Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Harder Type Questions Further...

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Higher Outcome 2

Higher Unit 3Higher Unit 3

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Further Differentiation Trig Functions

Harder Type Questions

Further Integration

Integration Exam Type Questions

Integrating Trig Functions

Differentiation The Chain Rule

Harder Type Questions

Differentiation Exam Type Questions

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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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Higher Outcome 2Example

2( ) 2cos sin '( )

3 . Find f x x x f x

2'( ) (2cos ) sin

f x x xdx dx

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

dx dx dx

22 (cos ) sin

d dx x

dx dx. ( )

d dgc g x c

2'( ) 2sin cos

3 f x x x

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4 cos Differentiate with respect to

4 cos4 cos

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

x x x xdx dx dx

12 ( sin )

12 sin

sin 12 x x

Simplify expression -

where possible

Restore the original form of expression

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Higher Outcome 2

The Chain Rule for Differentiating

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

dy dy du

dx du dx

Example 2 8( 3) Differentiate y x

2( 3) Let u x 8 then y u

dy dy du

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

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Higher Outcome 2

The Chain Rule for DifferentiatingTrig Functions

Trig functions can be differentiated using the same chain rule method.

dy dy du

dx du dx

Worked Example:

sin 4 Differentiate y x

4 Let u x sin Then y u

dy dy du

dx du dxcos 4 u 4cos4

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Higher Outcome 2

Example

5'( ) ( )

Find when f x f x

2( 3) Let u x4( ) 5 Then f x u

520 df

520 2 u x 2 520( 3) 2 x x

df df du

dx du dx

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Find when

(2 1) Let u x1

Then y u

dy dy du

dx du dx

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

3 2(3 2 1) Let u x x 5( ) Then f x u

2(9 4 ) du

df df du

dx du dx4 25 (9 4 ) u x x 3 2 4 25(3 2 1) (9 4 ) x x x x

'(0) 0 f

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Higher Outcome 2

Example

4cos Find when dy

cos Let u x 4 Then y u

sin du

dy dy du

dx du dx34 ( sin ) u x

34sin cos dy

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sin Find when dy

sin Let u x1

2 Then y u

cos du

dy dy du

dx du dx

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Higher Outcome 2Example1

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

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

Where x = 3:

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)

2 1 1 5( 3) (2 6)

25 5 25 25

1(2 1)

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:

( ) 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

( ) 40

C x xx

Re-arrange 2

251600

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Higher Outcome 2

xdx dx x

Using chain rule 2

25 251600 2 1

25 253200 1

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

25 253200 1 0 5

when dC

x xdx x x

5 5 Consider the sign of when and dC

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

1600 55

21600 10

160,000 £ Expensive components?

Aeroplane parts maybe ?

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Higher Outcome 2

Integrating Composite Functions

By “reversing” the Chain Rule we see that:

1( )( )

n ax bax b dx c

Worked Example

4(3 7) x dx5(3 7)

4 1(3 7)

3(4 1)

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Higher Outcome 2

1( )( )

n ax bax b dx c

Example : 4(4 2) Evaluate dt

4(4 2) t dt4 1(4 2)

4( 4 1)

3(4 2)

Re-arrangeCompare with standard form

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2( 3) Evaluate u du

Compare with standard form

2( 3) u du2 1( 3)

1( 2 1)

2 1( 3)

u du cu

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Higher Outcome 2Example 2 3

1(2 1) Evaluate x dx

3(2 1) x dx3 1(2 1)

( )2(3 1)

xc( ) F x

4(2 1)( )

1(2 1) x dx (2) (1) F F 4 41

(5 3 )8

2 2 2 21(5 3 )(5 3 )

1(16)(34)

1(2 1) 68 x dx

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1(3 4)

Evaluate x dx

32(3 4) x dx ( ) F x

(3 4)( )

c522(3 4)

1(3 4)

x dx (4) ( 1) F F

5 52 2

2(16 (1) )

(4 1)15

2(1024 1)

2046(3 4)

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Higher Outcome 2

(sin ) cos

(cos ) sin

cos sin

sin cos

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

From “reversing” the chain rule we can see that:

1cos( ) sin( )

1sin( ) cos( )

ax b dx ax b ca

Worked Example

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

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Higher Outcome 2

Example

sin5 cos( 2 ) Evaluate t t dt

Apply the standard form

(twice).sin(5 ) s( 2 ) t dt co t dt 1 1

cos5 sin 25 2

Break up into two easier integrals

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Higher Outcome 2Example 2

3cos 22

Evaluate x dx

3cos 22

x dx( ) F x 3sin 2 x dx

Re-arrange 3

cos2 ( )2

Apply the standard form

3cos 22

cos 2 cos 22 2 4

33cos 2

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Higher Outcome 2

Integrating Trig Functions (Area)

