1. Unit 3 Outcome 2
Further Calculus

2. Reminder on Derivative Graphs
The graph of y = g(x) is given below
y =g(x) X x 6 8
SPs occur at x = 6 & x = 8
g'(x)
New y-values

3. The graph of y = g '(x) is
y = g '(x) X

4. Derivative Graph of y = sinx
-/2
3/2
/2
5/2
SPs at x = -/2 , x = /2 , x = 3/2 & x = 5/2
x -/  /  3/2  5/2
f (x)
New y-values

5. Roller-coaster from 0 to 2 ie shape of y = cosx
/2
3/2
-/2 2
5/2
Roller-coaster from 0 to 2
ie shape of y = cosx
Hence
If y = sinx
then dy/dx = cosx

6. A similar argument shows that ….
If y = cosx
then 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

7. Example
If g(x) = cosx – sinx then find g  (/2) .
**********
g  (x) = -sinx – cosx
g  (/2) = -sin/2 - cos/2 =
= -1
Example
Prove that y = 7x – 2cosx is always increasing!
*********
dy/dx = sinx
since -2 < 2sinx < 2
then 5 < dy/dx < 9
Gradient is always positive so function is always increasing.

8. 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
= 2/2
= 2
Line
Using y – b = m(x – a)
we get y - 0 = 2(x - /4)
ie y = 2x - 1/4 2

9. Integrating sinx & cosx
If y = sinx then dy/dx = cosx
So  cosx dx = sinx + C
If y = -cosx then dy/dx = sinx
So  sinx dx = -cosx + C
Example
 (3cosx – 4sinx) dx =
3sinx + 4cosx + C

10.  3sinx dx Example = [-3cosx ] = (-3cos /3) - (-3cos0)
= (-3 X ½ ) – (-3 X 1)
= -11/2 + 3
= 11/2
Example
Find the total shaded area.
y = sinx 
3/2

11. 1st area =  sinx dx = [ -cosx ] = -cos - (- cos0) = -(-1) – ( -1)
= -(-1) – ( -1)
= 2
3/2
2nd area =  sinx dx
= [ -cosx ]
3/2
= -cos 3/2 - (- cos )   =
= -1
(actual area = 1)
So total area = = 3units2
alternatively
By symmetry,
2nd area = ½ of first = 1 etc.

12. Show that the shaded area = 22units2.
Example
Show that the shaded area = 22units2.
y = cosx
y = sinx
Limits
Curves meet when sinx = cosx
sinx = cosx cosx cosx
tanx = 1
Q1 & Q3
x = /4 or 5/4
tan-11 = /4

13. Between /4 and 5/4 the sin graph is higher
So shaded area = 
5/4
( sinx – cosx )dx
/4
= [ ]
5/4
-cosx - sinx
/4
= ( -cos5/4 - sin5/4 ) - (-cos/4 - sin/4 )
= (1/2 + 1/2 ) - (-1/2 - 1/2 )
= 1/ / / /2
= 4/2
= 4 X 2
2 X 2
= 42
= 22units2 2

14. 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: d = f(t)
speed: s = f´(t)
2nd derivative
acceleration: a = f´´(t)

15. 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 .

16. 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) f(t) = 2t2 – 5t
so f(10) = (2 X 102) – (5 X 10)
Speed is 150m/s =
= 150

17. Distance travelled = 4162/3 m = 4162/3
(ii) d =  f(t) dt 10
=  2t2 – 5t dt 10
= [ 2/3t3 - 5/2t2 ] 10
= (2000/3 – 250) - 0
Distance travelled = 4162/3 m
= 4162/3
(iii) f ´(t) = 4t - 5
so f ´(10) = (4 X 10) - 5
= 35
Acceleration = 35m/s/s

18. 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 = ie 5.
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

19. Similarly
If f(x) = 2cos3x ,
then to find f(/3)
firstly 3 X /3 = 
then 2cos  = -2
h(x) = 3x - inner
g(…) = 2cos(…) - outer
f(x) = g(h(x))

20. Chain Rule for Differentiation
Suppose that f(x) = (2x + 1)3 & we want to find f (x).
***********
Breaking brackets (2x + 1)3 = (2x + 1)(2x + 1)2
= (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 .

21. The Chain Rule offers a much quicker method and is defined as follows ..
If f(x) = g(h(x))
Then f (x) = g (h(x)) X h (x)
ie 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

22. Example
outer = (…)20
g(x) = (5x – 9)20
20(…)19
g (x) = 20(5x – 9)19
X 5
inner = 5x - 9
= 100(5x – 9)19
Example 5
f(x) = (6x - 5) . Find f (5)
********
outer = (…) 1/2
f(x) = (6x - 5) = (6x – 5)1/2
½(…)-½
f (x) = 1/2(6x – 5)- 1/2
X 6
inner = 6x - 5 = 6
(6x – 5)1/2
3 .
So f (5) =
= 3/5 =
25
(6x - 5)

23. Example V(t) = 2 = 2(4t – 3t2)-3 (4t – 3t2)3 so V  (t) =
X (4 – 6t)
= -6(4 – 6t)
common factor -2
(4t – 3t2)4
= 12(3t – 2)
(4t – 3t2)4

