IGCSE · Practice questions · Product and quotient rules

Differentiate, tidy, solve.

Six original Cambridge-style questions on the product and quotient rules: differentiating products and quotients of powers, exponentials, logarithms, roots and trigonometric functions, showing that a derivative has a given form, and using the factorised result to find stationary points.

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

The method

Four pieces, one rule, then tidy.

Spot the shape: a product \(uv\) or a quotient \(\dfrac{u}{v}\). A function inside another needs the chain rule as well.
List \(u\), \(u'\), \(v\), \(v'\), using the chain rule for brackets, roots and \(e^{kx}\).
Apply: product \(uv' + vu'\); quotient \(\dfrac{vu' - uv'}{v^{2}}\), bottom first, minus in the middle.
Tidy: use a common denominator for roots, take out every common factor, cancel brackets with \(v^{2}\).
Use it: set each factor of the numerator to zero (never divide by \(x\)), ignore \(e^{kx}\), and find \(y\) from the original equation.

Recall

[2 marks]

Differentiate \(y = x^{4}\ln x\), giving your answer in a factorised form.

\(u = x^{4}\), \(u' = 4x^{3}\); \(v = \ln x\), \(v' = \dfrac{1}{x}\). ✓
\(\dfrac{dy}{dx} = x^{4}\cdot\dfrac{1}{x} + 4x^{3}\ln x = x^{3} + 4x^{3}\ln x = x^{3}(1 + 4\ln x)\). ✓

Calculation

[3 marks]

Show that the derivative of \(y = \dfrac{3x - 1}{x + 2}\) can be written as \(\dfrac{k}{(x + 2)^{2}}\), stating the value of the constant \(k\).

\(u = 3x - 1\), \(u' = 3\); \(v = x + 2\), \(v' = 1\). ✓
\[\frac{dy}{dx} = \frac{3(x + 2) - (3x - 1)}{(x + 2)^{2}} = \frac{3x + 6 - 3x + 1}{(x + 2)^{2}}\] ✓
\(= \dfrac{7}{(x + 2)^{2}}\), so \(k = 7\). ✓

Calculation

[4 marks]

A curve has equation \(y = e^{2x}(\sin x + \cos x)\).

(a) Show that \(\dfrac{dy}{dx} = e^{2x}(\sin x + 3\cos x)\).

(b) Find the exact gradient of the curve at the point where \(x = \dfrac{\pi}{2}\).

(a) \(u = e^{2x}\), \(u' = 2e^{2x}\); \(v = \sin x + \cos x\), \(v' = \cos x - \sin x\). ✓
\[\frac{dy}{dx} = e^{2x}(\cos x - \sin x) + 2e^{2x}(\sin x + \cos x)\] ✓
\(= e^{2x}(\cos x - \sin x + 2\sin x + 2\cos x) = e^{2x}(\sin x + 3\cos x)\), as required. ✓
(b) At \(x = \dfrac{\pi}{2}\): \(\sin\dfrac{\pi}{2} = 1\), \(\cos\dfrac{\pi}{2} = 0\), so the gradient is \(e^{\pi}(1 + 0) = e^{\pi}\). ✓

Analysis

[5 marks]

The curve \(y = x^{2}\sqrt{x + 3}\) is defined for \(x \ge -3\).

(a) Show that \(\dfrac{dy}{dx} = \dfrac{x(5x + 12)}{2\sqrt{x + 3}}\).

(b) Hence find the \(x\)-coordinates of the stationary points of the curve.

(a) \(u = x^{2}\), \(u' = 2x\); \(v = (x + 3)^{\frac{1}{2}}\), \(v' = \tfrac{1}{2}(x + 3)^{-\frac{1}{2}}\). ✓
\[\frac{dy}{dx} = \frac{x^{2}}{2\sqrt{x + 3}} + 2x\sqrt{x + 3}\] ✓
Over the common denominator \(2\sqrt{x + 3}\): \[\frac{dy}{dx} = \frac{x^{2} + 4x(x + 3)}{2\sqrt{x + 3}} = \frac{5x^{2} + 12x}{2\sqrt{x + 3}} = \frac{x(5x + 12)}{2\sqrt{x + 3}}\] ✓
(b) The numerator is zero when \(x = 0\) or \(5x + 12 = 0\), so \(x = 0\) or \(x = -\tfrac{12}{5}\). ✓
Both are greater than \(-3\), so both are in the domain: \(x = 0\) and \(x = -2.4\). ✓

Analysis

[5 marks]

A curve has equation \(y = \dfrac{x^{2} + 5}{x - 2}\), \(x \ne 2\).

(a) Find \(\dfrac{dy}{dx}\), giving the numerator in factorised form.

(b) Find the coordinates of the stationary points of the curve.

(a) \(u = x^{2} + 5\), \(u' = 2x\); \(v = x - 2\), \(v' = 1\). ✓
\[\frac{dy}{dx} = \frac{2x(x - 2) - (x^{2} + 5)}{(x - 2)^{2}} = \frac{x^{2} - 4x - 5}{(x - 2)^{2}} = \frac{(x - 5)(x + 1)}{(x - 2)^{2}}\] ✓
(b) \(\dfrac{dy}{dx} = 0\) when \(x = 5\) or \(x = -1\). ✓
\(x = 5\): \(y = \dfrac{25 + 5}{3} = 10\). \(x = -1\): \(y = \dfrac{1 + 5}{-3} = -2\). ✓
The stationary points are \((5, 10)\) and \((-1, -2)\). ✓

Analysis

[6 marks]

A curve has equation \(y = \dfrac{(\ln x)^{2}}{x}\), \(x > 0\).

(a) Show that \(\dfrac{dy}{dx} = \dfrac{\ln x\,(2 - \ln x)}{x^{2}}\).

(b) Hence find the exact coordinates of the two stationary points of the curve.

(a) \(u = (\ln x)^{2}\), so by the chain rule \(u' = 2\ln x \cdot \dfrac{1}{x} = \dfrac{2\ln x}{x}\); \(v = x\), \(v' = 1\). ✓
\[\frac{dy}{dx} = \frac{x\cdot\dfrac{2\ln x}{x} - (\ln x)^{2}}{x^{2}} = \frac{2\ln x - (\ln x)^{2}}{x^{2}} = \frac{\ln x\,(2 - \ln x)}{x^{2}}\] ✓
(b) The numerator is zero when \(\ln x = 0\) or \(\ln x = 2\), so \(x = 1\) or \(x = e^{2}\). ✓
\(x = 1\): \(y = \dfrac{0^{2}}{1} = 0\). \(x = e^{2}\): \(y = \dfrac{2^{2}}{e^{2}} = \dfrac{4}{e^{2}}\). ✓
The stationary points are \((1, 0)\) and \(\left(e^{2}, \dfrac{4}{e^{2}}\right)\). ✓

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.