Example

cosy xsiny x

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

T0o180

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Higher Outcome 2

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

(cos ( sin ))

Area x x dx

(cos sin )

x x dx

( ) F x (cos sin ) x x dx sin cos ( ) x x c

(cos sin )

Area x x dx3

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Higher Outcome 2

3 3sin cos sin 0 cos0

sin cos 0 14 4

1 2 Area

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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 +

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Higher Outcome 2

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

3 1cos (cos3 3cos )

4 x x x

3 1cos (cos3 3cos )

4 x dx x x dx

1(cos3 3cos )

4 x x dx

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

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5( ) '( ) 5 3 (1) 32 Find given and f x f x x f

5( ) 5 3 f x x dx

Use the “reverse” chain rule

5 15 3

So we have: 6

5 1 3(1) 32

Giving:

32 1432 32

15 15 c 6

5 3 896( )

Integrating Functions

Differentiation

Higher Mathematics

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Calculus Revision

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Differentiate 24 3 7x x

Calculus Revision

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Differentiate

3 26 3 9x x x

x x x x

2 16 3 9x x x

Split up

Straight line form

22 6 9x x Differentiate

Calculus Revision

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Differentiate

2 22 5x x

2 23 12 2

2 5x x

Calculus Revision

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Differentiate2

2 23 2 1x x

Straight line form

Differentiate

Calculus Revision

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Differentiate (3 5)( 2)x x

23 6 5 10x x x Multiply out

Differentiate 6 1x

23 10x x

Calculus Revision

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Differentiate 2 8003 r

2 13 800r r Straight line form

Differentiate26 ( 1)800r r

Calculus Revision

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Differentiate2( 2 )x x x

multiply out3 22x x

differentiate23 4x x

Calculus Revision

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Differentiate3

1 3x x Straight line form

Differentiate2 4( 3)x x

2 43x x

Calculus Revision

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Differentiate 2x x x

22x x xStraight line form

multiply out

Differentiate

2 2x x3 1

2 25 3

2 2x x

Calculus Revision

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Differentiate 2 4x x

multiply out

Simplify

4 8 2x x x

Differentiate

6 8x x

Straight line form1

26 8x x 1

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

2(7 1)x

Chain Rule

Simplify

2(7 1) 7x

2(7 1)x

Calculus Revision

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Differentiate

2(2 5)x

Chain Rule

Simplify

2(2 5) 2x

2(2 5)x

Calculus Revision

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Differentiate 3 1x

Chain Rule

Simplify

Straight line form 1

23 1 3x

Calculus Revision

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Differentiate 42

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

Chain Rule

Simplify

Straight line form1

2(2 1)x

2(2 1) 2x

2(2 1)x

Calculus Revision

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Differentiate2

2cos sin3

22sin cos

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

2(sin )x1

2(sin ) cosx x

2cos (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

Chain Rule

Simplify

2cos5xStraight line form

2sin 5 5x

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 2 43200( )A x x

Chain Rule

Simplify

Straight line form2 1( ) 43200A x x x

2( ) 2 43200A x x x

43200( ) 2A x x

Calculus Revision

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Differentiate 2

2( )f x x

Differentiate

Straight line form1

22( ) 2f x x x

2( ) 4f x x x

Calculus Revision

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Differentiate 3 22 7 4 4y x x x

26 14 4y x x

Calculus Revision

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Differentiate 62siny x

Chain Rule

Simplify

62 cos 1

62 cos

Calculus Revision

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Differentiate3

multiply out

Differentiate

4A a a

Calculus Revision

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Differentiate

13 2( ) (8 )f x x

Chain Rule

Simplify

13 221

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

12 3 2( ) (8 )f x x x

Calculus Revision

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Differentiate16

, 0y x xx

Differentiate

Straight line form1

216y x x

21 8dy

Calculus Revision

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Differentiate 23 3 16( )

2A x x

Straight line form

Multiply out 23 3 3 3 16( )

2 2A x x

Differentiate

2 13 3( ) 24 3

2A x x x

2( ) 3 3 24 3A x x x

Calculus Revision

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Differentiate 1

2( ) 5 4f x x

Chain Rule

Simplify

2( ) 5 4 5f x x

2( ) 5 4f x x

Calculus Revision

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Differentiate 216( ) 240

3A x x x

32( ) 240

3A x x

Calculus Revision

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Differentiate 2( ) 3 (2 1)f x x x

