View
220
Download
1
Category
Preview:
Citation preview
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
Recommended