IGCSE · Practice questions · Differentiation and the chain rule

Outside, inside, multiply.

Six original Cambridge-style questions on differentiation: rewriting as powers, the standard results for trig, exponential and log functions, the chain rule, second derivatives and the gradient at a given point.

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

Rewrite, differentiate, then substitute.

Rewrite roots and fractions as powers: \(\sqrt{x} = x^{\frac{1}{2}}\), \(\dfrac{a}{x^{n}} = ax^{-n}\); split a fraction over a single term.
Differentiate term by term with the standard results; keep constant multiples.
Chain rule: outside derivative, inside unchanged, times the inside derivative.
Second derivative: differentiate your first derivative again.
Gradient at a point: substitute the \(x\)-value into \(\dfrac{dy}{dx}\), in radians for trig.

Recall

[3 marks]

Find \(\dfrac{dy}{dx}\) when \(y = 2x^{5} - \dfrac{3}{x} + 4\sqrt{x}\).

Rewrite: \(y = 2x^{5} - 3x^{-1} + 4x^{\frac{1}{2}}\). ✓
\(\dfrac{dy}{dx} = 10x^{4} + 3x^{-2} + 2x^{-\frac{1}{2}}\). ✓
\(\dfrac{dy}{dx} = 10x^{4} + \dfrac{3}{x^{2}} + \dfrac{2}{\sqrt{x}}\). ✓

Calculation

[4 marks]

Given that \(f(x) = \dfrac{x^{2} - 4}{2\sqrt{x}}\) for \(x > 0\), find \(f'(x)\) and hence find \(f'(4)\).

Split: \(f(x) = \dfrac{x^{2}}{2x^{\frac{1}{2}}} - \dfrac{4}{2x^{\frac{1}{2}}} = \tfrac{1}{2}x^{\frac{3}{2}} - 2x^{-\frac{1}{2}}\). ✓
\(f'(x) = \tfrac{3}{4}x^{\frac{1}{2}} + x^{-\frac{3}{2}}\). ✓
\(f'(4) = \tfrac{3}{4}(2) + \dfrac{1}{8}\). ✓
\(f'(4) = \dfrac{13}{8}\). ✓

Calculation

[3 marks]

Given that \(y = 3e^{2x} - 4\sin x + 5\ln x\), find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\).

\(\dfrac{dy}{dx} = 6e^{2x} - 4\cos x + \dfrac{5}{x}\). ✓
\(\dfrac{5}{x} = 5x^{-1} \to -5x^{-2}\) and \(-4\cos x \to +4\sin x\). ✓
\(\dfrac{d^{2}y}{dx^{2}} = 12e^{2x} + 4\sin x - \dfrac{5}{x^{2}}\). ✓

Calculation

[4 marks]

Differentiate with respect to \(x\):

(a) \((5 - 2x)^{-3}\) (b) \(\cos\left(4x - \dfrac{\pi}{3}\right)\) (c) \(\ln(3x^{2} + 1)\) (d) \(e^{x^{2}}\)

(a) \(-3(5 - 2x)^{-4} \times (-2) = 6(5 - 2x)^{-4}\). ✓
(b) \(-\sin\left(4x - \dfrac{\pi}{3}\right) \times 4 = -4\sin\left(4x - \dfrac{\pi}{3}\right)\). ✓
(c) \(\dfrac{1}{3x^{2} + 1} \times 6x = \dfrac{6x}{3x^{2} + 1}\). ✓
(d) \(e^{x^{2}} \times 2x = 2xe^{x^{2}}\). ✓

Analysis

[5 marks]

(a) A curve has equation \(y = (x^{3} - 19)^{\frac{2}{3}}\). Find the \(y\)-coordinate and the gradient of the curve at the point where \(x = 3\).

(b) Find the exact gradient of the curve \(y = \tan^{2} x\) at the point where \(x = \dfrac{\pi}{3}\).

(a) At \(x = 3\): \(x^{3} - 19 = 8\), so \(y = 8^{\frac{2}{3}} = 4\). ✓
\[\frac{dy}{dx} = \tfrac{2}{3}(x^{3} - 19)^{-\frac{1}{3}} \times 3x^{2} = \frac{2x^{2}}{(x^{3} - 19)^{\frac{1}{3}}}\] ✓
At \(x = 3\): \(\dfrac{2(9)}{8^{\frac{1}{3}}} = \dfrac{18}{2} = 9\). ✓
(b) \(\dfrac{dy}{dx} = 2\tan x\sec^{2} x\); at \(x = \dfrac{\pi}{3}\), \(\tan\dfrac{\pi}{3} = \sqrt{3}\) and \(\sec^{2}\dfrac{\pi}{3} = \dfrac{1}{(\frac{1}{2})^{2}} = 4\). ✓
Gradient \(= 2\sqrt{3} \times 4 = 8\sqrt{3}\). ✓

Analysis

[5 marks]

The curve \(y = ax^{2} + \dfrac{b}{x}\), where \(a\) and \(b\) are constants, has gradient 0 at the point where \(x = 1\) and gradient 7 at the point where \(x = 2\).

(a) Find the values of \(a\) and \(b\).

(b) Find the value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x = 1\).

(a) \(y = ax^{2} + bx^{-1}\), so \(\dfrac{dy}{dx} = 2ax - \dfrac{b}{x^{2}}\). ✓
\(x = 1\): \(2a - b = 0\), so \(b = 2a\). \(x = 2\): \(4a - \dfrac{b}{4} = 7\). ✓
Substitute: \(4a - \dfrac{a}{2} = 7 \Rightarrow \dfrac{7a}{2} = 7 \Rightarrow a = 2\), \(b = 4\). ✓
(b) \(\dfrac{d^{2}y}{dx^{2}} = 2a + \dfrac{2b}{x^{3}} = 4 + \dfrac{8}{x^{3}}\). ✓
At \(x = 1\): \(\dfrac{d^{2}y}{dx^{2}} = 12\). ✓

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.