Rate of change /
Differentiation (3)
•Differentiating
•Using
differentiating
The Key Bit
The general rule (very important) is :-
n
x
If y =
dy = nxn-1
dx
E.g. if y = x2
dy
= 2x
dx
E.g. if y = x3
dy
= 3x2
dx
E.g. if y = 5x4
dy
= 5 x 4x3
dx
dy
3
=
20x
dx
Warm-up
Find dy for these functions :dx
y=
dy
= 4x
dx
2x2
y=
2x5
dy
= 10x4
dx
y=
5x2
dy
= 10x + 10
dx
y=
x3
y=
2x4
+ 10x + 5
+
-
x2
+x
4x2
+7
Gradient at
x=-2
= -8
= 10x(-2)4=160
= -20 + 10 = -10
dy
= 3x2 + 2x +1 = 3(-2)2+2(-2)+1
dx
dy = 8x3 - 8x
dx
= 12 -4 +1 = 9
= 8(-2)3 -8(-2)
= 8 x-8 – 8 x-2
= -64 +16 = -48
A differentiating Problem
The gradient of y = ax3 + 4x2 – 12x is 2 when x=1
What is a?
dy
= 3ax2 + 8x -12
dx
When x=1
dy
dx
= 3a + 8 – 12 = 2
3a - 4 = 2
3a = 6
a=2
Try this
The gradient of y = 4x3 - ax2 + 10x is 6 when x=1
What is a?
dy
= 12x2 – 2ax + 10
dx
When x=1
dy
dx
= 12 -2a +10 = 6
22 - 2a = 6
16 = 2a
a=8
Rate of change /
Differentiation (3 pt2)
•Equations
of tangents
•Equations of normals
Function Notation
We have seen: if
y = x3 – 12x
dy = 3x2 - 12
dx
Instead of ‘y’ we may use the function notation f(x)
If
then
so
f(x)= x3 – 12x
dy
dx
is replace by f’(x)
f’(x)= 3x2 - 12
f’(x) represent the differential/gradient function
Linear graphs
y  mx  c
Gradient =
Increase in y
GradientIncrease in x
dy
dx
y - intercept
m and c will always be numbers
in your examples
y = 5x + 7
y = 2x - 1
Definition
Parallel lines are ones with the same slope/gradient.
i.e. the number in front of the ‘x’ is the same
y = 3x - 11.31
y = 3x
y = 3x + 8
y = 3x + 84
y = 3x - 21
y = 3x + 1
y = 3x - 1.5
y = 3x + 2/3
y = 3x - 3
y = 3x + 43
Equations of Tangents
y=x2
The tangent to
the curve is the
gradient at that
point
y
(3,9)
x
What is the equation of the tangent?
y = mx + c
y=x2
Substitute gradient:
dy
= 2x
9 = 6 x3 + c
dx
c = 9 - 18 = -9
dy
When x=3;
= 2 x3 = 6
y = 6x - 9
dx
Equations of Tangents - try me
y=3x2 + 4x + 1
y
(1,8)
x
-
differentiate
gradient at x =1
y = mx + c
find c
What is the equation of the tangent?
y = mx + c
y=3x2 + 4x + 1
Substitute gradient:
dy
= 6x + 4
8 = 10x1 + c
dx
c = 8 - 10 = -2
dy
When x=1;
= 6+4 = 10 y = 10x - 2
dx
Perpendicular Lines
If two lines with gradients
m1 and m2 are
perpendicular, then:
1
m2  
m1
m1m2  1
Equations of Normals
y=x2
The normal is always
perpendicular to the
tangent
y
normal
mT x mN = -1
(3,9)
x
What is the equation of the normal?
dy
y = mx + c
When x=3;
= 2x3 = 6
dx
Substitute gradient:
mT x mN = -1
9 = -1/6 x 3 + c
6 x mN = -1
c = 9 - -3/6 = 9 + 1/2
y = -1/6 x + 9 1/2
mN = -1/6
Equations of Normals
y=x2
The normal is always
perpendicular to the
tangent
y
normal
mT x mN = -1
(3,9)
x
What is the equation of the normal?
dy
y = mx + c
When x=3;
= 2x3 = 6
dx
Substitute gradient:
mT x mN = -1
9 = -1/6 x 3 + c
6 x mN = -1
c = 9 + 3/6 = 9 + 1/2
y = -1/6 x + 9 1/2
mN = -1/6
Equations of Normal - try me
y=3x2 + 4x + 1
y
(1,8)
normal
x
-
differentiate
gradient at x =1
mt x mn = -1
y = mnx + c
find c
What is the equation of the normal?
dy
y = mnx + c
= 6x + 4
dx
Substitute gradient:
dy
When x=1;
= 6+4 = 10 8 = -1/ x 1 + c
dx
10
c = 8 + 1/10 = 8 1/10
mT x mN = -1
1/
1/
y
=
x
+
8
1
10
10
10 x mN = -1 mN = - /10