The shape of the lesson
By the end of the lesson, learners can
- From chord to tangent: In the lab, drag P and Q onto the curve and read the gradient of the chord PQ.
- Multiply by the power, reduce it by one: Open the gradient function tab.
- Gradient at a point: \(\dfrac{dy}{dx}\) is a formula.
- Prepare: expand any brackets so \(y\) is a sum of terms \(ax^n\).
- Rule: \(ax^n \to anx^{n-1}\); \(ax \to a\); a constant \(\to 0\).
- Notation: write \(\dfrac{dy}{dx} = \dots\) for the gradient function.
- Gradient at a point: substitute the \(x\)-coordinate into \(\dfrac{dy}{dx}\).
- Point with a given gradient: solve \(\dfrac{dy}{dx} = k\), then find \(y\) from the original equation.
One picture every idea returns to
If \(y = ax^{n}\) then \(\dfrac{dy}{dx} = anx^{n-1}\): multiply by the power, then take one off the power. A constant term differentiates to 0. Substitute an \(x\)-value into \(\dfrac{dy}{dx}\) to get the gradient there.
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, A formula for the gradient.. Extended. 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: From chord to tangent | In the lab, drag P and Q onto the curve and read the gradient of the chord PQ. Project tab 4 of the lab (Functions and differentiation lab, gradient of a curve) and let a student drive it while the class predicts. Model the worked example Spotting the gradient function on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Multiply by the power, reduce it by one | Open the gradient function tab. Project tab 5 of the lab (Functions and differentiation lab, gradient function) and let a student drive it while the class predicts. Model the worked example A sum of terms on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Gradient at a point | \(\dfrac{dy}{dx}\) is a formula. Model the worked example Gradient at a given point on the board, one line at a time. Your turn: Find the gradient of \(y = 2x^3 - x^2\) when \(x = -1\). | Connect-Extend-Challenge |
| 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, 16 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 Gradient at a point, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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 |
|---|---|
| A tangent touches the curve at one point and has the same steepness there. | A chord joins two points of the curve and only approximates the gradient: do not stop at the chord. |
| Do not differentiate each bracket and multiply: for \((x - 4)(2x + 1)\) that gives \(1 \times 2 = 2\), which is wrong. | Expand first, then differentiate term by term. |
| Substitute into \(\dfrac{dy}{dx}\) for the gradient, but into \(y\) for the coordinate. | Putting \(x = 2\) into \(y = x^3 - 2x + 1\) gives 5, the height of the point, not its gradient. |
Differentiation: gradient functions: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\dfrac{dy}{dx}\) when (a) \(y = 6x^4\) (b) \(y = \tfrac{1}{2}x^2 - 3x + 4\).
Differentiate \(y = x^3 + 5x^2 - 7\).
Find the gradient of the curve \(y = 2x^3 - 4x\) at the point where \(x = -1\).
\(y = (x - 2)(x + 5)\). Find the gradient of the curve where it crosses the \(y\)-axis.
The curve \(y = x^2 + kx\) has gradient 10 at the point where \(x = 3\).
(a) Find \(k\). (b) Find the \(x\)-coordinate of the point where the gradient is 0.
A ball is thrown upwards. Its height, \(h\) metres, after \(t\) seconds is \(h = 30t - 5t^2\).
(a) Find \(\dfrac{dh}{dt}\), the rate of change of height.
(b) Find the rate of change of height when \(t = 1\) and when \(t = 4\), and explain the sign of each answer.
(c) Find the time when the ball is momentarily at rest, and its height then.
Answer key and marking notes
- (a) \(6 \times 4x^3 = 24x^3\).
- (b) \(\tfrac{1}{2} \times 2x - 3 + 0 = x - 3\).
\(\dfrac{dy}{dx} = 3x^2 + 10x\).
- \(\dfrac{dy}{dx} = 6x^2 - 4\).
- At \(x = -1\): \(6(-1)^2 - 4 = 6 - 4 = 2\).
- Expand: \(y = x^2 + 3x - 10\).
- \(\dfrac{dy}{dx} = 2x + 3\).
- The curve crosses the \(y\)-axis at \(x = 0\), so the gradient is \(2(0) + 3 = 3\).
- (a) \(\dfrac{dy}{dx} = 2x + k\). At \(x = 3\): \(6 + k = 10\), so \(k = 4\).
- (b) \(2x + 4 = 0\), so \(x = -2\).
- (a) \(\dfrac{dh}{dt} = 30 - 10t\).
- (b) \(t = 1\): \(30 - 10 = 20\) m/s, positive, so the ball is rising.
