The shape of the lesson
By the end of the lesson, learners can
- The product rule: Name the two factors \(u\) and \(v\), differentiate each (chain rule if needed), then combine: first times derivative of second, plus second times derivative of first.
- The quotient rule: For a fraction, \(u\) is the top and \(v\) is the bottom.
- Stationary points and tangents: The rules give \(\dfrac{dy}{dx}\) as a product or a fraction.
- Product: \((uv)' = uv' + vu'\).
- Quotient: \(\left(\dfrac{u}{v}\right)' = \dfrac{vu' - uv'}{v^2}\).
- Chain rule inside \(u\) or \(v\) (for example \(e^{3x}\), \(\sin 2x\)).
- Factorise the answer, especially common exponentials.
- Stationary points: a product is zero when a factor is; a fraction is zero when its numerator is.
One picture every idea returns to
Product rule: if \(y = uv\), then \(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\). Quotient rule: if \(y = \dfrac{u}{v}\), then \(\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}\). In the quotient rule the order matters: bottom times derivative of top comes first.
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, When two functions meet.. Papers 2 and 3. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The product rule | Name the two factors \(u\) and \(v\), differentiate each (chain rule if needed), then combine: first times derivative of second, plus second times derivative of first. Model the worked example Power times log on the board, one line at a time. Your turn: Differentiate \(y = x^3\cos 2x\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: The quotient rule | For a fraction, \(u\) is the top and \(v\) is the bottom. Model the worked example A linear fraction on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Stationary points and tangents | The rules give \(\dfrac{dy}{dx}\) as a product or a fraction. Model the worked example Maximum of xe⁻ˣ on the board, one line at a time. | Numbered Heads Together |
| 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, 24 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 Stationary points and tangents, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| \(\dfrac{d}{dx}(uv) \ne \dfrac{du}{dx} \times \dfrac{dv}{dx}\). | For \(y = x^2e^{3x}\), multiplying derivatives gives \(6xe^{3x}\), which is wrong; the product rule gives \(xe^{3x}(2 + 3x)\). |
| Swapping the terms in the numerator, \(u v' - v u'\), changes the sign of the whole answer. | For \(\dfrac{2x - 4}{3x + 2}\) it would give \(-\dfrac{16}{(3x + 2)^2}\). Also, never forget to square the denominator. |
| When solving \(e^{-x}(1 - x) = 0\), do not 'divide by zero' or try to solve \(e^{-x} = 0\): it has no solution. | Only the other factor gives stationary points. |
The product and quotient rules: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
Differentiate \(y = x^3e^{2x}\), simplifying your answer.
Differentiate \(y = \dfrac{3x - 1}{x^2 + 2}\), simplifying the numerator.
Find the \(x\)-coordinates of the stationary points of \(y = e^{-x}\sin 2x\) for \(0 < x < \pi\). Give your answers correct to 3 significant figures.
Use the quotient rule to show that (a) \(\dfrac{d}{dx}\sec x = \sec x\tan x\) (b) \(\dfrac{d}{dx}\cot x = -\operatorname{cosec}^2 x\).
Find the equation of the tangent to the curve \(y = x\ln x\) at the point where \(x = e\), and find where this tangent crosses the \(x\)-axis.
The concentration, \(C\) mg per litre, of a drug in a patient's blood \(t\) hours after an injection is modelled by \(C = 12te^{-0.5t}\).
(a) Find \(\dfrac{dC}{dt}\).
(b) Find the maximum concentration and the time at which it occurs.
(c) Find the rate of change of concentration when \(t = 4\), and interpret its sign.
Answer key and marking notes
- \(\dfrac{dy}{dx} = 3x^2e^{2x} + x^3 \times 2e^{2x}\).
- \(= x^2e^{2x}(3 + 2x)\).
- \(\dfrac{dy}{dx} = \dfrac{(x^2 + 2)(3) - (3x - 1)(2x)}{(x^2 + 2)^2}\).
- Numerator: \(3x^2 + 6 - 6x^2 + 2x = -3x^2 + 2x + 6\).
- \(\dfrac{dy}{dx} = \dfrac{6 + 2x - 3x^2}{(x^2 + 2)^2}\).
- \(\dfrac{dy}{dx} = e^{-x} \times 2\cos 2x + \sin 2x \times (-e^{-x}) = e^{-x}(2\cos 2x - \sin 2x)\).
