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. Look at how the gradient changes as we move along a curve: Explain that the direction of a curve changes as we move along it. Demonstrate this by moving the point along the curve. Observe where the gradient of the tangent at the point is positive, negative and 0.

Differentiation If we continued the process of differentiating from first principles we would obtain the following results: y x x2 x3 x4 x5 x6 1 2x 3x2 4x3 5x4 6x5 What pattern do you notice? In general: Establish that to differentiate a power of x we multiply by the power and subtract 1 from the original power. These results are expected to be used without proof at this stage. and when xn is preceded by a constant multiplier a we have:

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

Task 1 Differentiate y = x4 y = 3x4 y = 5x3 y = 2x8 y = -5x6 y = x12

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

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.

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

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)

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.

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

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)

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.