Multiply out

Differentiate

3 2( ) 6 3f x x x

2( ) 18 6f x x x

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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Integration

Higher Mathematics

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Calculus Revision

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Integrate 2 4 3x x dx 3 24

x xx c

3 212 3

3x x x c

Integrate term by term

simplify

Calculus Revision

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Find 3cos x dx3sin x c

Calculus Revision

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Integrate (3 1)( 5)x x dx 23 14 5x x dx

3 23 145

x xx c

3 27 5x x x c

Multiply out brackets

simplify

Calculus Revision

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Find 2sin d 2cos c

Calculus Revision

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Integrate2(5 3 )x dx

3(5 3 )

9(5 3 )x c

Standard Integral

(from Chain Rule)

Calculus Revision

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Find p, given

x dx 1

x dx 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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Evaluate

x dx 21

1x 1 12 1

Straight line form

Calculus Revision

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(2 3)x dx17

7 7(2 3) (0 3)

7 75 3

5580.36 156.21 (4sf)5424

Use standard Integral

(from chain rule)

Calculus Revision

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3sin cos2

x x dx1

3cos sin2

x x c Integrate term by term

Calculus Revision

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Integrate3

2x dxx

32 2x x dx 3 222

3x x c

Straight line form

Calculus Revision

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Integrate3 1x dx

13 2x x dx

42x x c

Straight line form

Calculus Revision

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Integrate

3 5x xdx

2 25x x dx 7 3

Straight line form

Calculus Revision

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Integrate

22x x dx

2 24 1

3 4x x c

Split into separate fractions

Calculus Revision

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Find 2

2 1x dx

4 44 1 2 1

4 2 4 2

4 45 3

(from chain rule)

Calculus Revision

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Find sin 2 cos 34

x x dx

2 3cos 2 sin 3

4x x c

Calculus Revision

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7cos t c

Calculus Revision

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Integrate

3 1x dx

Straight line form

2 23 1 0 1

2 24 1

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

22(4 )dv

2(4 )tv c

3v t c Starts at rest, so

v = 0, when t = 0 34

3 3(4 )v t

Calculus Revision

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A curve for which 3sin(2 )dy

12, 3passes through the point

Find y in terms of x. 3

2cos(2 )y x c

2 123 cos(2 ) c Use the point 5

2 63 cos( ) c 3 3

2 23 c

4 3 3 3

2 4cos(2 )y x

Calculus Revision

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Integrate 2 2

2 2, 0

x xdx x

x x 2 24x x dx

Split into

separate fractions

Multiply out brackets

Calculus Revision

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

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

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

31 cos 3

31 cos

3 21 c

6c 1 7

3 6cos(3 )y x

Calculus Revision

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Integrate2

(7 3 )dx

x2(7 3 )x dx

1(7 3 )

3(7 3 )x c

Straight line form

Calculus Revision

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The graph of

( )y g x passes through the point (1, 2).

express y in terms of x. If

x xy x c

simplify

xy x c

x Use the point

41 1 1

2 14 1 4

3c Evaluate c41 1

13y x x

Calculus Revision

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Integrate2 5x

2 5xdx

2 25x x dx

310x x c

Straight line form

Calculus Revision

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A curve for which 26 2dy

passes through the point (–1, 2).

Express y in terms of x.

3 26 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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Evaluate22

1x dxx

1x x dx

Cannot use standard integral

So multiply out

12x x x dx

25 2 1

5x x x 5 2 5 21 1 1 1

5 2 5 12 2 1 1

32 1 1

5 2 54 64 40 20 2

10 10 10 10 82

Calculus Revision

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Evaluate4

x dx 43

3 32 24 1

Straight line form

Calculus Revision

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Evaluate0

(2 3)x dx

(from chain rule)

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

Calculus Revision

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The curve ( )y f x ,112

passes through the point

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

2( ) sin 2f x x c

2 121 sin 2 c

use the given point ,112

2 61 sin c

2 21 c 3

4c 1 3

2 4( ) sin 2f x x

6 2sin

Calculus Revision

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2(6 cos )x x x dx Integrate

3 26sin

x xx c Integrate term by term

Calculus Revision

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Integrate 33 4x x dx4 23 4

42x x c

Calculus Revision

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Evaluate1

01 3x dx

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

3 32 24 1

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Higher Outcome 2 Higher Unit 3 Further Differentiation Trig Functions Harder Type Questions Further...

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