- \(e^{-x} \ne 0\), so \(2\cos 2x = \sin 2x\), giving \(\tan 2x = 2\).
- For \(0 < 2x < 2\pi\): \(2x = 1.1071\) or \(1.1071 + \pi = 4.2487\).
- \(x = 0.554,\ 2.12\).
- (a) \(\sec x = \dfrac{1}{\cos x}\): \(\dfrac{\cos x \times 0 - 1 \times (-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x}\).
- \(= \dfrac{1}{\cos x} \times \dfrac{\sin x}{\cos x} = \sec x\tan x\).
- (b) \(\cot x = \dfrac{\cos x}{\sin x}\): \(\dfrac{\sin x(-\sin x) - \cos x\cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}\).
- \(= -\dfrac{1}{\sin^2 x} = -\operatorname{cosec}^2 x\).
- At \(x = e\): \(y = e\ln e = e\).
- \(\dfrac{dy}{dx} = x \times \dfrac{1}{x} + \ln x = 1 + \ln x = 2\) at \(x = e\).
- Tangent: \(y - e = 2(x - e)\), so \(y = 2x - e\).
- Crosses the \(x\)-axis when \(y = 0\): \(x = \dfrac{e}{2}\).
- (a) \(\dfrac{dC}{dt} = 12e^{-0.5t} + 12t(-0.5)e^{-0.5t} = 6e^{-0.5t}(2 - t)\).
- (b) \(\dfrac{dC}{dt} = 0\) when \(t = 2\) (the exponential is never zero). \(\dfrac{dC}{dt} > 0\) just before and \(< 0\) just after, so this is a maximum.
- \(C = 24e^{-1} = 8.83\) mg per litre, after 2 hours.
- (c) At \(t = 4\): \(6e^{-2}(2 - 4) = -12e^{-2} = -1.62\) mg per litre per hour.
- Negative: the concentration is falling at that time.
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 \(y = x^3e^{2x}\) with respect to \(x\).
- A\(x^2e^{2x}(3 + 2x)\)
- B\(6x^2e^{2x}\)
- C\(x^2e^{2x}(3 + x)\)
- D\(3x^2e^{2x}\)
Differentiate \(y = \dfrac{x}{x + 3}\) with respect to \(x\).
- A\(\dfrac{3}{(x + 3)^2}\)
- B\(-\dfrac{3}{(x + 3)^2}\)
- C\(1\)
- D\(\dfrac{3}{x + 3}\)
Find the gradient of the curve \(y = x\ln x\) at the point where \(x = e\).
- A\(2\)
- B\(1\)
- C\(\dfrac{1}{e}\)
- D\(e\)
Differentiate \(y = \dfrac{\sin x}{x}\) with respect to \(x\).
- A\(\dfrac{x\cos x - \sin x}{x^2}\)
- B\(\dfrac{\sin x - x\cos x}{x^2}\)
- C\(\cos x\)
- D\(\dfrac{x\cos x + \sin x}{x^2}\)
Find the coordinates of the stationary points of the curve \(y = x^2e^{-x}\).
- A\((0, 0)\) and \(\left(2, \dfrac{4}{e^{2}}\right)\)
- B\(\left(2, \dfrac{4}{e^{2}}\right)\) only
- C\((0, 0)\) and \(\left(-2, 4e^{2}\right)\)
- D\((0, 0)\) and \(\left(2, 4e^{2}\right)\)
Find the coordinates of the stationary points of the curve \(y = \dfrac{x}{x^2 + 4}\).
- A\(\left(2, \dfrac{1}{4}\right)\) and \(\left(-2, -\dfrac{1}{4}\right)\)
- B\(\left(2, \dfrac{1}{4}\right)\) only
- Cthere are no stationary points
- D\(\left(2, \dfrac{1}{4}\right)\) and \(\left(-2, \dfrac{1}{4}\right)\)
Find the equation of the tangent to the curve \(y = x^2\ln x\) at the point where \(x = e\).
- A\(y = 3ex - 2e^2\)
- B\(y = 2ex - e^2\)
- C\(y = 2x + e^2 - 2e\)
- D\(y = ex\)
Find the gradient of the curve \(y = \dfrac{\cos x}{1 - \sin x}\) at the point where \(x = \dfrac{\pi}{6}\).
