The shape of the lesson
By the end of the lesson, learners can
- Any rational power: The rule \(\dfrac{d}{dx}x^{n} = nx^{n-1}\) works for every rational \(n\): negative, fractional, the lot.
- The chain rule: For a bracket raised to a power, let \(u\) be the inside.
- Second derivatives of composites: To find \(\dfrac{d^{2}y}{dx^{2}}\) for a bracket, apply the chain rule a second time.
- Rewrite: \(\sqrt[n]{x^{m}} = x^{\frac{m}{n}}\) and \(\dfrac{k}{x^{n}} = kx^{-n}\).
- Fractions over \(x^{p}\): divide each term, never differentiate top and bottom separately.
- Chain rule: \(\dfrac{d}{dx}\left[f(x)\right]^{n} = n\left[f(x)\right]^{n-1} \times f'(x)\).
- Linear inside: \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).
- Tidy: write negative and fractional powers back as fractions and roots if asked.
One picture every idea returns to
Write every term as \(kx^{n}\) before differentiating. For a function of a function, \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\): differentiate the outside leaving the inside alone, then multiply by the derivative of the inside. So \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).
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, Outside, then inside.. Roots and reciprocals are just powers in disguise, so one rule handles them all. 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: Any rational power | The rule \(\dfrac{d}{dx}x^{n} = nx^{n-1}\) works for every rational \(n\): negative, fractional, the lot. Model the worked example Roots and reciprocals on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The chain rule | For a bracket raised to a power, let \(u\) be the inside. Model the worked example A linear inside on the board, one line at a time. Your turn: Differentiate \(y = (4 - x^{2})^{3}\). | Pairs |
| 21 to 30 min | Teach and check: Second derivatives of composites | To find \(\dfrac{d^{2}y}{dx^{2}}\) for a bracket, apply the chain rule a second time. Model the worked example Chain rule twice on the board, one line at a time. | 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, 21 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 Second derivatives of composites, 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 |
|---|---|
| \(\dfrac{1}{2x^{2}}\) is \(\tfrac{1}{2}x^{-2}\), NOT \(2x^{-2}\): only the \(x^{2}\) moves up with a negative power; the 2 stays in the denominator as \(\tfrac{1}{2}\). | Return to Any rational power: ask a pair to state the correct version and show it on the lab before moving on. |
| The most common lost mark: forgetting to multiply by the derivative of the inside. \(\dfrac{d}{dx}(3x - 1)^{5}\) is \(15(3x - 1)^{4}\), not \(5(3x - 1)^{4}\). | Watch the sign too when the inside is something like \(4 - x^{2}\). |
| Do not expand \((2x - 3)^{4}\) before differentiating unless you have to. | It is slow, and a single slip in the binomial expansion costs every later mark. |
Differentiating powers and the chain rule: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\dfrac{dy}{dx}\) for \(y = 5x^{3} - \dfrac{2}{\sqrt{x}}\).
Differentiate \((5 - 2x)^{6}\) with respect to \(x\).
A curve has equation \(y = \dfrac{4}{(1 - 3x)^{2}}\). Find the gradient of the curve at the point where \(x = 1\).
\(f(x) = \sqrt{x^{2} + 9}\). Find \(f'(x)\) and the exact value of \(f'(4)\).
The curve \(y = (x^{2} - 1)^{3}\).
(a) Show that \(\dfrac{dy}{dx} = kx(x^{2} - 1)^{2}\), stating the value of \(k\).
(b) Find the \(x\)-coordinates of the points where the gradient is zero.
A curve has equation \(y = \dfrac{12}{\sqrt{2x + 1}}\) for \(x > -\tfrac{1}{2}\).
(a) Find \(\dfrac{dy}{dx}\).
(b) Find the coordinates of the point on the curve where the gradient is \(-\tfrac{4}{9}\).
(c) Explain why the gradient of the curve is negative at every point.
Answer key and marking notes
- \(y = 5x^{3} - 2x^{-\frac{1}{2}}\).
- \(\dfrac{dy}{dx} = 15x^{2} - 2 \times (-\tfrac{1}{2})x^{-\frac{3}{2}} = 15x^{2} + x^{-\frac{3}{2}}\).
- \(= 15x^{2} + \dfrac{1}{x\sqrt{x}}\)
- \(6(5 - 2x)^{5} \times (-2)\)
- \(= -12(5 - 2x)^{5}\)
- \(y = 4(1 - 3x)^{-2}\).
