Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Further Integration Integrating Trig

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Higher Outcome 2

Higher Unit 3Higher Unit 3

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Further Differentiation Trig Functions

Further Integration

Integrating Trig Functions

Differentiation The Chain Rule

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Higher Outcome 2

The Chain Rule for Differentiating

To differentiate composite functions (such as functions with brackets in them) we can

use:

dy dy du

dx du dx

Example 2 8( 3) Differentiate y x

2( 3) Let u x 8 then y u

78 dy

udu

2 du

xdx

dy dy du

dx du dx78 2 u x 2 78( 3) .2 x x 2 716 ( 3)

dyx x

dx

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Higher Outcome 2


Good News !

There is an easier way.

'( ) '( ( )) '( )h x g f x f x

2 8( ) ( 3)h x x Differentiate

2( ) ( 3)f x x Let 8( )g x x then

2 7'( ( )) 8( 3)g f x x 2 8( ( )) ( 3)g f x x '( ) 2f x x

2 7'( ) 16 ( 3)h x x x

You have 1 minute to come up with the rule.

1. Differentiate outside the bracket.

2. Keep the bracket the same.

3. Differentiate inside the bracket.

2 7'( ) 16 ( 3)h x x x

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Higher Outcome 2


Example

42

5'( ) ( )

3

Find when f x f x

x

2 4( ) 5( 3)f x x

2 5'( ) 20( 3)f x x

2 5

40'( )

( 3)

xf x

x




2 5'( ) 20( 3) 2f x x x

You are expected to do the chain rule all at once

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Higher Outcome 2Example

1

2 1

Find when

dyy

dx x

1

2(2 1)dy

xdx

3-2

12 -1 2

2

dyx

dx

3

2

1

(2 1)

dy

dxx

The Chain Rule for Differentiating1. Differentiate outside the bracket.



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3 2 5( ) (3 2 1) '(0) If , find the value of f x x x f

3 2 4 2'( ) 5(3 2 1) (9 - 4 )f x x x x x Let

'(0) 0 f


3 2 4 2'(0) 5(3 0 2 0 1) (9 0 - 4 0)f Let

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Higher Outcome 2Example1

32 1

Find the equation of the tangent to the curve where y xx

1

2 1

y

xRe-arrange: 1

2 1 y x

The slope of the tangent is given by the derivative of the equation.

Use the chain rule:

21 2 1 2

dy

xdx 2

2

2 1

x

Where x = 3:

1 2

5 25

and

dyy

dx

1 23,5 25

We need the equation of the line through with slope

The Chain Rule for Differentiating Functions

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Higher Outcome 2

1 2( 3)

5 25

y x

2 1 1 5( 3) (2 6)

25 5 25 25

y x x

1(2 1)

25

y x Is the required equation

Remember

y - b = m(x – a)


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In a small factory the cost, C, in pounds of assembling x components in a month is given by:

21000

( ) 40 , 0

C x x xx

Calculate the minimum cost of production in any month, and the corresponding number of

components that are required to be assembled.

0 Minimum occurs where , so differentiate dC

dx

225

( ) 40

C x xx

Re-arrange 2

251600

x

x


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Higher Outcome 2

225

1600

dC d

xdx dx x

225

1600

dx

dx x

Using chain rule 2

25 251600 2 1

x

x x 2

25 253200 1

x

x x

2

25 250 0 1 0

where or dC

xdx x x

2 225 25 i.e. where or x x

2 25No real values satisfy x 0 We must have x

5 We must have items assembled per month x


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Higher Outcome 2

2

25 253200 1 0 5

when dC

x xdx x x

5 5 Consider the sign of when and dC

x xdx

For x < 5 we have (+ve)(+ve)(-ve) = (-ve)

Therefore x = 5 is a minimum

Is x = 5 a minimum in the (complicated) graph?

Is this a minimum

?

For x > 5 we have (+ve)(+ve)(+ve) = (+ve)


For x = 5 we have (+ve)(+ve)(0) = 0x = 5

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Higher Outcome 2

The cost of production:2

25( ) 1600 5

, C x x x

x

225

1600 55

21600 10

160,000 £ Expensive components?

Aeroplane parts maybe ?


