The shape of the lesson
By the end of the lesson, learners can
- Exponentials and logarithms: \(e^{x}\) is the one function that is its own derivative.
- Trigonometric functions: These results are only true when \(x\) is in radians.
- The inverse tangent: Paper 3 only.
- Using the derivatives: Everything from Pure 1 still applies: the gradient gives tangents and normals, and \(\dfrac{dy}{dx} = 0\) gives stationary points.
- 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.
One picture every idea returns to
\(\dfrac{d}{dx}e^{x} = e^{x}\), \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}\), \(\dfrac{d}{dx}\sin x = \cos x\), \(\dfrac{d}{dx}\cos x = -\sin x\), \(\dfrac{d}{dx}\tan x = \sec^2 x\), \(\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}\) (x in radians). For a composite, multiply by the derivative of the inside: \(\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}\) and \(\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Beyond powers of x.. Papers 2 and 3 (the derivative of \(\tan^{-1} x\) is Paper 3 only). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Exponentials and logarithms | \(e^{x}\) is the one function that is its own derivative. Model the worked example Exponentials on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Trigonometric functions | These results are only true when \(x\) is in radians. Model the worked example Chain rule with trig on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: The inverse tangent | Paper 3 only. Model the worked example Tangent to y = tan⁻¹ x on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: Using the derivatives | Everything from Pure 1 still applies: the gradient gives tangents and normals, and \(\dfrac{dy}{dx} = 0\) gives stationary points. Model the worked example A stationary point on the board, one line at a time. Your turn: Find the gradient of \(y = e^{2x} + \ln x\) at the point where \(x = 1\). Give an exact answer. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 26 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Using the derivatives, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(\dfrac{d}{dx}\ln(4x) = \dfrac{4}{4x} = \dfrac{1}{x}\), NOT \(\dfrac{4}{x}\) and not \(\dfrac{1}{4x}\). | (Check: \(\ln 4x = \ln 4 + \ln x\), and \(\ln 4\) is a constant.) |
| The minus sign belongs to cosine: \(\dfrac{d}{dx}\cos x = -\sin x\), but \(\dfrac{d}{dx}\sin x = +\cos x\). | A memory aid: going round sin, cos, \(-\sin\), \(-\cos\), each step is one derivative. |
| \(\tan^{-1} x\) is the inverse function, not \(\dfrac{1}{\tan x} = \cot x\). | Their derivatives are completely different: \(\dfrac{1}{1 + x^2}\) versus \(-\operatorname{cosec}^2 x\). |
| When solving \(e^{2x} = 3\), take \(\ln\) of both sides: \(2x = \ln 3\). | Writing \(2x = e^{3}\) or \(x = \ln\dfrac{3}{2}\) are the two classic slips. |
Derivatives of exponentials, logarithms and trigonometric functions: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
Differentiate with respect to \(x\): (a) \(e^{4x}\) (b) \(\ln(5x - 2)\) (c) \(\sin(3x + 1)\) (d) \(\tan\dfrac{x}{2}\).
Differentiate \(y = \ln\dfrac{(2x + 1)^3}{x^2 + 4}\).
The curve \(y = 2\cos^2 x + \sin 2x\) is defined for \(0 < x < \pi\). Find the exact \(x\)-coordinates of its stationary points.
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 \(ax + by = c\).
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.
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.
Answer key and marking notes
- (a) \(4e^{4x}\)
- (b) \(\dfrac{5}{5x - 2}\)
- (c) \(3\cos(3x + 1)\)
- (d) \(\dfrac{1}{2}\sec^2\dfrac{x}{2}\)
- Log laws: \(y = 3\ln(2x + 1) - \ln(x^2 + 4)\).
- \(\dfrac{dy}{dx} = 3 \times \dfrac{2}{2x + 1} - \dfrac{2x}{x^2 + 4}\).
- \(= \dfrac{6}{2x + 1} - \dfrac{2x}{x^2 + 4}\).
- \(\dfrac{dy}{dx} = 2 \times 2\cos x \times (-\sin x) + 2\cos 2x = -4\sin x\cos x + 2\cos 2x\).