- \(t = 4\): \(30 - 40 = -10\) m/s, negative, so the ball is falling.
- (c) \(30 - 10t = 0\) gives \(t = 3\) s; \(h = 90 - 45 = 45\) m (its greatest height).
6 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.
\(y = 6x^4 - x^3 + 2x - 9\). Find \(\dfrac{dy}{dx}\).
- A\(24x^3 - 3x^2 + 2\)
- B\(24x^3 - 3x^2 + 2x\)
- C\(6x^3 - x^2 + 2\)
- D\(24x^3 - 3x^2 + 2 - 9\)
\(y = x^2(3x - 4)\). Find \(\dfrac{dy}{dx}\).
- A\(9x^2 - 8x\)
- B\(6x\)
- C\(6x^2 - 8x\)
- D\(9x^2 - 4\)
Find the gradient of the curve \(y = 2x^3 - 5x^2 + 4\) at the point where \(x = -1\).
- A\(16\)
- B\(-4\)
- C\(4\)
- D\(-3\)
Find the values of \(x\) at which the gradient of \(y = x^3 - 6x^2 + 5\) is 15.
- A\(x = -1\) or \(x = 5\)
- B\(x = 5\) only
- C\(x = 1\) or \(x = -5\)
- D\(x = 0\) or \(x = 4\)
The graph of \(y = x^2\) is drawn with the chord \(PQ\), from \(P(1, 1)\) to \(Q(3, 9)\), and the tangent at \(P\). What is the gradient of the curve at \(P\)?
- A\(2\)
- B\(4\)
- C\(1\)
- D\(8\)
A ball is thrown upwards. Its height after \(t\) seconds is \(h = 30t - 5t^2\) metres. Find the rate at which the height is changing when \(t = 2\).
- A\(10\) m/s
- B\(40\) m/s
- C\(50\) m/s
- D\(20\) m/s
Quiz key
\(6 \times 4x^3 = 24x^3\), \(-1 \times 3x^2 = -3x^2\), \(2x = 2x^1\) gives \(2x^0 = 2\), and the constant \(-9\) gives 0. So \(\dfrac{dy}{dx} = 24x^3 - 3x^2 + 2\). Writing \(2x\) keeps the power the same for the \(x\) term; \(6x^3 - x^2 + 2\) lowers the powers but forgets to multiply by them; keeping \(-9\) forgets that a constant has zero gradient.
Expand: \(y = 3x^3 - 4x^2\). Then \(\dfrac{dy}{dx} = 9x^2 - 8x\). The answer \(6x\) differentiates each part separately (\(2x\) and 3) and multiplies them, which is not allowed. \(6x^2 - 8x\) is \(2x(3x - 4)\): it differentiates the \(x^2\) and leaves the bracket alone. \(9x^2 - 4\) comes from expanding wrongly as \(3x^3 - 4x\).
\(\dfrac{dy}{dx} = 6x^2 - 10x\). At \(x = -1\): \(6(-1)^2 - 10(-1) = 6 + 10 = 16\). The answer \(-4\) comes from \(6 - 10\), forgetting that \(-10 \times (-1) = +10\). The answer 4 takes \((-1)^2\) as \(-1\). \(-3\) is the \(y\)-coordinate, \(2(-1)^3 - 5(-1)^2 + 4\), found by substituting into \(y\) instead of \(\dfrac{dy}{dx}\).
\(\dfrac{dy}{dx} = 3x^2 - 12x = 15\), so \(3x^2 - 12x - 15 = 0\), \(x^2 - 4x - 5 = 0\), \((x - 5)(x + 1) = 0\), giving \(x = 5\) or \(x = -1\). Check: \(3(25) - 60 = 15\) and \(3 + 12 = 15\). \(x = 0\) or 4 sets the gradient equal to 0 instead of 15. \(x = 1\) or \(-5\) has the signs of the roots flipped, and \(x = 5\) only drops a solution.
\(\dfrac{dy}{dx} = 2x\), so at \(x = 1\) the gradient is 2. This is the gradient of the tangent \(y = 2x - 1\) drawn at \(P\). The chord has gradient \(\dfrac{9 - 1}{3 - 1} = 4\): it is steeper than the tangent and only approximates the gradient. 1 is the \(y\)-coordinate of \(P\) and 8 is the rise of the chord without dividing by the run.
\(\dfrac{dh}{dt} = 30 - 10t\). At \(t = 2\): \(30 - 20 = 10\) m/s, still rising. 40 is the height \(h = 60 - 20\), not its rate of change. 50 adds instead of subtracting. 20 comes from \(30 - 5t\), forgetting to multiply \(5t^2\) by the power 2.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.