Unit 3 Outcome 2
Further Calculus
Reminder on Derivative Graphs
The graph of
y = g(x) is given below
y =g(x)
5
SPs occur at
x=6 &x=8
6
7
8
9
x
6
g'(x) +
0
X
10
8
-
0
+
New y-values
The graph of y = g '(x) is
y = g '(x)
6
8
X
Derivative Graph of y = sinx
y = sinx
-/
2
3/
2
/
2
5/
2
SPs at x = -/2 , x = /2 , x = 3/2 & x = 5/2
x
f (x)
-/
2
0

/
+
0
2

3/
2

5/
2
-
0
+
0
New y-values
/
2
3/
2
-/
2
2
Roller-coaster from 0 to 2
Hence
If
then
ie shape of y = cosx
y = sinx
dy/
dx
= cosx
5/
2
A similar argument shows that ….
If
then
y = cosx
dy/
dx
= -sinx
If we keep differentiating the basic trig functions we get the
following cycle
f(x)
f  (x)
sinx
cosx
cosx
-sinx
-sinx
-cosx
-cosx
sinx
Example
If g(x) = cosx – sinx
then find
**********
g  (x) = -sinx – cosx
g  (/2) = -sin/2 - cos/2 =
g  (/2) .
-1 - 0 = -1
Example
Prove that y = 7x – 2cosx is always increasing!
*********
dy/
dx = 7 + 2sinx
since -2 < 2sinx < 2
then 5 < dy/dx < 9
Gradient is always positive so function is always increasing.
Example
Find the equation of the tangent to y = sinx – cosx
at the point where x = /4 .
********
Point of contact
If x = /4 then y = sin/4 – cos/4 =
1/
2
- 1/2 = 0
(/4 ,0)
Gradient
dy/
dx
= cosx + sinx
when x = /4 then
dy/
dx
= cos/4 + sin/4 =
1/
2
+ 1/2 =
Line
Using y – b = m(x – a)
we get y - 0 = 2(x - /4)
ie
y = 2x - 1/4 2
2/
2
= 2
Integrating sinx & cosx
If
y = sinx
So
If
y = -cosx
So
then
dy/
dx
= cosx
 cosx dx = sinx + C
then
dy/
dx
= sinx
 sinx dx = -cosx + C
Example
 (3cosx – 4sinx) dx = 3sinx + 4cosx + C
Example

/
0
3
3sinx dx
=
/
[-3cosx ]0
3
= (-3cos /3) - (-3cos0)
= (-3 X ½ ) – (-3 X 1)
= -11/2 + 3
= 11/2
Example
y = sinx
Find the total
shaded area.

3/
2

1st
area =
0
sinx dx =

[ -cosx ] 0
= -cos - (- cos0)
= -(-1) – ( -1)
= 2
3/
2nd area =
  sinx dx
2
=
[ -cosx ]  /
3
2
= -cos 3/2 - (- cos )
=
0 - 1
= -1
So
alternatively
By symmetry,
(actual area = 1)
total area = 2 + 1 = 3units2
2nd area = ½ of first = 1
etc.
Example
Show that the shaded area = 22units2.
y = cosx
y = sinx
Limits
Curves meet when
sinx = cosx
sinx = cosx
cosx
cosx
tanx = 1
x = /4 or 5/4
Q1 & Q3
tan-11 = /4
Between
/
4
and
5/
 / ( sinx – cosx )dx
/
[ -cosx - sinx ] /
5/
So shaded area =
the sin graph is higher
4

