1. Rate of change / Differentiation (3)
Differentiating
Using differentiating

2. If y = xn = nxn-1 The Key Bit dy dx
The general rule (very important) is :-
If y = xn dy dx
= nxn-1
E.g. if y = x2
= 2x dy dx
E.g. if y = x3
= 3x2 dy dx
E.g. if y = 5x4
= 5 x 4x3
= 20x3 dy dx

3. Find for these functions :- Gradient at x=-2
Warm-up dy dx
Find for these functions :-
Gradient at
x=-2 dy dx
y= 2x2
y = 2x5
y = 5x2 + 10x + 5
y = x3 + x2 + x
y = 2x4 - 4x2 + 7
= 4x
= 10x4
= 10x + 10
= 3x2 + 2x +1
= 8x3 - 8x
= -8
= 10x(-2)4=160
= = -10
= 3(-2)2+2(-2)+1
= = 9
= 8(-2)3 -8(-2)
= 8 x-8 – 8 x-2
= = -48

4. A differentiating Problem
The gradient of y = ax3 + 4x2 – 12x is 2 when x=1
What is a? dy dx
= 3ax2 + 8x -12
When x=1 dy dx
= 3a + 8 – 12 = 2
3a - 4 = 2
3a = 6
a = 2

5. Try this The gradient of y = 4x3 - ax2 + 10x is 6 when x=1 What is a?dy dx
= 12x2 – 2ax + 10
When x=1 dy dx
= 12 -2a +10 = 6
22 - 2a = 6
16 = 2a
a = 8

6. Rate of change / Differentiation (3 pt2)
Equations of tangents
Equations of normals

7. Function Notation y = x3 – 12x f(x)= x3 – 12x f’(x)= 3x2 - 12
We have seen: if dy dx
= 3x2 - 12
Instead of ‘y’ we may use the function notation f(x)
f(x)= x3 – 12x If dy dx
then
is replace by f’(x)
f’(x)= 3x2 - 12 so
f’(x) represent the differential/gradient function

8. Linear graphs y = 5x + 7 y = 2x - 1 y - intercept Gradient
Increase in y
Increase in x
Gradient = dy dx
y - intercept
Gradient
m and c will always be numbers
in your examples
y = 5x + 7
y = 2x - 1

9. Definition y = 3x - 11.31 y = 3x + 2/3 y = 3x y = 3x + 8 y = 3x + 84
Parallel lines are ones with the same slope/gradient.
i.e. the number in front of the ‘x’ is the same
y = 3x
y = 3x + 2/3
y = 3x
y = 3x + 8
y = 3x + 84
y = 3x - 3
y = 3x - 21
y = 3x + 1
y = 3x + 43
y = 3x - 1.5

10. Equations of Tangents y=x2 What is the equation of the tangent?
The tangent to the curve is the gradient at that point
(3,9) y x
What is the equation of the tangent?
y=x2
y = mx + c
Substitute gradient:
9 = 6x3 + c
c = = -9
y = 6x - 9
= 2x dy dx
When x=3; = 2x3 = 6 dy dx

11. Equations of Tangents - try me
y=3x2 + 4x + 1
- differentiate
- gradient at x =1
- y = mx + c
- find c
(1,8) x
What is the equation of the tangent?
y=3x2 + 4x + 1
y = mx + c
Substitute gradient:
8 = 10x1 + c
c = = -2
y = 10x - 2
= 6x + 4 dy dx
When x=1; = 6+4 = 10 dy dx

12. Perpendicular Lines
If two lines with gradients m1 and m2 are perpendicular, then:

13. Equations of Normals normal mT x mN = -1
y=x2
The normal is always perpendicular to the tangent
normal
mT x mN = -1 y
(3,9) x
What is the equation of the normal?
When x=3; = 2x3 = 6 dy dx
y = mx + c
Substitute gradient:
9 = -1/6 x 3 + c
c = /6 = 9 + 1/2
y = -1/6 x + 9 1/2
mT x mN = -1
6 x mN = -1
mN = -1/6

14. Equations of Normals normal mT x mN = -1
y=x2
The normal is always perpendicular to the tangent
normal
mT x mN = -1 y
(3,9) x
What is the equation of the normal?
When x=3; = 2x3 = 6 dy dx
y = mx + c
Substitute gradient:
9 = -1/6 x 3 + c
c = 9 + 3/6 = 9 + 1/2
y = -1/6 x + 9 1/2
mT x mN = -1
6 x mN = -1
mN = -1/6

15. Equations of Normal - try me
y=3x2 + 4x + 1
- differentiate
- gradient at x =1
- mt x mn = -1
- y = mnx + c
- find c
(1,8)
normal x
What is the equation of the normal?
= 6x + 4 dy dx
y = mnx + c
Substitute gradient:
8 = -1/10 x 1 + c
c = 8 + 1/10 = 8 1/10
y = -1/10 x + 8 1/10
When x=1; = 6+4 = 10 dy dx
mT x mN = -1
10 x mN = -1
mN = -1/10