- A\(2\)
- B\(-2\)
- C\(-4\)
- D\(1\)
Quiz key
\(\dfrac{du}{dx} = 3x^2\), \(\dfrac{dv}{dx} = 2e^{2x}\). So \(\dfrac{dy}{dx} = x^3 \cdot 2e^{2x} + e^{2x} \cdot 3x^2 = x^2e^{2x}(3 + 2x)\). The tempting \(6x^2e^{2x}\) multiplies the two derivatives, which is not the product rule. \(x^2e^{2x}(3 + x)\) forgets the chain rule factor 2 in the derivative of \(e^{2x}\), and \(3x^2e^{2x}\) keeps only one of the two terms.
\(\dfrac{dy}{dx} = \dfrac{(x + 3)(1) - x(1)}{(x + 3)^2} = \dfrac{3}{(x + 3)^2}\). The tempting \(-\dfrac{3}{(x + 3)^2}\) swaps the order in the numerator, which flips the sign. 1 divides the derivative of the top by the derivative of the bottom, and \(\dfrac{3}{x + 3}\) forgets to square the denominator.
\(\dfrac{dy}{dx} = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \ln x\). At \(x = e\): \(1 + 1 = 2\). The tempting 1 keeps only one term of the product rule. \(\dfrac{1}{e}\) multiplies the derivatives \(1 \times \dfrac{1}{x}\), and \(e\) is the value of \(y\) at \(x = e\), not the gradient.
\(\dfrac{dy}{dx} = \dfrac{v u' - u v'}{v^2} = \dfrac{x\cos x - \sin x \cdot 1}{x^2}\). The tempting \(\dfrac{\sin x - x\cos x}{x^2}\) has the numerator the wrong way round, giving the negative of the answer. \(\cos x\) divides \(\cos x\) by the derivative of \(x\), and \(\dfrac{x\cos x + \sin x}{x^2}\) uses a plus sign, mixing up the quotient and product rules.
\(\dfrac{dy}{dx} = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2 - x)\). This is zero when \(x = 0\) or \(x = 2\) (\(e^{-x} \ne 0\)). At \(x = 0\), \(y = 0\); at \(x = 2\), \(y = 4e^{-2} = \dfrac{4}{e^2}\). The tempting "\(\left(2, \dfrac{4}{e^2}\right)\) only" divides by \(x\) and loses the root \(x = 0\). \((-2, 4e^2)\) comes from \(\dfrac{d}{dx}e^{-x} = +e^{-x}\), dropping the chain rule minus sign, and \((2, 4e^2)\) substitutes \(e^{+2}\) for \(e^{-2}\).
\(\dfrac{dy}{dx} = \dfrac{(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = \dfrac{4 - x^2}{(x^2 + 4)^2}\). Zero when \(x^2 = 4\), \(x = \pm 2\). \(y(2) = \dfrac{2}{8} = \dfrac{1}{4}\) and \(y(-2) = -\dfrac{1}{4}\). The tempting "\(\left(2, \dfrac{1}{4}\right)\) only" takes just the positive square root. "No stationary points" uses a plus sign in the numerator, \(x^2 + 4 + 2x^2\), which is never zero, and the last option forgets that \(y\) is negative when \(x = -2\).
\(y = e^2\). \(\dfrac{dy}{dx} = 2x\ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x\), which is \(2e + e = 3e\) at \(x = e\). Tangent: \(y - e^2 = 3e(x - e)\), so \(y = 3ex - 2e^2\). The tempting \(y = 2ex - e^2\) keeps only the \(2x\ln x\) term (gradient \(2e\)). \(y = 2x + e^2 - 2e\) multiplies the derivatives (\(2x \times \dfrac{1}{x} = 2\)), and \(y = ex\) keeps only the \(x\) term (gradient \(e\)).
\(\dfrac{dy}{dx} = \dfrac{(1 - \sin x)(-\sin x) - \cos x(-\cos x)}{(1 - \sin x)^2} = \dfrac{-\sin x + \sin^2 x + \cos^2 x}{(1 - \sin x)^2} = \dfrac{1 - \sin x}{(1 - \sin x)^2} = \dfrac{1}{1 - \sin x}\). At \(x = \dfrac{\pi}{6}\): \(\dfrac{1}{1 - 1/2} = 2\). The tempting \(-2\) reverses the order of the numerator. \(-4\) adds the two terms as in the product rule, and 1 forgets to square the denominator.
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.