Higher Unit 3Higher Unit 3

Further Differentiation Trig Functions

Further Integration Integrating Trig Functions

Differentiation The Chain Rule

The Chain Rule for Differentiating

To differentiate composite functions (such as functions with brackets in them) use:

dxdu

dudy

dxdy

82 3 ateDifferenti xy 3Let 2 xu 8 and uy

78ududy

xdxdu

2

dxdu

dudy

dxdy

xudxdy

28 7 xx 23872

72 316 xx

The Chain Rule for Differentiating

In practice we do this as follows:

82 3 ateDifferenti xy

72 316 xx

1. Differentiate the bracket:

72 38 x

2. Differentiate inside the bracket: x2

3. Multiply the answers together

72 316 xxdxdy

8 78

Differentiate 14)( xxf

Use indices to get rid of root sign

2

1

)14()( xxf

Chain Rule 2

1

)14(21

)(

xxf 4

Simplify 2

1

)14(2)(

xxf )14(2

x

Differentiate23

3)(

xxf

Use indices to get variable on working line

2

1

)23(3)(

xxf

Chain Rule 2

3

)23(321

)(

xxf 3

Simplify 2

3

)23(29

)(

xxf 3)23(2

9

x

Differentiate 14)( xxf

Use indices to get rid of root sign

2

1

)14()( xxf

Chain Rule 2

1

)14(21

)(

xxf 4

Simplify 2

1

)14(2)(

xxf )14(2

x

First bring denominator on to the working line (indices)

43 1

5)( when )( Find

xxfxf

43 15)(

xxf

1. Differentiate the bracket: 53 154

x

2. Differentiate inside the bracket: 23x

532 160

xx3. Multiply the answers together

53

2

1

60)(

x

xxf

251

when Find

x

ydxdy

First use indices to get rid of square root and bring denominator on to the working line

21

25 xy

1. Differentiate the bracket: 23

2521 x

2. Differentiate inside the bracket: 5

23

2525 x3. Multiply the answers together

3)25(2

5

xdxdy

Differentiate (5x – 1)3

Chain Rule

Simplify

3(5x – 1)2 × 5

15(5x – 1)2

Differentiate (2x3 – x + 2)4

Chain Rule 4 (2x3 – x + 2)3 (6x2 – 1)

Differentiate 14)( xxf

Use indices to get rid of root sign

2

1

)14()( xxf

Chain Rule 2

1

)14(21

)(

xxf 4

Simplify 2

1

)14(2)(

xxf )14(2

x

Differentiate23

3)(

xxf

Use indices to get variable on working line

2

1

)23(3)(

xxf

Chain Rule 2

3

)23(321

)(

xxf 3

Simplify 2

3

)23(29

)(

xxf 3)23(2

9

x

2 where 53

1 totangent the of equation the Find

x

xy

153 xy

1. 2231 x 2. 3 2533 x3.

253

3

xdx

dy

xdxdy

m of values allfor tan

2tan523

3

m 3tan m

523

1 2 When

y x 1 y

)-(- From axmby )2-(31- xy 73 xy

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

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

required to be assembled.

01000

40)(2

x

xxxC

SP’s C’(x) = 0 21100040)( xxxC

21 1000401000402)( xxxxC

2

100040

1000402)(

xxxxC

SP’s C’(x) = 0

2

100040

1000402)(

xxxxC

C’(x) = 0 01000

40

xx 0

100040 2

x

0100040 2 x 0100040 2 x

100040 2 x 100040 2 x

252 x 252 x 55 orx

Number of components must be

positive

ShapeShape

→→55→→xx

– – 0 0 + +

min5xWhen x = 5

2

51000

540)5(

C

000,160£)5( C

C’(x)

Derivatives of Trig Functions

f(x) = sin x f ’(x) = cos x

f(x) = cos x f ’(x) = – sin x

To calculate the value of any derivative the To calculate the value of any derivative the angles must be measured in radiansangles must be measured in radians

The basic derivatives are given in a formula The basic derivatives are given in a formula list in the examlist in the exam

Differentiate xxxf cos5sin2)(

)sin(5cos2)( xxxf xx sin5cos2

Differentiate xxxf 4sin3cos)(

Put in brackets use Chain Rule

)4sin()3cos()( xxxf

4)4cos()3()3sin()( xxxf

Simplify xxxf 4cos43sin3)(

Find f ’(π/3) when xxf 3sin)(

Put in brackets use Chain Rule

3)(sin)( xxf

Simplify xxxf cossin3)( 2

2)(sin3)( xxf xcossinsin33xx is the same is the same as (sin as (sin xx))33

21

23

3)3

(2

f21

43

3 89

Differentiate xxf cos)(

Put in brackets use Chain Rule

21

)(cos)( xxf

Simplify xxxf sin)(cos21

)( 2

1

2

1

)(cos21

)(

xxf xsin

xx

xfcos2sin

)(

Differentiate )1sin()( 2 xxf

Use Chain Rule

Simplify

)1cos)( 2 xxf x2

)1cos2)( 2 xxxf

Differentiate

xxf4

cos)(

Use Chain Rule

Simplify

1

xxf4

sin)(

xxf4

sin)(

Differentiate xxxf 22 cossin)(

Use Chain Rule

Simplify

xxf sin2)(

xxxxxf sincos2cossin2)(

0)( xf

xcos

xcos2 xsin

Integrating Composite FunctionsIntegrating Composite Functions

cna

baxdxbax

nn

)1()(

)(1

dxx 6)14( (4x – 1)7

4

(6 + 1)

