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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
The Chain Rule for Differentiating
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
The Chain Rule for Differentiating
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
1. Differentiate outside the bracket.
2. Keep the bracket the same.
3. Differentiate inside the bracket.
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.
2. Keep the bracket the same.
3. Differentiate inside the bracket.
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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 4 2'( ) 5(3 2 1) (9 - 4 )f x x x x x Let
'(0) 0 f
The Chain Rule for Differentiating
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)
The Chain Rule for Differentiating Functions
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Higher Outcome 2Example
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
The Chain Rule for Differentiating Functions
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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
The Chain Rule for Differentiating Functions
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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)
The Chain Rule for Differentiating Functions
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 ?
The Chain Rule for Differentiating Functions
Calculus Revision
Back NextQuit
Differentiate5( 2)x
Chain rule45( 2) 1x
45( 2)x Simplify
Calculus Revision
Back NextQuit
Differentiate3(5 1)x
Chain Rule23(5 1) 5x
215(5 1)x Simplify
Calculus Revision
Back NextQuit
Differentiate2 4(5 3 2)x x
Chain Rule 2 34(5 3 2) 10 3x x x
Calculus Revision
Back NextQuit
Differentiate
5
2(7 1)x
Chain Rule
Simplify
3
25
2(7 1) 7x
3
235
2(7 1)x
Calculus Revision
Back NextQuit
Differentiate
1
2(2 5)x
Chain Rule
Simplify
3
21
2(2 5) 2x
3
2(2 5)x
Calculus Revision
Back NextQuit
Differentiate 3 1x
Chain Rule
Simplify
Straight line form 1
23 1x
1
21
23 1 3x
1
23
23 1x
Calculus Revision
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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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Higher Outcome 2Example
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
Trig Function Differentiation
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Higher Outcome 2Example
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
Trig Function Differentiation
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
1. Differentiate outside the bracket.
2. Keep the bracket the same.
3. Differentiate inside the bracket.
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Higher Outcome 2
The Chain Rule for DifferentiatingTrig Functions
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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Higher Outcome 2Example
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
The Chain Rule for DifferentiatingTrig Functions
Calculus Revision
Back NextQuit
Differentiate2
2cos sin3
x x
22sin cos
3x x
Calculus Revision
Back NextQuit
Differentiate 3cos x
3sin x
Calculus Revision
Back NextQuit
Differentiate cos 4 2sin 2x x
4sin 4 4cos 2x x
Calculus Revision
Back NextQuit
Differentiate 4cos x
34cos sinx x
Calculus Revision
Back NextQuit
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
Back NextQuit
Differentiate sin(3 2 )x
Chain Rule
Simplify
cos(3 2 ) ( 2)x
2cos(3 2 )x
Calculus Revision
Back NextQuit
Differentiatecos5
2
x
Chain Rule
Simplify
1
2cos5xStraight line form
1
2sin 5 5x
5
2sin 5x
Calculus Revision
Back NextQuit
Differentiate ( ) cos(2 ) 3sin(4 )f x x x
( ) 2sin(2 ) 12cos(4 )f x x x
Calculus Revision
Back NextQuit
Differentiate 62siny x
Chain Rule
Simplify
62cos 1dy
xdx
62cosdy
xdx
Calculus Revision
Back NextQuit
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
Integrating Composite Functions
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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Higher Outcome 2Example
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.
2. Divide by new power.
3. Compensate for bracket.
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
Integrating Composite Functions
4 4(2 2 1) (1 1 1)
8 8
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Higher Outcome 2Example
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
Integrating Composite Functions
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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Higher Outcome 2Example
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.
2. Divide by new power.
3. Compensate for bracket.
Calculus Revision
Back NextQuit
Integrate2(5 3 )x dx
3(5 3 )
3 3
xc
31
9(5 3 )x c
Standard Integral
(from Chain Rule)
Calculus Revision
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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
Use standard Integral
(from chain rule)
Calculus Revision
Back NextQuit
Evaluate0
2
3
(2 3)x dx
Use standard Integral
(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
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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
Integrating Trig Functions
(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
Integrating Trig Functions
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
2. Keep the bracket the same
3. Compensate for inside 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
2. Keep the bracket the same
3. Compensate for inside 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
Integrating Trig Functions
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
Integrating Trig Functions
Calculus Revision
Back NextQuit
Find2sin7
t dt2
7cos t c
Calculus Revision
Back NextQuit
Find 3cos x dx3sin x c
Calculus Revision
Back NextQuit
Find 2sin d 2cos c
Calculus Revision
Back NextQuit
2(6 cos )x x x dx Integrate
3 26sin
3 2
x xx c Integrate term by term
Calculus Revision
Back NextQuit
Find1
3sin cos2
x x dx1
3cos sin2
x x c Integrate term by term
Calculus Revision
Back NextQuit
Find sin 2 cos 34
x x dx
1 1
2 3cos 2 sin 3
4x x c
Calculus Revision
Back NextQuit
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
Back NextQuit
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
Back NextQuit
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.