further differentiation and integration f ’(x) x
TRANSCRIPT
Further Differentiation and Integration
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
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90
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270
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360
360
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
– 2
– 2
siny x
f ’(x)
x
Further Differentiation and Integration
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
90
90
180
180
270
270
360
360
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
– 2
– 2
siny x
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
90
90
180
180
270
270
360
360
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
– 2
– 2
'( ) cosf x x
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
90
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0.5
0.5
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1
1.5
1.5
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– 0.5
– 0.5
– 1
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– 1.5
– 2
– 2
cosy x
f ’(x)
x
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
90
90
180
180
270
270
360
360
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
– 2
– 2
cosy x
0.5 1 1.5 2 – 0.5 – 1 – 1.5 – 2
y
x
90
90
180
180
270
270
360
360
0.5
0.5
1
1
1.5
1.5
2
2
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
– 2
– 2
siny x
sin cosd
x xdx
cos sind
x xdx
For these results to be true, x must be measured in radians.
1. Find '( ) when ( ) 2sinf x f x x
'( ) 2cosf x x
2. Find (3sin 5cos )d
x xdx
3cos ( 5sin )x x
3cos 5sinx x
Integrating Sin x and Cos x
cos sin sin cosx dx x c x dx x c For these results to be true, x must be measured in radians.
3sin( ) 4cos( )x x dx 3(-cos( )) 4sin( )x x c
-3cos( ) 4sin( )x x c
sin cosd
x xdx
cos sind
x xdx
Since and
4
0 Evaluate 4cos 2x sin x dx
4
04sin 2 cosx x
4sin 2 cos 4sin 0 2 cos04 4
4 20 2
2 2
41 2
2
3 2 1
Derivative of (ax + b)n
If ( )ny x a 1then, ( )ndyn x a
dx
Taking this a step further:
If ( )ny ax b 1then, ( )ndyan ax b
dx
This is known as the chain rule.
My rule:- Differentiate outside, differentiate inside then multiply.
3 3 2
1. Differentiate:
(a) 3 5 ( ) 2 6 ( ) 2 6x b x c x
3( ) (3 5)d
a xdx
23 3 5 .3x 2
9 3 5x
12( ) 2 6
db x
dx
12
12 6 .2
2x
122 6x
1
2 6x
1
2 3( ) 2 6d
c xdx
2
2 31
2 6 .43
x x
2
2 34
2 63x
x
223
4
3 2 6
x
x
52 6y x
Let 2 6u x 5y u
45dy
udu
But we require dydx
dy dy dudx du dx
2dudx
45 2u 410u But 2 6u x
410 2 6
dyx
dx
Another Approach
32. Differentiate 3sin x 33Remember: sin sinx x
33sind
xdx
23 3 sin .cosx x 29cos sinx x
1( ) ( )n ndax b an ax b
dx
Let u = sin x cosdu
xdx
33y u 29dy
udu
dy dy dudx du dx
2 29 cos 9cos sinu x x x
Applications
The chain rule allows us to investigate applications involving composite functions.
2
Ex 1. Find the equation of the tangent to the graph of
5 at the point where 2.y x x
Remember: to find the equation of a line we need a point and a gradient.
when 2, 4 5 3x y (2,3)
1
2 25d
xdx
1
2 21
5 .22
x x
1
2 25x x
12When 2, 2(4 5)
dyx
dx
2
3
2We have the point (2,3) and know
3m
Using ( )y b m x a 2
3 ( 2)3
y x
3 9 2 4y x
3 2 5y x
Integrating (ax+b)n
Here we can use the chain rule.
4(3 5)x dx5(3 5)x
5 × 3+ C
My rule: Integrate outside, differentiate inside then divide.
2(2 3)x dx3(2 3)x
3 × 2+ C
3(2 3)6
xC
5(3 5)15
xC
1
( 1)
nn ax b
ax b dx Ca n
Integrating sin( ) and cos( )ax b ax b
1cos( ) sin( )ax b dx ax b C
a
1sin( ) cos( )ax b dx ax b C
a
Find cos 34
x dx
1
sin 33 4
x C
Find the area enclosed by the graph of sin 4 ,6
the x axis and the lines 0 and8
y x
x x
1 2 – 1 – 2
y
x
8
0Area = sin 4
6x dx
8
0
1cos 4
4 6x
1 2 1cos cos
4 3 4 6
1 38 8
1 3
8