Rate of change / Rate of change / Differentiation (3)Differentiation (3)
•DifferentiatingDifferentiating•Using differentiatingUsing differentiating
The Key Bit
The general rule (very important) is :-
If y = xn
dydx
= nxn-1
E.g. if y = x2
= 2xdydx
E.g. if y = x3
= 3x2dydx
E.g. if y = 5x4
= 5 x 4x3
= 20x3
dydxdydx
Find for these functions :- dydx
y= 2x2
y = 2x5
y = 5x2 + 10x + 5
y = x3 + x2 + x
y = 2x4 - 4x2 + 7
dydx
dydx
dydx
dydx
dydx
= 4x
= 10x4
= 10x + 10
= 3x2 + 2x +1
= 8x3 - 8x
Gradient at x=-2
= -8
= 10x(-2)4=160
= -20 + 10 = -10
= 3(-2)2+2(-2)+1 = 12 -4 +1 = 9 = 8(-2)3 -8(-2) = 8 x-8 – 8 x-2= -64 +16 = -48
Warm-up
A differentiating ProblemA differentiating Problem
The gradient of y = ax3 + 4x2 – 12x is 2 when x=1
What is a?dydx
= 3ax2 + 8x -12
When x=1dydx
= 3a + 8 – 12 = 2
3a - 4 = 23a = 6 a = 2
Try thisTry this
The gradient of y = 4x3 - ax2 + 10x is 6 when x=1
What is a?dydx
= 12x2 – 2ax + 10
When x=1dydx
= 12 -2a +10 = 6
22 - 2a = 616 = 2a a = 8
Rate of change / Rate of change / Differentiation (3 pt2)Differentiation (3 pt2)
•Equations of tangentsEquations of tangents•Equations of normalsEquations of normals
Function NotationFunction Notation
f(x)= x3 – 12x
dydx
= 3x2 - 12
Instead of ‘y’ we may use the function notation f(x)
dydx is replace by f’(x)
If
then
so f’(x)= 3x2 - 12
y = x3 – 12xWe have seen: if
f’(x) represent the differential/gradient function
Linear graphsLinear graphs
cmxy Gradient
y - intercept
m and c will always be numbers in your examplesy = 5x + 7 y = 2x - 1
Increase in yIncrease in x
Gradient =
dydx
DefinitionParallel linesParallel lines are ones with the same slope/gradient. are ones with the same slope/gradient.
i.e.i.e. the number in front of the ‘x’ is the samethe number in front of the ‘x’ is the same
y = 3x + 8
y = 3x - 1.5
y = 3x + 2/3
y = 3x + 84y = 3x - 21
y = 3x + 43y = 3x + 1
y = 3x - 3
y = 3x - 11.31y = 3x
y
x
Equations of Tangentsy=x2 The tangent
to the curve is the gradient at that point
(3,9)
What is the equation of the tangent?
y=x2
= 2x dydx
When x=3; = 2x3 = 6dydx
y = mx + cSubstitute gradient:9 = 6x3 + cc = 9 - 18 = -9y = 6x - 9
y
x
Equations of Tangents - try mey=3x2 + 4x + 1
(1,8)
What is the equation of the tangent?
y=3x2 + 4x + 1= 6x + 4 dy
dx
When x=1; = 6+4 = 10dydx
y = mx + cSubstitute gradient:8 = 10x1 + cc = 8 - 10 = -2y = 10x - 2
- differentiate- gradient at x =1- y = mx + c- find c
Perpendicular LinesPerpendicular Lines
12
1
mm 121 mm
If two lines with gradients m1 and m2 are perpendicular, then:
y
x
Equations of Normalsy=x2
(3,9)
What is the equation of the normal?
When x=3; = 2x3 = 6dydx
y = mx + cSubstitute gradient:9 = -1/6 x 3 + cc = 9 - -3/6 = 9 + 1/2 y = -1/6 x + 9 1/2
mT x mN = -16 x mN = -1mN = -1/6
The normal is always perpendicular to the tangentnormal
mT x mN = -1
y
x
Equations of Normalsy=x2
(3,9)
What is the equation of the normal?
When x=3; = 2x3 = 6dydx
y = mx + cSubstitute gradient:9 = -1/6 x 3 + cc = 9 + 3/6 = 9 + 1/2 y = -1/6 x + 9 1/2
mT x mN = -16 x mN = -1mN = -1/6
The normal is always perpendicular to the tangentnormal
mT x mN = -1
y
x
Equations of Normal - try mey=3x2 + 4x + 1
(1,8)
What is the equation of the normal?
= 6x + 4 dydx
When x=1; = 6+4 = 10dydx
y = mnx + c
- differentiate- gradient at x =1- mt x mn = -1- y = mnx + c- find c
normal
mT x mN = -110 x mN = -1 mN = -1/10
Substitute gradient:8 = -1/10 x 1 + cc = 8 + 1/10 = 8 1/10 y = -1/10 x + 8 1/10
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