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C1: Simple Differentiation

C1: Simple Differentiation. Learning Objective: to calculate the gradient of a curve by applying the rules of differentiation. The gradient of a curve. The gradient of a curve at a point is given by the gradient of the tangent at that point.

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C1: Simple Differentiation

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  1. C1: Simple Differentiation Learning Objective: to calculate the gradient of a curve by applying the rules of differentiation

  2. The gradient of a curve The gradient of a curve at a point is given by the gradient of the tangent at that point. Look at how the gradient changes as we move along a curve:

  3. Differentiation y x x2 x3 x4 x5 x6 In general: If we continued the process of differentiating from first principles we would obtain the following results: 1 2x 3x2 4x3 5x4 6x5 What pattern do you notice? and when xn is preceded by a constant multiplier a we have:

  4. The gradient function represents the derivative of y with respect to x. represents the derivative of s with respect to t. then: So the gradient of the tangent to the curve can be calculated by differentiation. So if y = x3 This notation can be adapted for other variables so, for example: If the curve is written using function notation as y = f(x), then the derived function can be written as f ′(x).

  5. Task 1 Differentiate • y = x4 • y = 3x4 • y = 5x3 • y = 2x8 • y = -5x6 • y = x12 • y = -0.5x2 • y = 2.5x4 • y = 15x8 • y = 2/3 x3

  6. Differentiation Differentiating: Find the gradient of the curve y = 3x4 at the point (–2, 48). At the point (–2, 48) x = –2 so: The gradient of the curve y = 3x4 at the point (–2, 48) is –96.

  7. Task 2 • Find the gradient of the tangent of y = x4 at the point (3, 81). • Find the gradient of the curve whose equation is y = 3x2 at the point (2, 12). • Find the gradients of the curve y = 2x2 at the points C and D where the curve meets the line y = x + 3.

  8. y = x-2 y = x1/2 y = 3x2 + 5x - 6 dy/dx = -2x-3 dy/dx = ½ x-1/2 dy/dx = 6x + 5 You can apply the rules of differentiation to functions involving negative or fractional powers and for polynomials.Examples:

  9. Task 3: Find dy/dx when y equals: • 2x2 -6x + 3 • ½ x2 + 12x • 4x2 - 6 • 8x2 + 7x +12 • 5 + 4x – 5x2 • x3 + x2 – x1/2 • 2x-3 • 1/3 x1/2 + 4x-2 • 5√x • x3(3x + 1)

  10. Task 4: • Find the gradient of the curve whose equation is y = 2x2 – x – 1 at the point (2,5). • Find the y co-ordinate and the value of the gradient at the point P with x co-ordinate 1 on the curve with equation y = 3 + 2x - x2. • Find the co-ordinates of the point on the curve with equation y = x2 + 5x – 4 where the gradient is 3. • Find the point or points on the curve with equation f(x), where the gradient is zero: a) f(x) = x3/2 – 6x + 1 b) f(x) = x-1 + 4x.

  11. Second Order Derivatives You can repeat the process of differentiation to give a second order derivative. The notation d2y/dx2 or f’’(x) is used. Example: y = 3x5 + 4/x2 y = 3x5 + 4x-2 dy/dx = 15x4 -8x-3 d2y/dx2 = 60x3 + 24x-4 = 60x3 + 24/x4

  12. Task 5: Find dy/dx and d2y/dx2 when y equals: • 12x2 + 3x + 8 • 15x + 6 +3/x • 9√x – 3/x2 • (5x + 4)(3x – 2) • (3x+ 8)/(x2)

  13. Task 6: Using different notation • Find dθ/dt where θ = t2 – 3t • Find dA/dr where A = 2πr • Given that r = 12/t, find the value of dr/dt when t = 3. • The surface area, A cm2, of an expanding sphere of radius r cm is given by A = 4πr2. Find the rate of change of the area with respect to the radius at the instant when the radius is 6cm. • The displacement, s metres, of a car from a fixed point at time t seconds is given by s = t2 + 8t. Find the rate of change of the displacement with respect to time at the instant when t = 5.

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