Rates of Change and the Gradient Function

1. 1 Show a curved distance-time graph. 'The gradient of a chord gives average speed. As the chord gets shorter, we approach the instantaneous speed at a point.' Draw chords getting progressively shorter. 'At the limit, we get the tangent — and its gradient is the instantaneous rate of change.'
2. 2 For f(x) = xⁿ, f'(x) = nxⁿ⁻¹. 'Multiply by the power, then reduce the power by 1.' Demonstrate: f(x) = x³ → f'(x) = 3x². f(x) = x⁵ → f'(x) = 5x⁴. Constants: f(x) = 7 → f'(x) = 0. Linear: f(x) = 4x → f'(x) = 4.
3. 3 Differentiate term by term: f(x) = 3x⁴ − 2x² + 5x − 1. f'(x) = 12x³ − 4x + 5. Students differentiate five polynomial expressions. Include fractional and negative powers: f(x) = x^(1/2), f(x) = x^(−1).
4. 4 Find the gradient of y = x³ − 4x at x = 2. Differentiate: dy/dx = 3x² − 4. Substitute x = 2: 3(4) − 4 = 8. 'The gradient of the tangent at x = 2 is 8.' Find the equation of the tangent: passes through (2, 0) with gradient 8. y = 8x − 16.
5. 5 At a stationary point, f'(x) = 0. For y = x³ − 3x: f'(x) = 3x² − 3 = 0. x² = 1. x = ±1. Find y-values: (1, −2) and (−1, 2). To classify: second derivative test. f''(x) = 6x. At x=1: f''(1)=6>0 → minimum. At x=−1: f''(−1)=−6<0 → maximum.
6. 6 A ball is thrown. Height: h = 20t − 5t². Velocity = dh/dt = 20 − 10t. When is the ball at maximum height? (velocity = 0: t = 2s). What is the maximum height? h(2) = 40 − 20 = 20m. Students solve a similar projectile problem.
7. 7 Students sketch f(x) = x³ − 3x using only the stationary points and the information f'(x) gives about increasing/decreasing intervals. f'(x) > 0 when increasing. f'(x) < 0 when decreasing.

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