- \(\dfrac{dy}{dx} = -8(1 - 3x)^{-3} \times (-3) = \dfrac{24}{(1 - 3x)^{3}}\).
- At \(x = 1\): \(\dfrac{24}{(-2)^{3}} = -3\).
- \(f(x) = (x^{2} + 9)^{\frac{1}{2}}\), so \(f'(x) = \tfrac{1}{2}(x^{2} + 9)^{-\frac{1}{2}} \times 2x = \dfrac{x}{\sqrt{x^{2} + 9}}\).
- \(f'(4) = \dfrac{4}{\sqrt{25}} = \dfrac{4}{5}\).
- (a) \(\dfrac{dy}{dx} = 3(x^{2} - 1)^{2} \times 2x = 6x(x^{2} - 1)^{2}\), so \(k = 6\).
- (b) \(6x(x^{2} - 1)^{2} = 0\) when \(x = 0\) or \(x^{2} = 1\).
- \(x = 0,\ x = 1,\ x = -1\).
- (a) \(y = 12(2x + 1)^{-\frac{1}{2}}\), so \(\dfrac{dy}{dx} = 12 \times (-\tfrac{1}{2})(2x + 1)^{-\frac{3}{2}} \times 2 = -\dfrac{12}{(2x + 1)^{\frac{3}{2}}}\).
- (b) \(\dfrac{12}{(2x + 1)^{\frac{3}{2}}} = \dfrac{4}{9}\) gives \((2x + 1)^{\frac{3}{2}} = 27\).
- So \(2x + 1 = 27^{\frac{2}{3}} = 9\), \(x = 4\), and \(y = \dfrac{12}{\sqrt{9}} = 4\). The point is \((4, 4)\).
- (c) For \(x > -\tfrac{1}{2}\), \((2x + 1)^{\frac{3}{2}} > 0\), so \(\dfrac{dy}{dx} = -\dfrac{12}{\text{positive}}\) is always negative.
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 \(y = \dfrac{4}{x^{2}}\).
- A\(-\dfrac{8}{x^{3}}\)
- B\(\dfrac{8}{x^{3}}\)
- C\(-\dfrac{8}{x}\)
- D\(\dfrac{2}{x}\)
Find the gradient of the curve \(y = 4x^{\frac{3}{2}}\) at the point where \(x = 9\).
- A\(18\)
- B\(162\)
- C\(108\)
- D\(54\)
Differentiate \(y = (5 - 2x)^{4}\).
- A\(-8(5 - 2x)^{3}\)
- B\(4(5 - 2x)^{3}\)
- C\(8(5 - 2x)^{3}\)
- D\(-2(5 - 2x)^{3}\)
Write \(\dfrac{3}{2\sqrt{x}}\) in the form \(kx^{n}\).
- A\(\tfrac{3}{2}x^{-\frac{1}{2}}\)
- B\(6x^{-\frac{1}{2}}\)
- C\(\tfrac{3}{2}x^{\frac{1}{2}}\)
- D\(\tfrac{3}{2}x^{-2}\)
Find the gradient of the curve \(y = \sqrt{x^{2} + 9}\) at the point where \(x = 4\).
- A\(\dfrac{4}{5}\)
- B\(\dfrac{1}{10}\)
- C\(\dfrac{8}{5}\)
- D\(1\)
Given \(y = \dfrac{x + 4}{\sqrt{x}}\), find the value of \(\dfrac{dy}{dx}\) when \(x = 4\).
- A\(0\)
- B\(\dfrac{5}{4}\)
- C\(-\dfrac{15}{4}\)
- D\(4\)
Given \(y = \sqrt{4x + 1}\), find the value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x = 2\).
- A\(-\dfrac{4}{27}\)
- B\(-\dfrac{1}{27}\)
- C\(-\dfrac{1}{108}\)
- D\(\dfrac{2}{3}\)
Find all the values of \(x\) at which the curve \(y = (x^{2} - 3)^{3}\) has zero gradient.
