AS & A Level · Practice questions · Derivatives of exponentials, logarithms and trigonometric functions

Six derivatives, endless functions.

Six original Cambridge-style questions on the standard derivatives of exponential, logarithmic and trigonometric functions, at Paper 2 and Paper 3 standard (question 5 is Paper 3 only).

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

Standard result, times the inside derivative.

Exponential: \(\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}\).
Logarithm: \(\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}\); simplify with log laws first.
Trig (radians): \(\sin \to \cos\), \(\cos \to -\sin\), \(\tan \to \sec^2\).
Paper 3: \(\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}\).
Stationary points: set \(\dfrac{dy}{dx} = 0\) and use \(\ln\) to solve exponential equations exactly.

Recall

[4 marks]

Differentiate with respect to \(x\): (a) \(e^{4x}\) (b) \(\ln(5x - 2)\) (c) \(\sin(3x + 1)\) (d) \(\tan\dfrac{x}{2}\).

(a) \(4e^{4x}\) ✓
(b) \(\dfrac{5}{5x - 2}\) ✓
(c) \(3\cos(3x + 1)\) ✓
(d) \(\dfrac{1}{2}\sec^2\dfrac{x}{2}\) ✓

Calculation

[3 marks]

Differentiate \(y = \ln\dfrac{(2x + 1)^3}{x^2 + 4}\).

Log laws: \(y = 3\ln(2x + 1) - \ln(x^2 + 4)\). ✓
\(\dfrac{dy}{dx} = 3 \times \dfrac{2}{2x + 1} - \dfrac{2x}{x^2 + 4}\). ✓
\(= \dfrac{6}{2x + 1} - \dfrac{2x}{x^2 + 4}\). ✓

Calculation

[5 marks]

The curve \(y = 2\cos^2 x + \sin 2x\) is defined for \(0 < x < \pi\). Find the exact \(x\)-coordinates of its stationary points.

\(\dfrac{dy}{dx} = 2 \times 2\cos x \times (-\sin x) + 2\cos 2x = -4\sin x\cos x + 2\cos 2x\). ✓
\(= -2\sin 2x + 2\cos 2x\). ✓
Set to zero: \(\sin 2x = \cos 2x\), so \(\tan 2x = 1\). ✓
For \(0 < 2x < 2\pi\): \(2x = \dfrac{\pi}{4}\) or \(\dfrac{5\pi}{4}\). ✓
\(x = \dfrac{\pi}{8},\ \dfrac{5\pi}{8}\). ✓

Calculation

[4 marks]

Find the equation of the normal to the curve \(y = 3e^{2x - 4} + x\) at the point where \(x = 2\). Give your answer in the form \(ax + by = c\).

At \(x = 2\): \(y = 3e^{0} + 2 = 5\). ✓
\(\dfrac{dy}{dx} = 6e^{2x - 4} + 1 = 7\) at \(x = 2\). ✓
Normal gradient \(= -\dfrac{1}{7}\): \(y - 5 = -\dfrac{1}{7}(x - 2)\). ✓
\(7y - 35 = -x + 2\), so \(x + 7y = 37\). ✓

Analysis

[5 marks]

Paper 3 only. The curve \(y = \tan^{-1} x - \dfrac{1}{2}x\) has two stationary points. Find their exact coordinates and determine the nature of each.

\(\dfrac{dy}{dx} = \dfrac{1}{1 + x^2} - \dfrac{1}{2} = 0\), so \(1 + x^2 = 2\) and \(x = \pm 1\). ✓
\(x = 1\): \(y = \dfrac{\pi}{4} - \dfrac{1}{2}\). \(x = -1\): \(y = -\dfrac{\pi}{4} + \dfrac{1}{2}\). ✓
\(\dfrac{d^2y}{dx^2} = -\dfrac{2x}{(1 + x^2)^2}\). ✓
At \(x = 1\): \(-\dfrac{1}{2} < 0\), maximum at \(\left(1, \dfrac{\pi}{4} - \dfrac{1}{2}\right)\). ✓
At \(x = -1\): \(\dfrac{1}{2} > 0\), minimum at \(\left(-1, \dfrac{1}{2} - \dfrac{\pi}{4}\right)\). ✓

Analysis

[5 marks]

The temperature, \(T\) °C, of a cup of coffee \(t\) minutes after it is made is modelled by \(T = 20 + 70e^{-0.05t}\).

(a) Find the rate of change of temperature when \(t = 10\).

(b) Find the time at which the coffee is cooling at 1 °C per minute.

(a) \(\dfrac{dT}{dt} = 70 \times (-0.05)e^{-0.05t} = -3.5e^{-0.05t}\). ✓
At \(t = 10\): \(-3.5e^{-0.5} = -2.12\) °C per minute (falling at 2.12 °C per minute). ✓
(b) \(3.5e^{-0.05t} = 1\), so \(e^{-0.05t} = \dfrac{1}{3.5}\). ✓
\(-0.05t = \ln\dfrac{1}{3.5}\), so \(t = 20\ln 3.5 = 25.1\) minutes. ✓

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.