Calculus Revision

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Differentiate5( 2)x

Chain rule45( 2) 1x

45( 2)x Simplify

Calculus Revision

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Differentiate3(5 1)x

Chain Rule23(5 1) 5x

215(5 1)x Simplify

Calculus Revision

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Differentiate2 4(5 3 2)x x

Chain Rule 2 34(5 3 2) 10 3x x x

Calculus Revision

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Differentiate

5

2(7 1)x

Chain Rule

Simplify

3

25

2(7 1) 7x

3

235

2(7 1)x

Calculus Revision

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Differentiate

1

2(2 5)x

Chain Rule

Simplify

3

21

2(2 5) 2x

3

2(2 5)x

Calculus Revision

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Differentiate 3 1x

Chain Rule

Simplify

Straight line form 1

23 1x

1

21

23 1 3x

1

23

23 1x

Calculus Revision

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Differentiate

13 2( ) (8 )f x x

Chain Rule

Simplify

13 221

2( ) (8 ) ( 3 )f x x x

12 3 23

2( ) (8 )f x x x

Calculus Revision

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Differentiate 1

2( ) 5 4f x x

Chain Rule

Simplify

1

21

2( ) 5 4 5f x x

1

25

2( ) 5 4f x x

Calculus Revision

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Differentiate 42

5

3x

Chain Rule

Simplify

Straight line form2 45( 3)x

2 520( 3) 2x x

2 540 ( 3)x x

Calculus Revision

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Differentiate1

2 1x

Chain Rule

Simplify

Straight line form1

2(2 1)x

3

21

2(2 1) 2x

3

2(2 1)x

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Higher Outcome 2

Trig Function Differentiation

The Derivatives of sin x & cos x

(sin ) cos (cos ) sin d d

x x x xdx dx

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2( ) 2cos sin '( )

3 . Find f x x x f x

2'( ) (2cos ) sin

3

d d

f x x xdx dx

[ ( ) ( )]d df dgf x g x

dx dx dx

22 (cos ) sin

3

d dx x

dx dx. ( )

d dgc g x c

dx dx

2'( ) 2sin cos

3 f x x x


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3

3

4 cos Differentiate with respect to

x xx

x

33

3

4 cos4 cos

x x

x xx

3 3(4 cos ) 4 (cos ) d d d

x x x xdx dx dx

4

4

12 ( sin )

12 sin

x x

x x

4

4

sin 12 x x

x


Simplify expression -

where possible

Restore the original form of expression

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Higher Outcome 2

The Chain Rule for DifferentiatingTrig Functions

Worked Example:

sin(4 )y x Differentiate

cosdy

dx cos4

dyx

du cos4 4

dyx

dx

4cos4 dy

xdx




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Higher Outcome 2


Example

4cos Find when dy

y xdx

4(cos )dy

xdx

34(cos ) (-sin )dy

x xdx

34sin cos dy

x xdx

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sin Find when dy

y xdx

1

2(sin )y x 1

21(sin ) cos2

dyx x

dx

cos

2 sin

dy x

dx x


Calculus Revision

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Differentiate2

2cos sin3

x x

22sin cos

3x x

Calculus Revision

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Differentiate 3cos x

3sin x

Calculus Revision

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Differentiate cos 4 2sin 2x x

4sin 4 4cos 2x x

Calculus Revision

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Differentiate 4cos x

34cos sinx x

Calculus Revision

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Differentiate sin x

Chain Rule

Simplify

Straight line form

1

2(sin )x1

21

2(sin ) cosx x

1

21

2cos (sin )x x

1

2

cos

sin

x

x

Calculus Revision

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Differentiate sin(3 2 )x

Chain Rule

Simplify

cos(3 2 ) ( 2)x

2cos(3 2 )x

Calculus Revision

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Differentiatecos5

2

x

Chain Rule

Simplify

1

2cos5xStraight line form

1

2sin 5 5x

5

2sin 5x

Calculus Revision

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Differentiate ( ) cos(2 ) 3sin(4 )f x x x

( ) 2sin(2 ) 12cos(4 )f x x x

Calculus Revision

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Differentiate 62siny x

Chain Rule

Simplify

62cos 1dy

xdx

62cosdy

xdx

Calculus Revision

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Differentiate2 2( ) cos sinf x x x

( ) 2 cos sin 2sin cosf x x x x x Chain Rule

Simplify ( ) 4 cos sinf x x x

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Higher Outcome 2

Integrating Composite Functions

Harder integration

1( )( )

( 1)

n

n ax bax b dx c

a n

4(3 7) x dx5(3 7)

15

xc

4 1(3 7)

3(4 1)

x

c

we get

You have 1 minute to come up with the rule.