24. Chain Rule with Trig Expressions
Example
outer = 6sin(…)
y = 6sin3
inner = 3
dy/d = 6cos3
X 3
= 18cos3
Example
Find f´(/2) when f(x) = 3cos(2x - /4 )
outer = 3cos(…)
inner = 2x - /4
f (x) = -3sin(2x - /4 )
X 2
= -6sin(2x - /4 )
So f (/2) =
-6sin( - /4)
= -6sin3/4
= -6sin135° .
= -6 X 1/2
= -6/2 X 2/2
= -32

25. Example
P() = cos2
show that P´() = 2tan cos2
***********
P() = cos2
= 1/3(cos)-2
outer = 1/3(….)-2
inner = cos
P´() = -2/3(cos)-3
X -sin
= sin (cos)3
= sin cos(cos)2
= 2tan cos2

26. Applications of Chain Rule
Example
Find the equation of the tangent to the curve y = (3x – 1)2
(x  1/3)
at the point where x = 1.
*********
Point Contact
If x = 1 then y = (3 – 1)2
= -2/4
= -1/2
(1, -1/2)
Gradient
y = -2(3x – 1)-2
outer = -2(…)-2
inner = 3x -1
dy/dx = 4(3x – 1) -3
X 3

27. dy/dx = 12(3x – 1) -3 = 12 (3x – 1)3 when x = 1, dy/dx = 12 23 = 3/2
ctd
dy/dx = 12(3x – 1) -3 =
(3x – 1)3
when x = 1, dy/dx =
= 3/2
gradient
Using 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 X2
or 2y = 3x - 4
or 3x - 2y = 4

28. Example
Find the coordinates of the stationary points on the curve y = 2sinx – cos2x and determine their nature.
**********
SPs occur when dy/dx = 0
2cosx + 2sin2x = 0
2cosx + 4sinxcosx = 0
2cosx(1 + 2sinx) = 0
2cosx = 0 or sinx = 0
1 + 2sinx = 0
cosx = (graph)
sinx = -1/2
Q3 or Q4
x = /2 or 3/2
sin-1(1/2) = /6
Q3: x =  + /6 = 7/6
Q4: x = 2 - /6 = 11/6

29. (/2 ,3) & (3/2 ,-1) are maximum turning points.
using y = 2sinx – cos2x
(/2 ,3)
x = /2
y = 2sin/2 – cos
= 2 – (-1)
= 3
(3/2 ,-1)
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
(11/6 , -3/2)
x  /2  7/6  3/2  11/6 
2cosx
1 + 2sinx
dy/dx
(/2 ,3) & (3/2 ,-1) are maximum turning points.
(7/6 , -3/2) & (11/6 , -3/2) are minimum turning points.

30. Graph looks like ….

31. Integrating (ax + b)n
NB: we do this by reversing the chain rule as follows
Consider
f(x) = (ax + b)n 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 a(n + 1)
+ C

32.   30(5x – 2)2 dx Example = 30(5x – 2)3 + C = 2(5x – 2)3 + C 3 X 5
dt (4t + 1)  6
=  6
= [ ]
(4t + 1)1/2 6
(4t + 1)-1/2 dt 2
½ X 4 2 2
= [ ] 6
½ (4t + 1) 2
= ( ½ X 5) – ( ½ X 3)
= 1

33. Example Find the shaded area ! ********* Limits (3x + 2)2 – 4 = 0
y = (3x + 2)2 - 4
Find the shaded area !
*********
Limits
(3x + 2)2 – 4 = 0
(3x + 2)2 = 4
3x = -2 or 2
3x = or 0
x = -4/3 or 0

34. Shaded area =  (3x + 2)2 – 4 dx = [ ] (3x + 2)3 – 4x 3 X 3 = [ ]
Shaded area = 
(3x + 2)2 – 4 dx
-4/3
= [ ]
(3x + 2)3 – 4x
3 X 3
-4/3
= [ ]
1/9(3x + 2)3 – 4x
-4/3
= 8/9 - ( -8/9 + 16/3 )
= 16/ /3
= -35/9
Negative sign indicates area under X-axis.
Actual area = 35/9 units2

35. 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
= cos(ax + b)
If 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

36.  4sin(2x + ) dx  cos(4 - ) d  -6sin(3 + /2) d Example
= 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
/2
= [ ] 
–2cos(2x + )
/2
= (-2cos3) - (-2cos2)
= 2 – (-2)
= 4

37. Example
By firstly rearranging the formula cos2 = 2cos2 - 1
Find  cos2 d
/2
***********
 cos2 d
/2
cos2 = 2cos2 - 1
/2
2cos2 - 1 = cos2
=  (1/2cos2 + 1/2) d
2cos2 = cos2 + 1
= [ ½ X 1/2sin2 + 1/2 ]
/2
cos2 = 1/2cos2 + 1/2
= [1/4sin2 + 1/2 ]
/2
= (/4 + 1/4sin) - ( sin0)
= /4

38. Example Find the shaded area !! Curves cross when sin2x = cosx
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 3/2
x = /6 or 5/6 B
A C

39. First area =  (sin2x – cosx) dx
/2
/6
= [ ]
/2
-1/2cos2x - sinx
/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