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
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 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
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 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
4 du
dxcos
dyu
du
dy dy du
dx du dxcos 4 u 4cos4
dyx
dx
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Higher Outcome 2
The Chain Rule for Differentiating
Example
42
5'( ) ( )
3
Find when f x f x
x
2( 3) Let u x4( ) 5 Then f x u
520 df
udu
2 du
xdx
520 2 u x 2 520( 3) 2 x x
2 5
40
( 3)
df x
dx x
df df du
dx du dx
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Higher Outcome 2Example
1
2 1
Find when
dyy
dx x
(2 1) Let u x1
2
Then y u
2du
dx
3
21
2
dyu
du
dy dy du
dx du dx
3
21
22
u
3
2
u
3
2
1
(2 1)
dy
dxx
The Chain Rule for Differentiating
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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
x xdx
45 df
udu
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
The Chain Rule for Differentiating
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Higher Outcome 2
The Chain Rule for DifferentiatingTrig Functions
Example
4cos Find when dy
y xdx
cos Let u x 4 Then y u
sin du
xdx
34 dy
udu
dy dy du
dx du dx34 ( sin ) u x
34sin cos dy
x xdx
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Higher Outcome 2Example
sin Find when dy
y xdx
sin Let u x1
2 Then y u
cos du
xdx
1
21
2
dyu
du
dy dy du
dx du dx
1
21
cos2
u x
cos
2 sin
dy x
dx x
The Chain Rule for DifferentiatingTrig Functions
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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
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Higher Outcome 2
Integrating Composite Functions
By “reversing” the Chain Rule we see that:
1( )( )
( 1)
n
n ax bax b dx c
a n
Worked Example
4(3 7) x dx5(3 7)
15
xc
4 1(3 7)
3(4 1)
x
c
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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 dt4 1(4 2)
4( 4 1)
tc
3(4 2)
12
tc
Re-arrangeCompare with standard form
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Higher Outcome 2Example
2( 3) Evaluate u du
Compare with standard form
2( 3) u du2 1( 3)
1( 2 1)
uc
1( 3)
1
uc
2 1( 3)
( 3)
u du cu
Integrating Composite Functions
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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)( )
8
xc
2 3
1(2 1) x dx (2) (1) F F 4 41
(5 3 )8
2 2 2 21(5 3 )(5 3 )
8
1(16)(34)
8
2 3
1(2 1) 68 x dx
Integrating Composite Functions
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Higher Outcome 2Example
32
4
1(3 4)
Evaluate x dx
32(3 4) x dx ( ) F x
32
1
32
(3 4)( )
3( 1)
x
c522(3 4)
( )15
xc
32
4
1(3 4)
x dx (4) ( 1) F F
5 52 2
2(16 (1) )
15 52
(4 1)15
2(1024 1)
15
2046
15
32
4
1
2046(3 4)
5x dx
Integrating Composite Functions
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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
From “reversing” the chain rule we can see that:
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 Functions
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Higher Outcome 2
Integrating Trig Functions
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
t t c
Break up into two easier integrals
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Higher Outcome 2Example 2
4
3cos 22
Evaluate x dx
3cos 22
x dx( ) F x 3sin 2 x dx
Re-arrange 3
cos2 ( )2
x c
Apply the standard form
2
4
3cos 22
x dx
2 4
F F3
cos 2 cos 22 2 4
31 0
2
2
4
33cos 2
2 2
x dx
Integrating Trig Functions
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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
( ) F x (cos sin ) x x dx sin cos ( ) x x c
3
4
0
(cos sin )
Area x x dx3
(0)4
F F
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Higher Outcome 2
Integrating Trig Functions (Area)
3(0)
4
F F
3 3sin cos sin 0 cos0
4 4
sin cos 0 14 4
1 11
2 2
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
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
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
Integrating Trig Functions
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Higher Outcome 2Example
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
5 5 1
xc 6
5 3
30
xc
So we have: 6
5 1 3(1) 32
30
f c
62
30 c
64
30 c
Giving:
32
15 c
32 1432 32
15 15 c 6
5 3 896( )
30 30
xf x
Integrating Functions
Differentiation
Higher Mathematics
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5580.36 156.21 (4sf)5424
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2
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24
(4 )3
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344
3v t c Starts at rest, so
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340 4
3c
320
3c
32
3c
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24 32
3 3(4 )v t
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4
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4x c
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If
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( )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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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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The graph of
32
1 1
4
dyx
dx x
( )y g x passes through the point (1, 2).
express y in terms of x. If
3 2 1
4
dyx x
dx
4 1 1
4 1 4
x xy x c
simplify
4 1 1
4 4
xy x c
x Use the point
41 1 1
2 14 1 4
c
3c Evaluate c41 1
4 4
13y x x
x
Calculus Revision
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Integrate2 5x
dxx x
3
2
2 5xdx
x
3 3
2 2
2 5xdx
x x
1 3
2 25x x dx
3 1
2 2
3 1
2 2
5x xc
3 1
2 22
310x x c
Straight line form
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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Evaluate22
2
1
1x dxx
1
222
1x x dx
Cannot use standard integral
So multiply out
4 22
12x x x dx
25 2 1
1
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
10 1
58
Calculus Revision
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Evaluate4
1
x dx1
2
4
1
x dx 43
2
1
2
3x
4
1
32
3x
3 32 24 1
3 3
16 2
3 3
14
3
2
34
Straight line form
Calculus Revision
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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
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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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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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Integrate 33 4x x dx4 23 4
4 2
x xc
4 23
42x x c
Integrate term by term
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
Quit
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