4(3 7) x dx5(3 7)

15

xc

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Higher Outcome 2


1( )( )

( 1)

n

n ax bax b dx c

a n

Example :4(4 2) Evaluate

dt

t

4(4 2) t dt

3(4 2)t 3(4 2)

12

tc

1. Add one to the power.

2. Divide by new power.

3. Compensate for bracket.

3(4 2)

3

t

3(4 2)

4 ( 3)

t

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2( 3) Evaluate u du

1( 3)u 1( 3)

1

u

12 ( 3) 1

( 3)1 ( 3)

uu du c c

u

Integrating Composite Functions1. Add one to the power.



1( 3)

1 ( 1)

uc

You are expected to do the integration rule all at once

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Higher Outcome 2Example 2 3

1(2 1) Evaluate x dx

24

1

(2 1)

8

x

68


4 4(2 2 1) (1 1 1)

8 8

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32

4

1(3 4)

Evaluate x dx

32

4

1(3 4)x dx

45

2

1

(3 4)5

32

x

682

5


45

2

1

2(3 4)

15

x

5 5

2 22(3 4 4) 2(3 ( 1) 4)

15 15

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3( ) '( ) 2 1 (1) 2f x f x x f Find given and

3( ) 2 1f x x dx

Integrating

42 1

8

xc

So we have:

42 1 1(1) 2

8f c

41

8c

1

8c

Giving:

1 1528 8

c 42 1 15( )

8 8

xf x

Integrating Functions1. Add one to the power.



Calculus Revision

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Integrate2(5 3 )x dx

3(5 3 )

3 3

xc

31

9(5 3 )x c

Standard Integral

(from Chain Rule)

Calculus Revision

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Integrate2

1

(7 3 )dx

x2(7 3 )x dx

1(7 3 )

1 3

xc

11

3(7 3 )x c

Straight line form

Calculus Revision

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Find

1

6

0

(2 3)x dx17

0

(2 3)

7 2

x

7 7(2 3) (0 3)

14 14

7 75 3

14 14

5580.36 156.21 (4sf)5424

Use standard Integral

(from chain rule)

Calculus Revision

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Integrate

1

2

1

0 3 1

dx

x

1

2

1

0

3 1x dx

Straight line form

1

2

1

0

13

2

3 1x

1

0

23 1

3x

2 23 1 0 1

3 3

2 24 1

3 3

4 2

3 3

2

3

Calculus Revision

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Find 2

3

1

2 1x dx

24

1

2 1

4 2

x

4 44 1 2 1

4 2 4 2

4 45 3

8 8

68


(from chain rule)

Calculus Revision

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Evaluate0

2

3

(2 3)x dx


(from chain rule)

03

3

(2 3)

3 2

x

3 3(2(0) 3) (2( 3) 3)

6 6

27 27

6 6

27 27

6 6

54

6 9

Calculus Revision

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Evaluate1

01 3x dx

1

21

01 3x dx

13

2

0

3

2

1 3

3

x

13

2

0

21 3

9x

13

0

21 3

9x

3 32 21 3(1) 1 3(0)

9 9

3 32 24 1

9 9

16 2

9 9

14

9 5

91

Calculus Revision

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Find p, given

1

42p

x dx 1

2

1

42p

x dx 3

2

1

2

342

p

x

3 3

2 22 2

3 3(1) 42p

2 233 3

42p 32 2 126p

3 64p 3 2 1264 2p 1

12 32 16p

Calculus Revision

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A curve for which 26 2dy

x xdx

passes through the point (–1, 2).

Express y in terms of x.

3 26 2

3 2

x xy c 3 22y x x c

Use the point3 22 2( 1) ( 1) c 5c

3 22 5y x x

Calculus Revision

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Given the acceleration a is: 1

22(4 ) , 0 4a t t

If it starts at rest, find an expression for the velocity v where dv

adt

1

22(4 )dv

tdt

3

2

31

2

2(4 )tv c

3

24(4 )3

v t c

344

3v t c Starts at rest, so

v = 0, when t = 0 34

0 43

c

340 4

3c

320

3c

32

3c

3

24 32

3 3(4 )v t

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Higher Outcome 2


(sin ) cos

(cos ) sin

dx x

dxd

x xdx

cos sin

sin cos

x dx x c

x dx x c

Integration is opposite of

differentiation

Worked Example

123sin cosx x dx 3cos x 1

2 sin x c

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Higher Outcome 2

Special Trigonometry Integrals are

1cos( ) sin( )