- \(= -2\sin 2x + 2\cos 2x\).
- Set to zero: \(\sin 2x = \cos 2x\), so \(\tan 2x = 1\).
- For \(0 < 2x < 2\pi\): \(2x = \dfrac{\pi}{4}\) or \(\dfrac{5\pi}{4}\).
- \(x = \dfrac{\pi}{8},\ \dfrac{5\pi}{8}\).
- 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 \(x + 7y = 37\).
- \(\dfrac{dy}{dx} = \dfrac{1}{1 + x^2} - \dfrac{1}{2} = 0\), 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 \(\left(-1, \dfrac{1}{2} - \dfrac{\pi}{4}\right)\).
- (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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Differentiate \(e^{3x - 1}\) with respect to \(x\).
- A\(3e^{3x - 1}\)
- B\(e^{3x - 1}\)
- C\((3x - 1)e^{3x - 2}\)
- D\(\dfrac{1}{3}e^{3x - 1}\)
Differentiate \(\ln(5x^2)\) with respect to \(x\), where \(x > 0\).
- A\(\dfrac{2}{x}\)
- B\(\dfrac{1}{5x^2}\)
- C\(10x\)
- D\(\dfrac{2}{5x}\)
Differentiate \(3\tan 2x\) with respect to \(x\).
- A\(6\sec^2 2x\)
- B\(3\sec^2 2x\)
- C\(\dfrac{3}{2}\sec^2 2x\)
- D\(6\sec 2x\tan 2x\)
Find the gradient of the curve \(y = 2\sin x - \cos x\) at the point where \(x = \dfrac{\pi}{3}\).
- A\(1 + \dfrac{\sqrt{3}}{2}\)
- B\(1 - \dfrac{\sqrt{3}}{2}\)
- C\(\sqrt{3} - \dfrac{1}{2}\)
- D\(-1 + \dfrac{\sqrt{3}}{2}\)
(Paper 3 only) Find the gradient of the curve \(y = \tan^{-1}(3x)\) at the point where \(x = \dfrac{1}{3}\).
- A\(\dfrac{3}{2}\)
- B\(\dfrac{1}{2}\)
- C\(\dfrac{27}{10}\)
- D\(-3\operatorname{cosec}^2 1\)
Find the exact coordinates of the stationary point of the curve \(y = e^{2x} - 6x\).
- A\(\left(\dfrac{1}{2}\ln 3,\ 3 - 3\ln 3\right)\)
- B\(\left(\ln 3,\ 9 - 6\ln 3\right)\)
- C\(\left(\dfrac{1}{2}\ln 6,\ 6 - 3\ln 6\right)\)
- D\(\left(\ln\dfrac{3}{2},\ \dfrac{9}{4} - 6\ln\dfrac{3}{2}\right)\)
Find the equation of the normal to the curve \(y = \ln(2x - 1)\) at the point where \(x = 1\).
- A\(x + 2y = 1\)
- B\(y = 2x - 2\)
- C\(x - 2y = 1\)
- D\(x + y = 1\)
Find the values of \(x\), for \(0 < x < \pi\), at which the curve \(y = \sin 2x + 2\cos x\) has a stationary point.
- A\(x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)
- B\(x = \dfrac{\pi}{6}\) only
- C\(x = \dfrac{\pi}{2}\)
- D\(x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{3\pi}{2}\)
Quiz key
The inside is \(3x - 1\), with derivative 3, so \(\dfrac{d}{dx}e^{3x - 1} = 3e^{3x - 1}\). The tempting \(e^{3x - 1}\) forgets the chain rule factor. \((3x - 1)e^{3x - 2}\) treats the exponential like a power \(x^n\), which it is not, and \(\dfrac{1}{3}e^{3x - 1}\) is the integral, not the derivative.
\(\dfrac{10x}{5x^2} = \dfrac{2}{x}\). Or: \(\ln(5x^2) = \ln 5 + 2\ln x\), and \(\ln 5\) is a constant, so the derivative is \(\dfrac{2}{x}\). The tempting \(\dfrac{1}{5x^2}\) forgets to multiply by the derivative of the inside. \(10x\) forgets to divide by \(f(x)\), and \(\dfrac{2}{5x}\) keeps the 5, which belongs to a constant term.
