v Δv Change in area = vΔu + uΔv + ΔuΔv the corner is tiny and vanishes
Fig. 1: Why the product rule has two terms. A rectangle of sides u and v has area uv. When x changes a little, u grows by Δu and v by Δv, so the area gains a strip vΔu (teal), a strip uΔv (blue) and a tiny corner ΔuΔv (orange). Divide by Δx and let Δx tend to zero: the corner vanishes, leaving d(uv)/dx = u dv/dx + v du/dx. It is never just du/dx times dv/dx
Section 01

The product rule: two terms, not one.

When \(y\) is one function multiplied by another, call them \(u\) and \(v\). The derivative is NOT \(u' \times v'\): Fig. 1 shows the change in \(uv\) comes from two strips, one for each function changing while the other stays put. So keep one function, differentiate the other, then swap and add. Set your work out the same way every time: write \(u\), \(u'\), \(v\) and \(v'\) first, using the chain rule from the last lesson wherever a bracket or \(e^{kx}\) appears.

\[ y = uv \quad\Longrightarrow\quad \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \]

Worked example A power times an exponential
Differentiate \(y = x^{2}e^{3x}\), giving your answer in a fully factorised form.
  1. \(u = x^{2}\), so \(u' = 2x\). \(v = e^{3x}\), so by the chain rule \(v' = 3e^{3x}\).
  2. \[\frac{dy}{dx} = x^{2}\cdot 3e^{3x} + e^{3x}\cdot 2x\]
  3. Both terms contain \(x\) and \(e^{3x}\): \[\frac{dy}{dx} = xe^{3x}(3x + 2)\]
Answer: \(\dfrac{dy}{dx} = xe^{3x}(3x + 2)\)
Worked example A gradient at a point
Find the exact gradient of the curve \(y = x^{2}\ln x\) at the point where \(x = e\).
  1. \(u = x^{2}\), \(u' = 2x\); \(v = \ln x\), \(v' = \dfrac{1}{x}\).
  2. \[\frac{dy}{dx} = x^{2}\cdot\frac{1}{x} + \ln x\cdot 2x = x + 2x\ln x = x(1 + 2\ln x)\]
  3. At \(x = e\), \(\ln e = 1\): gradient \(= e(1 + 2) = 3e\).
Answer: \(3e\)
Your turn
Differentiate \(y = x\cos x\).
Examiner trap

Multiplying the two derivatives is the classic error: for \(x^{2}e^{3x}\) it gives \(2x \times 3e^{3x} = 6xe^{3x}\), which is wrong. The second classic error is dropping the chain rule inside \(v\): the derivative of \(e^{3x}\) is \(3e^{3x}\), not \(e^{3x}\), and the derivative of \(\sin 2x\) is \(2\cos 2x\).

Section 02

The quotient rule: bottom first.

For \(y = \dfrac{u}{v}\), the numerator is a difference, so here the order DOES matter. A good way to remember it: start with the bottom, \(v\), times the derivative of the top, then subtract the top times the derivative of the bottom, and divide everything by the bottom squared. Leave \(v^{2}\) as a square, do not expand it; it often cancels with a factor in the numerator.

\[ y = \frac{u}{v} \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}} \]

Worked example A logarithm over x
Find \(\dfrac{dy}{dx}\) for \(y = \dfrac{\ln x}{x}\).
  1. \(u = \ln x\), \(u' = \dfrac{1}{x}\); \(v = x\), \(v' = 1\).
  2. \[\frac{dy}{dx} = \frac{x\cdot\dfrac{1}{x} - \ln x\cdot 1}{x^{2}}\]
  3. \(x\cdot\dfrac{1}{x} = 1\), so \(\dfrac{dy}{dx} = \dfrac{1 - \ln x}{x^{2}}\).
Answer: \(\dfrac{dy}{dx} = \dfrac{1 - \ln x}{x^{2}}\)
Worked example An identity tidies it
Show that the derivative of \(y = \dfrac{\sin x}{1 + \cos x}\) is \(\dfrac{1}{1 + \cos x}\).
  1. \(u = \sin x\), \(u' = \cos x\); \(v = 1 + \cos x\), \(v' = -\sin x\).
  2. \[\frac{dy}{dx} = \frac{(1 + \cos x)\cos x - \sin x(-\sin x)}{(1 + \cos x)^{2}}\]
  3. Numerator: \(\cos x + \cos^{2}x + \sin^{2}x = \cos x + 1\), using \(\sin^{2}x + \cos^{2}x = 1\).
  4. \[\frac{dy}{dx} = \frac{1 + \cos x}{(1 + \cos x)^{2}} = \frac{1}{1 + \cos x}\]
Answer: \(\dfrac{dy}{dx} = \dfrac{1}{1 + \cos x}\), as required.
Your turn
Differentiate \(y = \dfrac{e^{x}}{x + 2}\) and factorise the numerator.
Examiner trap

Writing \(u v' - v u'\) on top puts the terms the wrong way round and gives the negative of the answer: for \(\dfrac{\ln x}{x}\) it gives \(\dfrac{\ln x - 1}{x^{2}}\). Also, \(\dfrac{u'}{v'}\) is not the derivative of \(\dfrac{u}{v}\): differentiating the top and bottom separately is never allowed.

