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Higher Maths 1 3 Differentiation 1

Higher Maths 1 3 Differentiation

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Higher Maths 1 3 Differentiation. 1. OUTCOME. SLIDE. UNIT. Higher Maths 1 3 Differentiation. 2. OUTCOME. SLIDE. UNIT. NOTE. The History of Differentiation. - PowerPoint PPT Presentation

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Page 1: Higher Maths          1            3     Differentiation

Higher Maths 1 3 Differentiation

1

Page 2: Higher Maths          1            3     Differentiation

Higher Maths 1 3 Differentiation

The History of DifferentiationDifferentiation is part of the science of Calculus, and was first developed in the 17th century by two different Mathematicians.

Gottfried Leibniz(1646-1716)

Germany

Sir Isaac Newton(1642-1727)

England

2

Differentiation, or finding the instantaneous rate of change, is an essential part of: • Mathematics and Physics

• Chemistry

• Biology

• Computer Science

• Engineering

• Navigation and Astronomy

Page 3: Higher Maths          1            3     Differentiation

Higher Maths 1 3 Differentiation

Calculating Speed

2 4 6 8

4

8

2

6

10

00

Time (seconds)

Dis

tan

ce (m

)

DS T×

÷÷

Example

Calculate the speed for each section of the journey opposite.

A

B

C

speed in A =

43

speed in B = 51 5

m/s

=

speed in C =

25 0.4

m/s

=

average speed =

9 1.22 m/s

11≈

1.33 m/s

Notice the following things:• the speed at each instant is not the same as the average• speed is the same as gradientD

TS

=

yx

= m=

3

Page 4: Higher Maths          1            3     Differentiation

Instantaneous Speed

Higher Maths 1 3 Differentiation

Time (seconds)

Dis

tan

ce (m

)

Time (seconds)

Dis

tan

ce (m

)

In reality speed does not often change instantly. The graph on the right is more realistic as it shows a gradually changing curve.The journey has the same average speed, but the instantaneous speed is different at each point because the gradient of the curve is constantly changing. How can we find the instantaneous speed?

DT

S

=

yx

= m=

4

Page 5: Higher Maths          1            3     Differentiation

Introduction to Differentiation

Higher Maths 1 3 Differentiation

Differentiate means

D

Tspeed

=

‘rate of change ofdistance with respect to time’

S

Tacceleration

=

‘find out how fast something is changing in comparison with something else at any one instant’.

gradient

=

yx

‘rate of change ofspeed with respect to time’‘rate of change of -coordinate with respect to -coordinate’y x

5

Page 6: Higher Maths          1            3     Differentiation

Estimating the Instantaneous Rate of Change

Higher Maths 1 3 Differentiation

The diagrams below show attempts to estimate the instantaneous gradient (the rate of change of with respect to ) at the point A.

A yAxx

yy

A

x

Notice that the accuracy improves as gets closer to zero.

x

y x

The instantaneous rate of change is written as:

yx

dydx = as approaches 0.

x

6

Page 7: Higher Maths          1            3     Differentiation

Basic Differentiation

The instant rate of change of y with respect to x is written as .

Higher Maths 1 3 Differentiation

dydx

By long experimentation, it is possible to prove the following:

dydx =

y = x n

nxn–1

If

then • multiply by the power

• reduce the power by

one

How to Differentiate:

Note that describes both the rate of change and the gradient.

dydx

7

Page 8: Higher Maths          1            3     Differentiation

Differentiation of Expressions with Multiple Terms

Higher Maths 1 3 Differentiation

y =ax m + bx

n + …

• multiply every x-term by the power

How to Differentiate:

amx m + bnx

n + …–1 –1dy

dx =

• reduce the power of every x-term by one

The basic process of differentiation can be applied to

every x-term in an algebraic expession.

ImportantExpressions must be written as the sum of individual terms before differentiating.

8

Page 9: Higher Maths          1            3     Differentiation

Examples of Basic Differentiation

Higher Maths 1 3 Differentiation

Find for

dydx

y = 3 x 4 – 5 x

3 + + 9

7x

2

y = 3 x 4 – 5 x

3 + 7 x -2 +

9dydx = 12 x

3 – 15 x 2 – 14

x -3

9 =9 x 0

this disappears because

= 12 x 3 – 15 x

2 –14x

3

9

(multiply by zero)

Example 1

Page 10: Higher Maths          1            3     Differentiation

Examples of Basic Differentiation

Higher Maths 1 3 Differentiation

Find the gradient of the curve

Example 2

10

y =( x + 3 ) ( x – 5 )

y =x ² – 2 x – 15

x 2

= 1 2x – 15

x 2

x 2

= 1 – 2 x -1 – 15 x

- 2

dydx = 2 x

- 2 + 30 x

-

3

disappears (multiply by zero)

=2 + 30

x 3x

2

At x =

5,

at the point

(5,0).

dydx =

2 + 3025 125

=8

25

Page 11: Higher Maths          1            3     Differentiation

The Derived Function

The word ‘derived’ means ‘produced from’, for example orange juice is derived from oranges.