\(\dfrac{d}{dx}\tan 2x = 2\sec^2 2x\), so \(\dfrac{d}{dx}3\tan 2x = 6\sec^2 2x\). The tempting \(3\sec^2 2x\) forgets the factor 2 from the inside. \(\dfrac{3}{2}\sec^2 2x\) divides by 2 instead of multiplying (an integration habit), and \(6\sec 2x\tan 2x\) is the derivative of \(3\sec 2x\), not \(3\tan 2x\).
\(\dfrac{dy}{dx} = 2\cos x + \sin x\). At \(x = \dfrac{\pi}{3}\): \(2 \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} = 1 + \dfrac{\sqrt{3}}{2}\). The tempting \(1 - \dfrac{\sqrt{3}}{2}\) uses \(\dfrac{d}{dx}\cos x = \sin x\), losing the minus sign that makes \(-\cos x\) become \(+\sin x\). \(\sqrt{3} - \dfrac{1}{2}\) is the value of \(y\), not the gradient, and \(-1 + \dfrac{\sqrt{3}}{2}\) gets both trigonometric derivatives with the wrong sign.
\(\dfrac{dy}{dx} = \dfrac{3}{1 + (3x)^2} = \dfrac{3}{1 + 9x^2}\). At \(x = \dfrac{1}{3}\): \(\dfrac{3}{1 + 1} = \dfrac{3}{2}\). The tempting \(\dfrac{1}{2}\) forgets the chain rule factor 3. \(\dfrac{27}{10}\) squares only \(x\), using \(\dfrac{3}{1 + x^2}\), and \(-3\operatorname{cosec}^2 1\) treats \(\tan^{-1}(3x)\) as \(\cot 3x\), the reciprocal, which is a different function.
\(2e^{2x} = 6\), so \(e^{2x} = 3\), \(2x = \ln 3\), \(x = \dfrac{1}{2}\ln 3\). Then \(y = e^{\ln 3} - 6 \times \dfrac{1}{2}\ln 3 = 3 - 3\ln 3\). The tempting \(x = \ln 3\) forgets to halve after taking logs. \(\dfrac{1}{2}\ln 6\) solves \(e^{2x} = 6\), forgetting the factor 2 in the derivative, and \(\ln\dfrac{3}{2}\) divides inside the log instead of dividing \(\ln 3\) by 2.
At \(x = 1\), \(y = \ln 1 = 0\). \(\dfrac{dy}{dx} = \dfrac{2}{2x - 1} = 2\), so the normal has gradient \(-\dfrac{1}{2}\): \(y = -\dfrac{1}{2}(x - 1)\), that is \(x + 2y = 1\). The tempting \(y = 2x - 2\) is the tangent, not the normal. \(x - 2y = 1\) uses gradient \(+\dfrac{1}{2}\), forgetting the minus sign, and \(x + y = 1\) uses \(\dfrac{dy}{dx} = \dfrac{1}{2x - 1} = 1\), missing the chain rule factor 2.
\(\dfrac{dy}{dx} = 2\cos 2x - 2\sin x = 2(1 - 2\sin^2 x) - 2\sin x\). Setting this to zero: \(2\sin^2 x + \sin x - 1 = 0\), so \((2\sin x - 1)(\sin x + 1) = 0\). \(\sin x = \dfrac{1}{2}\) gives \(\dfrac{\pi}{6}, \dfrac{5\pi}{6}\); \(\sin x = -1\) gives \(\dfrac{3\pi}{2}\), which is outside \(0 < x < \pi\). The tempting last option forgets to check the range. \(\dfrac{\pi}{6}\) only misses the second-quadrant angle, and \(\dfrac{\pi}{2}\) comes from \(\dfrac{d}{dx}\cos x = +\sin x\), which leads to \(\sin x = 1\) or \(-\dfrac{1}{2}\).
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.