Stage 1 · Learn

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

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Find \(\dfrac{dy}{dx}\) for \(y = \dfrac{x}{x + 1}\).

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Differentiate \(y = x\sin 2x\).

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For \(y = xe^{-2x}\), \(\dfrac{dy}{dx} = e^{-2x}(1 - 2x)\). Where is the stationary point?

Section 03

Simplify, factorise, then use it.

Exam questions rarely stop at 'differentiate'. They ask you to show that \(\dfrac{dy}{dx}\) has a given form, such as \(\dfrac{ax + b}{\sqrt{2x + 1}}\), and then 'hence' find the stationary point, where \(\dfrac{dy}{dx} = 0\). Two habits make this easy. First, combine terms over a common denominator: a term with \((2x + 1)^{-\frac{1}{2}}\) and a term with \((2x + 1)^{\frac{1}{2}}\) combine by multiplying the second by \(\dfrac{\sqrt{2x + 1}}{\sqrt{2x + 1}}\). Second, take out every common factor, because a fraction is zero only when its numerator is zero, and a product is zero only when one factor is.

Worked example Show that, then hence
The curve \(y = x\sqrt{2x + 1}\) is defined for \(x \ge -\tfrac{1}{2}\). (a) Show that \(\dfrac{dy}{dx} = \dfrac{3x + 1}{\sqrt{2x + 1}}\). (b) Hence find the exact coordinates of the stationary point.
  1. (a) \(u = x\), \(u' = 1\); \(v = (2x + 1)^{\frac{1}{2}}\), \(v' = \tfrac{1}{2}(2x + 1)^{-\frac{1}{2}} \times 2 = (2x + 1)^{-\frac{1}{2}}\).
  2. \[\frac{dy}{dx} = \frac{x}{\sqrt{2x + 1}} + \sqrt{2x + 1} = \frac{x + (2x + 1)}{\sqrt{2x + 1}} = \frac{3x + 1}{\sqrt{2x + 1}}\]
  3. (b) \(\dfrac{dy}{dx} = 0\) when \(3x + 1 = 0\), so \(x = -\tfrac{1}{3}\), which is in the domain.
  4. \(y = -\tfrac{1}{3}\sqrt{\tfrac{1}{3}} = -\dfrac{1}{3\sqrt{3}} = -\dfrac{\sqrt{3}}{9}\).
Answer: (a) shown   (b) \(\left(-\tfrac{1}{3}, -\tfrac{\sqrt{3}}{9}\right)\)
Worked example Cancel a common bracket
Find \(\dfrac{dy}{dx}\) for \(y = \dfrac{x + 1}{(2x - 3)^{2}}\), simplifying fully, and find the coordinates of the stationary point.
  1. \(u = x + 1\), \(u' = 1\); \(v = (2x - 3)^{2}\), \(v' = 2(2x - 3) \times 2 = 4(2x - 3)\).
  2. \[\frac{dy}{dx} = \frac{(2x - 3)^{2} - 4(x + 1)(2x - 3)}{(2x - 3)^{4}}\]
  3. Take out \((2x - 3)\) on top: \((2x - 3)\big[(2x - 3) - 4(x + 1)\big] = (2x - 3)(-2x - 7)\). Cancel one \((2x - 3)\): \[\frac{dy}{dx} = -\frac{2x + 7}{(2x - 3)^{3}}\]
  4. \(\dfrac{dy}{dx} = 0\) when \(2x + 7 = 0\), so \(x = -\tfrac{7}{2}\). Then \(y = \dfrac{-\tfrac{5}{2}}{(-10)^{2}} = -\dfrac{1}{40}\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{2x + 7}{(2x - 3)^{3}}\); stationary point \(\left(-\tfrac{7}{2}, -\tfrac{1}{40}\right)\)
Your turn
Find the coordinates of the stationary points of \(y = x^{2}e^{-x}\).
Examiner trap

In \(xe^{-x}(2 - x) = 0\), dividing both sides by \(x\) throws away the solution \(x = 0\); set each factor to zero instead. And never set the DENOMINATOR to zero: in Example 2, \(x = \tfrac{3}{2}\) makes \(\dfrac{dy}{dx}\) undefined, not zero, because the curve is not defined there.

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Find \(\dfrac{dy}{dx}\) for \(y = \dfrac{\ln x}{x^{2}}\), simplifying your answer.

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Given that \(y = (x - 1)\sqrt{4x + 1}\), it can be shown that \(\dfrac{dy}{dx} = \dfrac{ax + b}{\sqrt{4x + 1}}\). Find \(a\) and \(b\).

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Find the coordinates of the stationary points of \(y = \dfrac{x}{x^{2} + 4}\).

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Find the \(x\)-coordinates of the stationary points of \(y = x^{2}(3 - x)^{4}\).

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Products and quotients differentiated, factorised and used.

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Six original Cambridge-style questions
Product rule and quotient rule practice with the chain rule inside, show-that forms, factorising the derivative and finding stationary points.