4
4
5
=

4
4
= ( -cos5/4 - sin5/4 ) - (-cos/4 - sin/4 )
= (1/2 + 1/2 ) - (-1/2 - 1/2 )
=
1/
2
=
4/
2
+ 1/2 + 1/2 + 1/2
= 4 X 2
2 X 2
= 42
2
=
22units2
Distance,Speed,Acceleration!
Speed = change in distance
change in time
Acceleration = change in speed
change in time
Distance/Time Functions
Suppose a distance/time function is given by d = f(t)
then
distance:
speed:
acceleration:
d = f(t)
s = f´(t)
a = f´´(t)
2nd
derivative
This now means that if we have a “speed/time” function
Ie
S = f(t)
then
acceleration = f´(t)
and
distance =  f(t)dt
NB: Integration gives area under curve.
In Physics area under speed/time graph = distance .
Example
The speed of an object(in m/s) is given by the formula
f(t) = 2t2 – 5t
(i)
Find its speed after10 seconds.
(ii)
How far did it travel in the first 10 seconds?
(iii)
Find its rate of acceleration 10s into the journey.
*********
(i)
so
f(t) = 2t2 – 5t
f(10) = (2 X 102) – (5 X 10)
= 200 - 50
= 150
Speed is 150m/s
(ii)
d = 0 f(t) dt
10
=  0 2t2 – 5t dt
10
=
[
2/
3
t3
-
5/
2
t2
]
10
0
= (2000/3 – 250) - 0
=
(iii)
so
4162/
3
Distance travelled = 4162/3 m
f ´(t) = 4t - 5
f ´(10) = (4 X 10) - 5
= 35
Acceleration = 35m/s/s
CHAIN RULE
Inner & Outer Functions - think back to composite functions.
Suppose that f(x) = (3x – 1)4 .
To evaluate f(2) we first find the value of 3x –1 when x = 2
Then we find 54 ie 625.
We can think of f(x) as having inner and outer parts.
or f(x) = g(h(x))
g(…) = (….)4 - outer
h(x) = (3x – 1) - inner
ie 5.
Similarly
If
f(x) = 2cos3x ,
firstly 3 X /3 = 
h(x) = 3x - inner
then to find
f(/3)
then 2cos  = -2
g(…) = 2cos(…) - outer
f(x) = g(h(x))
Chain Rule for Differentiation
f(x) = (2x + 1)3 & we want to find f (x).
***********
Breaking brackets (2x + 1)3 = (2x + 1)(2x + 1)2
Suppose that
= (2x + 1)(4x2 + 4x + 1)
= 2x(4x2 + 4x + 1) + 1(4x2 + 4x + 1)
= 8x3 + 8x2 + 2x + 4x2 + 4x + 1
= 8x3 + 12x2 + 6x + 1
So
f (x) = 24x2 + 24x + 6
= 6(4x2 + 4x + 1)
= 6(2x + 1)(2x + 1)
= 6(2x + 1)2
This has taken a lot of work and would be really time
consuming if we wanted the derivative of say (2x + 1)10 .
The Chain Rule offers a much quicker method and is
defined as follows ..
ie
If
f(x) = g(h(x))
Then
f (x) = g (h(x)) X h (x)
differentiate the outer part
then multiply by derivative of inner part.
Back to
f(x) = (2x + 1)3
Outer = (…)3 & inner = 2x + 1
So
f (x) = 3(2x + 1)2 X 2
= 6(2x + 1)2
Example
outer = (…)20
20(…)19
inner = 5x - 9
g(x) = (5x – 9)20
g (x) = 20(5x – 9)19 X 5
= 100(5x – 9)19
Example
f(x) = (6x - 5) .
5
Find f (5)
********
f(x) = (6x - 5) = (6x – 5)1/2
f (x) = 1/2(6x – 5)- 1/2 X 6
=
3
(6x – 5)1/2
=
3
(6x - 5)
outer = (…) 1/2
½(…)-½
inner = 6x - 5
6
So f (5) = 3 . =
25
3/
5
Example
V(t) =
2
(4t – 3t2)3
=
2(4t – 3t2)-3
so
V  (t) =
=
-6(4t – 3t2)-4 X (4 – 6t)
-6(4 – 6t)
(4t – 3t2)4
= 12(3t – 2)
(4t – 3t2)4
common factor -2
Chain Rule with Trig Expressions
Example
y = 6sin3
dy/
d = 6cos3
outer = 6sin(…)
X3
inner = 3
= 18cos3
Example
Find f´(/2) when f(x) = 3cos(2x - /4 )
f (x) = -3sin(2x - /4 ) X 2
= -6sin(2x - /4 )
outer = 3cos(…)
inner = 2x - /4
.
= -6 X 1/2
So f (/2) = -6sin( - /4) = -6sin3/4 = -6sin135°
= -6/2 X 2/2
= -32
Example
P() =
1
3cos2
show that P´() = 2tan
3cos2
***********
P() =
1
3cos2
=
1/
-2
3(cos)
outer = 1/3(….)-2
inner = cos
P´() = -2/3(cos)-3
=
.
=
X -sin
2sin
3(cos)3
2sin .
3cos(cos)2
= 2tan
. 3cos2
Applications of Chain Rule
Example
Find the equation of the tangent to the curve y =
-2
1/ )
(x

3
(3x – 1)2
at the point where x = 1.
*********
Point Contact
If x = 1 then y =
-2
=
(3 – 1)2
-2/
4
=
-1/
2
(1, -1/2)
Gradient
y = -2(3x – 1)-2
dy/
dx
= 4(3x – 1) -3 X 3
outer = -2(…)-2
inner = 3x -1
ctd
dy/
dx
= 12(3x – 1) -3 =
when x = 1,
Using
dy/
dx
= 12
23
12
(3x – 1)3
=
3/
2
y – b = m(x – a)
we get y – (-1/2) = 3/2(x – 1)
or
y + 1/2 = 3 /2x – 3/2
or
y = 3 /2x – 2
or
2y = 3x - 4
or
3x - 2y = 4
X2
gradient
Example Find the coordinates of the stationary points on the
curve y = 2sinx – cos2x and determine their nature.
**********
dy/
dx = 0
SPs occur when
2cosx + 2sin2x = 0
2cosx + 4sinxcosx = 0
2cosx(1 + 2sinx) = 0
2cosx = 0 or 1 + 2sinx = 0
cosx = 0 (graph)
x = /2 or 3/2
1 + 2sinx = 0
sinx = -1/2
sin-1(1/2) = /6
Q3: x =  + /6 = 7/6
Q4: x = 2 - /6 = 11/6
Q3 or Q4
y = 2sinx – cos2x
using
/
(/2 ,3)
(3/2 ,-1)
2sin/
x= 2 y =
2 – cos = 2 – (-1) = 3
x = 3/2 y = 2sin3/2 – cos3 = -2 – (-1) = -1
x = 7/6 y = 2sin7/6 – cos7/3 = -1 – 1/2 = -3/2
(7/6 , -3/2)
x = 11/6 y = 2sin11/6 – cos11/3 = -1 – 1/2 = -3/2