1sin( ) cos( )

ax b dx ax b ca

ax b dx ax b ca

Worked Example

sin(2 2) x dx 1cos(2 2)2

x c

Integrating Trig Functions1. Integrate outside the bracket

2. Keep the bracket the same

3. Compensate for inside the bracket.

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Higher Outcome 2


Example

sin5 cos( 2 ) Evaluate t t dt

Integrate

sin(5 ) s( 2 )t dt co t dt 1 1cos5 sin 25 2

t t c

Break up into two easier integrals

1. Integrate outside the bracket



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Higher Outcome 2Example 2

4

3cos 22

Evaluate x dx

2

4

3cos 22

x dx

2

4

3sin 2x dx

Re-arrange

2

4

3cos22

x

Integrate

3cos 2 cos 2

2 2 4

3

1 02

3

2

Integrating Trig Functionscos( ) cos cos sin sin 1. Integrate outside the bracket



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Higher Outcome 2

Integrating Trig Functions (Area)

Example

cosy xsiny x

1

x

y

1

0

AThe diagram shows the

graphs of y = -sin x and y = cos x

a) Find the coordinates of A

b) Hence find the shaded area cos sin The curves intersect where x x

tan 1 x

0 We want x

1tan ( 1) x C

AS

T0o180

o

270o

90o

3

2

2

3 7

4 4

and

3

4

x

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Higher Outcome 2

Integrating Trig Functions (Area)

( ) The shaded area is given by top curve bottom curve dx

3

4

0

(cos ( sin ))

Area x x dx

3

4

0

(cos sin )

x x dx

3

40

sin cosx x

3 3sin cos sin0 cos0

4 4

1 11

2 2

1 2

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Higher Outcome 2Example 3cos3 cos(2 ) cos3 4cos 3cos a) By writing as show that x x x x x x

3cos b) Hence find x

cos(2 ) cos2 cos sin 2 sin x x x x x x

2 2(cos sin )cos (2sin cos )sin x x x x x x

2 2 2(cos sin )cos 2sin cos x x x x x

3 2cos 3sin cos x x x

3 2cos 3 1 cos cos x x x

3cos3 4cos 3cos x x x


Remember cos(x +

y) =

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Higher Outcome 2

3cos3 4cos 3cos As shown above x x x

3 1cos (cos3 3cos )

4 x x x

1 1 1(cos3 3cos ) sin3 3sin

4 4 3 x x dx x x c

3 1cos sin3 9sin

12 x dx x x c


Calculus Revision

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Find2sin7

t dt2

7cos t c

Calculus Revision

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Find 3cos x dx3sin x c

Calculus Revision

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Find 2sin d 2cos c

Calculus Revision

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2(6 cos )x x x dx Integrate

3 26sin

3 2

x xx c Integrate term by term

Calculus Revision

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Find1

3sin cos2

x x dx1

3cos sin2

x x c Integrate term by term

Calculus Revision

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Find sin 2 cos 34

x x dx

1 1

2 3cos 2 sin 3

4x x c

Calculus Revision

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The curve ( )y f x ,112

passes through the point

( ) cos 2f x x Find f(x)

1

2( ) sin 2f x x c

1

2 121 sin 2 c

use the given point ,112

1

2 61 sin c

1 1

2 21 c 3

4c 1 3

2 4( ) sin 2f x x

1

6 2sin

Calculus Revision

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If

( ) sin(3 )f x x passes through the point 9, 1

( )y f x express y in terms of x.

1

3( ) cos(3 )f x x c Use the point 9

, 1

1

31 cos 3

9c

1

31 cos

3c

1 1

3 21 c

7

6c 1 7

3 6cos(3 )y x

Calculus Revision

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A curve for which 3sin(2 )dy

xdx

5

12, 3passes through the point

Find y in terms of x. 3

2cos(2 )y x c

3 5

2 123 cos(2 ) c Use the point 5

12, 3

3 5

2 63 cos( ) c 3 3

2 23 c

3 33

4c

4 3 3 3

4 4c

3

4c

3 3

2 4cos(2 )y x

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Higher Outcome 2

Are you on Target !

• Update you log book

• Make sure you complete and correct

ALL of the Calculus questions in the

past paper booklet.

Documents

Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Further Integration Integrating Trig