cx

dxx

28)14(

)14(7

6

+ c

Integrating Composite FunctionsIntegrating Composite Functions

cna

baxdxbax

nn

)1()(

)(1

3)23( x

dx

Use indices to get denominator on

working line

dxx 3)23( cx

)13(3)23( 2

cx

6)23( 2

cx

2)23(6

1

cna

baxdxbax

nn

)1()(

)(1

3)23( x

dx

Use indices to get denominator on

working line

dxx 3)23( cx

)13(3)23( 2

cx

6)23( 2

cx

2)23(6

1

cna

baxdxbax

nn

)1()(

)(1

2

13)12( dxx

2

1

4

42)12(

x

83

85 44

68

Evaluate 4

12

3

)43( dxx

4

1

2

5

4

12

3

25

3

)43()43(

xdxx

4

1

5

215

)43(

x

2151

21516

55

4136

Evaluate f(x) given f ’(x) = (2x – 1)3 and f(1) = 2

dxxxf 3)12()( cx

42

)12( 4

cx

8

)12( 4

Since f(1) = 2 28

)112( 4

c

281

c87

1c87

18

)12()(

4

x

xf

Find p, given 421

dxxp

dxxp

1

dxxp

1

2

1

p

x

1

2

3

23

1

p

x

1

2

3

32

312

32 2

3

2

3

p

Find p, given 421

dxxp

dxxp

1

312

32 2

3

2

3

p

32

32 3

2

p

4232

32 3

2

p

3

12622 2

3

p 642

3

p

Reverse process

23 64 p

16 p

A curve for which passes through (–1, 2).

Express y in terms of x.

xxdxdy

26 2

dxxxy )26( 2c

xx

22

36 23

cxxy 232

Curve passes through (–1 , 2) c 23 )1()1(22

c 122

5 c

52 23 xxy

Given the acceleration a is:

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

40,)4(2 2

1

tta

dtdv

a 2

1

)4(2 tdtdv

dttv 2

1

)4(2

ct

v

23

1

)4(2 2

3

ct

3

)4(4 2

3

When t = 0 , v = 0 c

3

)4(40

2

3

c

332

0

332

c

332

3)4(4 2

3

t

v

Integrating Trig FunctionsIntegrating Trig Functions

Integration is opposite of Integration is opposite of differentiationdifferentiation

cxdxxxdx

xdsincoscos

)(sin

cxdxxxdx

xdcossinsin

)(cos

Special Trigonometry IntegralsSpecial Trigonometry Integrals

cbaxa

dxbax )sin(1

)cos(

cbaxa

dxbax )cos(1

)sin(

dttt ))3cos(5(sin

t5cos51

)3sin(31

t

c

ctt )3sin(31

5cos51

dxx

2

4

22

cos32

4

22

sin321

x

4

22

sin23

22

2sin

23

0sin

23

2sin

23

0123

23

Area between Trig CurvesArea between Trig Curves cosy xsiny x

1

x

y

1

0

A

The diagram shows the graphs of y = –sin x and y = cos x

a) Find the coordinates of A

b)Hence find the shaded area

CC

AASS

TT00oo180180oo

270270oo

9090oo

3

2

2

Curves intersect where y = y, ie – sin x = cos x

Divide through by – cos x

tan x = – 1

tan-1 (1) = 45o (π/4)

x = 3π/4 or 7π/4

A( 3π/4 , ?) A( 3π/4 , –1/√2)

135o

315o

cosy xsiny x

1

x

y

1

0

A

b) Hence find the shaded area

( 3π/4 , –1/√2)

Area =Area =

∫∫(top curve – bottom curve)(top curve – bottom curve)

43

0))sin((cos dxxxA

43

0)sin(cos dxxxA 4

3

0cossin

xx

)10())2

1(

21

(

12

2 12 2

22

2

222

2

By writing cos 3x as cos(2x + x), show that cos 3x = 4cos3 x – 3cos x

Hence find ∫cos3 x dx

From cos(A + B) = cos A cos B – sin A sin B

cos(2x + x) = cos 2x cos x – sin 2x sin x

= (2cos2x – 1)cos x – (2sin x cos x) sin x

= 2cos3x – cos x – 2sin2 x cos x

= 2cos3x – cos x – 2(1 – cos2 x) cos x

= 2cos3x – cos x – 2cos x + 2cos3 x

= 4cos3x – 3cos x

Hence find ∫cos3 x dx

cos 3x= 4cos3x – 3cos x )cos33(cos41

cos3 xxx

dxxxdxx )cos33(cos41

cos3

cxxdxx )sin33sin31

(41

cos3

cxxdxx )sin43

3sin121

cos3

dxxxx )cos6( 2

36 3x

2

2x xsin c

dxxx )sin3cos2( xsin2 )cos(3 x c

cxx cos3sin2

The curve y = f(x) passes through the point (π/12 , 1).

f ’(x) = cos 2x. Find f(x).

dxxxf 2cos)( cx 2sin21

1)12

(

f c

6

sin21

1 c21

21

1 c41

1

43

c

43

2sin21

)( xxf