Higher Maths 1 3 Differentiation

It is also possible to express differentiation using function notation.

=

f (x)

nxn–1

If

then f ′(x)

= x n f ′(x) dy

dxand

mean exactly the same thing written in different ways.

LeibnizNewton

The derived function is the rate of change of the function with respect to .

f ′(x) f (x)x

11

Page 12: Higher Maths          1            3     Differentiation

A tangent to a function is a straight line which intersects

the function in only one place, with the same gradient as

the function.

Tangents to Functions

Higher Maths 1 3 Differentiation

12

The gradient of any tangent to the function can be

found by substituting the x-coordinate of intersection into .

f (x)

A

B

f ′(x)

f (x)

mAB = f ′(x)

Page 13: Higher Maths          1            3     Differentiation

f ′(x)

Equations of Tangents

Higher Maths 1 3 Differentiation

13

y – b = m ( x – a ) To find the equation of a

tangent:

• substitute gradient and point of intersection intoy – b = m ( x – a )

• substitute -coordinate to find gradient at point of intersection

x

• differentiateStraight Line Equation

Example

Find the equation of the tangent to the function

at the point (2,4).

f (x)= x 31

2

= x 23

2

f ′(2) = ×

(2)

2

32

= 6

m =

y – 4 = 6 ( x – 2 )

6 x – y – 8 = 0

substitute:

Page 14: Higher Maths          1            3     Differentiation

Increasing and Decreasing Curves

Higher Maths 1 3 Differentiation

The gradient at any point on a curve can be found by differentiating.

14

uphill slopePositive

downhill slopeNegative

Gradient

Ifdydx

> 0 then y is

increasing.

Ifdydx

< 0 then y is

decreasing. Alternatively,If > 0f ′(x)is increasing.

f (x)then

If < 0f ′(x)is decreasing.

f (x)then

dydx < 0

dydx > 0

dydx < 0

Page 15: Higher Maths          1            3     Differentiation

If a function is neither increasing or decreasing, the gradient is zero and the function can be described as stationary.

Stationary Points

Higher Maths 1 3 Differentiation

15

There are two main types of stationary point.

Minimum

Maximum

Turning Points

Points of Inflection

At any stationary point, or alternatively

dydx

= 0 f ′(x)= 0

Rising

Falling

Page 16: Higher Maths          1            3     Differentiation

Investigating Stationary Points

Higher Maths 1 3 Differentiation

16

ExampleFind the stationary

point off (x) = x 2 – 8 x + 3

Stationary point

given by

f ′(x) = 2 x – 8

2 x – 8= 0

and determine its

nature.

x= 4

f ′(x) = 0

x 4 – 4+4

slope

The stationary point atis a minimum turning point.

– 0 +f ′(x)

x= 4

Use a nature table to reduce the amount of working.

‘slightly less than

four’

‘slightly more’

gradientis positive

Page 17: Higher Maths          1            3     Differentiation

Investigating Stationary Points

Higher Maths 1 3 Differentiation

17

Example 2

Investigate the stationary

points ofy =4 x 3 – x

4

dydx

=12 x 2 – 4 x

3 = 0

4 x 2 (3 – x ) = 0

4 x 2= 0

x = 0

3 – x= 0

x = 3

or

y = 0 y = 27

x 0 – 0+0

slope

dydx

rising point of inflection

+ 0 +

stationary point at (0,0):

x 3 – 3+3

slope

dydx

maximum turning point

+ 0 –

stationary point at (3, 27):

Page 18: Higher Maths          1            3     Differentiation

Positive and Negative Infinity

Higher Maths 1 3 Differentiation

18

Example

∞The symbol is used for infinity.+∞–∞

‘positive infinity’

‘negative infinity’

The symbol means ‘approaches’.

y =5 x 3 + 7 x

2

For very large , the value

of

becomes insignificant

compared with the value of

.

5 x 3

± x

as x ,∞ y 5 x 3

5×( )3+∞ = +∞5×( )3–∞ = –∞

as x +∞

∞+1 = ∞

y +∞and as x –∞

y –∞

7 x 2

Page 19: Higher Maths          1            3     Differentiation

x 7 – 7+7

slope

– 0 +dydx

Curve Sketching

Higher Maths 1 3 Differentiation

19

To sketch the graph of any function,

the following basic information is

required:• the stationary points and their nature

• the x-intercept(s) and y-intercept

• the value of y as x

approachespositive and negative

infinity

solve for y = 0 and

x = 0

solve for = 0 and use nature table

dydx

y = x 3 – 2 x

4

–∞y +∞

as x ∞y -2 x

4

+∞y –∞

as

x

and as x

Example

Page 20: Higher Maths          1            3     Differentiation

Graph of the Derived Function

Higher Maths 1 3 Differentiation

20

f ′(x)

f (x)The graph of can be thought of

as the graph of the gradient of .

Example

y = f (x)

y = f ′(x)f ′(x) = 0

f ′(x) = 0

Positive

Negative

Gradient

f ′(x) > 0

f ′(x) < 0

The roots of

are

f (x)

f ′(x)

given by the

stationarypoints of

.