- A\(x = 0,\ x = \sqrt{3},\ x = -\sqrt{3}\)
- B\(x = \sqrt{3},\ x = -\sqrt{3}\) only
- C\(x = 0\) only
- D\(x = 0,\ x = 3,\ x = -3\)
Quiz key
\(y = 4x^{-2}\), so \(\dfrac{dy}{dx} = 4 \times (-2)x^{-3} = -\dfrac{8}{x^{3}}\). \(\dfrac{8}{x^{3}}\) loses the minus sign from the power; \(-\dfrac{8}{x}\) moves the power from \(-2\) to \(-1\), but reducing by one takes \(-2\) to \(-3\); \(\dfrac{2}{x}\) comes from differentiating the denominator \(x^{2}\) as if it were on top.
\(\dfrac{dy}{dx} = 4 \times \tfrac{3}{2}x^{\frac{1}{2}} = 6\sqrt{x}\). At \(x = 9\): \(6 \times 3 = 18\). \(162 = 6 \times 9^{\frac{3}{2}}\) keeps the old power; \(108\) is \(y\) at \(x = 9\); \(54 = 6 \times 9\) forgets the square root.
Outside: \(4(5 - 2x)^{3}\). Inside: \(\dfrac{d}{dx}(5 - 2x) = -2\). Product: \(-8(5 - 2x)^{3}\). \(4(5 - 2x)^{3}\) forgets the derivative of the inside; \(8(5 - 2x)^{3}\) uses \(+2\) for it, losing the sign; \(-2(5 - 2x)^{3}\) multiplies by the inside derivative but drops the power 4.
\(\dfrac{3}{2\sqrt{x}} = \dfrac{3}{2} \times \dfrac{1}{x^{\frac{1}{2}}} = \tfrac{3}{2}x^{-\frac{1}{2}}\). \(6x^{-\frac{1}{2}}\) multiplies by the 2 instead of dividing; \(\tfrac{3}{2}x^{\frac{1}{2}}\) forgets that a root in the denominator gives a negative power; \(x^{-2}\) confuses a square root with a square.
\(\dfrac{dy}{dx} = \tfrac{1}{2}(x^{2} + 9)^{-\frac{1}{2}} \times 2x = \dfrac{x}{\sqrt{x^{2} + 9}}\). At \(x = 4\): \(\dfrac{4}{\sqrt{25}} = \dfrac{4}{5}\). \(\dfrac{1}{10}\) forgets to multiply by the derivative of the inside, \(2x\); \(\dfrac{8}{5}\) multiplies by \(2x\) but drops the \(\tfrac{1}{2}\); \(1\) comes from the false step \(\sqrt{x^{2} + 9} = x + 3\).
\(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\), so \(\dfrac{dy}{dx} = \tfrac{1}{2}x^{-\frac{1}{2}} - 2x^{-\frac{3}{2}}\). At \(x = 4\): \(\tfrac{1}{2} \times \tfrac{1}{2} - 2 \times \tfrac{1}{8} = \tfrac{1}{4} - \tfrac{1}{4} = 0\). \(\dfrac{5}{4}\) writes \(\dfrac{4}{\sqrt{x}}\) as \(4x^{\frac{1}{2}}\); \(-\dfrac{15}{4}\) reduces the power \(-\tfrac{1}{2}\) to \(+\tfrac{1}{2}\) instead of \(-\tfrac{3}{2}\); \(4\) divides the derivative of the top by the derivative of the bottom, which is not a rule.
\(\dfrac{dy}{dx} = \tfrac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = 2(4x + 1)^{-\frac{1}{2}}\). Then \(\dfrac{d^{2}y}{dx^{2}} = 2 \times (-\tfrac{1}{2})(4x + 1)^{-\frac{3}{2}} \times 4 = -4(4x + 1)^{-\frac{3}{2}}\). At \(x = 2\): \(-4 \times 9^{-\frac{3}{2}} = -\dfrac{4}{27}\). \(-\dfrac{1}{27}\) misses the factor 4 at one of the two steps; \(-\dfrac{1}{108}\) misses it at both; \(\dfrac{2}{3}\) is the first derivative at \(x = 2\).
\(\dfrac{dy}{dx} = 3(x^{2} - 3)^{2} \times 2x = 6x(x^{2} - 3)^{2}\). This is zero when \(x = 0\) or \(x^{2} = 3\), i.e. \(x = \pm\sqrt{3}\). "\(\pm\sqrt{3}\) only" forgets the factor \(2x\) from the inside; "\(0\) only" misses the squared bracket; \(\pm 3\) solves \(x^{2} = 3\) without taking the square root.
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.