2cosx
+
1 + 2sinx +
dy/
+
dx
x
/
2
0
+
0

+
-
7/
0
0
6

+
3/
2
0
0
(11/6 , -3/2)

+
-
11/
6
+
0
0
(/2 ,3) & (3/2 ,-1) are maximum turning points.
(7/6 , -3/2) & (11/6 , -3/2) are minimum turning points.

+
+
+
Graph looks like ….
Integrating (ax + b)n
NB: we do this by reversing the chain rule as follows
Consider
f(x) = (ax + b)n+1
a(n + 1)
f´(x) = (n + 1)(ax + b)n X a
a(n + 1)
= (ax + b)n
so it follows that

(ax + b)n dx
= (ax + b)n+1 + C
a(n + 1)
Example
 30(5x – 2)2 dx = 30(5x – 2)3 + C = 2(5x – 2)3 + C
3X5
Example

6
2
dt .
(4t + 1)
=

6
2
(4t +
1)-1/2
dt =
=
(4t + 1)1/2
½X4
]
6
[ ½ (4t + 1) ]
6
[
2
2
= ( ½ X 5) – ( ½ X 3)
=
1
Example
y = (3x + 2)2 - 4
y
x
*********
Limits
(3x + 2)2 – 4 = 0
(3x + 2)2 = 4
3x + 2 = -2 or 2
3x = -4 or 0
x =
-4/
3
or 0
Find the shaded area !
0
Shaded area =
/
-4
3
=
[
=
[
=
(3x + 2)2 – 4 dx
(3x + 2)3 – 4x
3X3
1/
8/
9
9(3x
+
2)3
– 4x
0
]
-4/
3
0
]
-4/
3
- ( -8/9 + 16/3 )
= 16/9 = -35/9
16/
3
Actual area = 35/9 units2
Negative sign indicates
area under X-axis.
Integration of sin(ax + b) & cos(ax + b)
Consider
If
f(x) = 1/asin(ax + b)
then f´(x) = 1/acos(ax + b) X a
If
= cos(ax + b)
g(x) = -1/acos(ax + b)
then g´(x) = 1/asin(ax + b) X a
= sin(ax + b)
It now follows that
 cos(ax + b)dx = 1/asin(ax + b) + C
(supplied)
 sin(ax + b)dx = -1/acos(ax + b) + C
Example
 cos(4 - ) d = 1/4sin(4 - ) + C
Example
 -6sin(3 + /2) d = 1/3 X 6cos(3 + /2) + C = 2cos(3 + /2) + C
Example

/

4sin(2x + ) dx
= [½ X –4cos(2x + )
2
=

]/


[ –2cos(2x + ) ] /

2
= (-2cos3) - (-2cos2)
= 2 – (-2)
= 4
2
Example
cos2 = 2cos2 - 1
By firstly rearranging the formula
0
/
Find
2
cos2 d
***********
cos2 =
2cos2
2cos2
2cos2 = cos2 + 1
cos2 = 1/2cos2 + 1/2
2
/
2
0
-1
- 1 = cos2
/
cos2 d
=
0
=
[½
=
[
=
(/4 + 1/4sin) - (0 + sin0)
=
/
1/
4
(1/2cos2 + 1/2) d
X
1/
2sin2 +
4sin2 +
1/
1/
/
2

]
2
0
/
2

]
2
0
Example
Find the shaded area !!
A
y = sin2x
B
y = cosx
C
Curves cross when
sin2x = cosx
2sinxcosx – cosx = 0
cosx(2sinx – 1) = 0
cosx = 0 or sin x = 1/2
x = /2 or
B
3/
2
x = /6 or
5/
A
C
6
First area =
=

/
(sin2x – cosx) dx
2
/
6
[
-1 /
2cos2x - sinx
/
]/

2
6
= ( -1/2cos - sin/2) - (-1/2cos /3 - sin/6)
= (1/2 – 1) - (-1/4 – ½)
= ½ - 1+ ¼ + ½
=
1/
4
The diagram has ½ turn symmetry about point B
So the total area = 2 X ¼ = 1/2unit2