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