C1: Simple Differentiation

Learning Objective: to calculate the gradient of a curve by applying

the rules of differentiation

The gradient of a curveThe gradient of a curve at a point is given by

the gradient of the tangent at that point.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:

DifferentiationIf we continued the process of differentiating from first principles we would obtain the following results:

ydy

dx

x x2 x3 x4 x5 x6

What pattern do you notice?

In general:

1If then n ndyy x nx

dx

and when xn is preceded by a constant multiplier a we have:

1 2x 3x2 4x3 5x4 6x5

1If = then =n ndyy ax anx

dx

The gradient functionSo the gradient of the tangent to the curve can be calculated by differentiation.

If the curve is written using function notation as y = f(x), then the derived function can be written as f ′(x).

represents the derivative of y with respect to x.dy

dx

So if y = x3 2= 3dy

xdx

then:

This notation can be adapted for other variables so, for example:

represents the derivative of s with respect to t.ds

dt

Task 1

Differentiate 1. y = x4

2. y = 3x4

3. y = 5x3

4. y = 2x8

5. y = -5x6

6. y = x12

7. y = -0.5x2

8. y = 2.5x4

9. y = 15x8

10. y = 2/3 x3

DifferentiationFind the gradient of the curve y = 3x4 at the point (–2, 48).

Differentiating: 3=12dy

xdx

At the point (–2, 48) x = –2 so:

3=12( 2)dy

dx

=12 × 8

= 96

The gradient of the curve y = 3x4 at the point (–2, 48) is –96.

Task 2

1. Find the gradient of the tangent of y = x4

at the point (3, 81).

2. Find the gradient of the curve whose equation is y = 3x2 at the point (2, 12).

3. Find the gradients of the curve y = 2x2 at the points C and D where the curve meets the line y = x + 3.

You can apply the rules of differentiation to functions involving negative or fractional powers and for polynomials.

Examples:

1. y = x-2

2. y = x1/2

3. y = 3x2 + 5x - 6

1. dy/dx = -2x-3

2. dy/dx = ½ x-1/2

3. dy/dx = 6x + 5

Task 3: Find dy/dx when y equals:

1. 2x2 -6x + 32. ½ x2 + 12x3. 4x2 - 64. 8x2 + 7x +125. 5 + 4x – 5x2

6. x3 + x2 – x1/2

7. 2x-3

8. 1/3 x1/2 + 4x-2

9. 5√x10. x3(3x + 1)

Task 4:

1. Find the gradient of the curve whose equation is y = 2x2 – x – 1 at the point (2,5).

2. 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.

3. Find the co-ordinates of the point on the curve with equation y = x2 + 5x – 4 where the gradient is 3.

4. 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:

1. 12x2 + 3x + 8

2. 15x + 6 +3/x

3. 9√x – 3/x2

4. (5x + 4)(3x – 2)

5. (3x+ 8)/(x2)

Task 6: Using different notation

1. Find dθ/dt where θ = t2 – 3t2. Find dA/dr where A = 2πr3. Given that r = 12/t, find the value of dr/dt when t = 3.4. 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.

5. 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.