1. Differentiation
Calculus was developed in the 17th century by Sir Issac Newton and Gottfried Leibniz who disagreed fiercely over who originated it.
Calculus provides a tool for solving problems involving motion. For example, methods obtained from calculus are used to study the orbits of planets, to calculate the flight path of a rocket and to predict the path of a particle through a magnetic field.
In fact calculus can be used to study any situation where a rate of increase or decrease is involved. (rate of change)
If we remember from standard grade distance time graphs. The speed was obtained from the gradient of the graph.
Time
Distance d t
This is the gradient of the line.

2. To determine the speed we needed two points on the graph.
This gave us the average speed between these two points.
But what about the speed at a certain time? The instantaneous speed. x y
f(x+h)
f(x) h
For an instantaneous gradient we require ‘h’ to be zero.
Let us examine this gradient as ‘h’ gets smaller and smaller. x
x+h
In other words, let ‘h’ approach zero.
We label this gradient

3. The derived function represents: The rate of change of the function x y h
x+h
f(x)
f(x+h)
The derived function represents:
The rate of change of the function
The gradient of the tangent to the graph of the function

4. Using the formula for f `(x) is called differentiating from first principles.

5. Numerous functions were derived in this way and a pattern was noticed.
In general terms:

7. The derivative is also called the ‘rate of change’ or the ‘gradient of the tangent to the curve’. They all mean find the derived function.
1. Find the gradient of the tangent to the curve f(x) = x2 at x = 5.

8. 2. Smoke form a factory chimney travels in t hours
2. Smoke form a factory chimney travels in t hours. Calculate the speed (rate of change) of the smoke after 4 hours.
The distance travelled by the smoke is
After 4 hours, t = 4.
The speed of the smoke after 4 hours is ¼ km / h.

9. Page 92 Exercise 6E
We can also derive functions with more than one term.

10. Derivatives of products and quotients
We can also derive more complicated expressions. Before differentiating a function we have to express it as the sum of individual terms.

11. Now there will be times when you have a correct answer but it will not match the one shown in the back of the book. However they could well be the same. Look at our previous answer.
Always check for equality when checking your answers.

12. The gradient of the slope 4 metres after the start of the slope is -1.
1. A designer of an artificial ski slope describes the shape of the slope by the function
h(d) d
d is the horizontal distance in metres and h(d) is the height in metres. Calculate the gradient of the slope 4 metres horizontally from the start of the slope.
The gradient of the slope 4 metres after the start of the slope is -1.

13. This is a geometric notation.
Instead of writing the derivative as f `(x) we can use the Leibniz notation.
This is a geometric notation.
The Greek letter delta Δ means ‘change’. So going back to our gradient diagram, x y Δy
This is normally written Δx

14. 1. Find the derivative of y = 3x4 -5x2 – 3x + 6
At t = 9,

16. Equations of Tangents
Hence to find the equation of a tangent we need to determine:
the coordinates (a,b) of the point
the gradient, (rate of change) at that point.

17. Hence (9,27) is the point.
Now we need the gradient at the point (9,27).

18. Increasing and decreasing functions
This function is increasing
This function is decreasing
It has a positive gradient
It has a negative gradient
If f(x) increases as x increases then the function is increasing.
If f(x) decreases as x increases then the function is decreasing.

19. Let us look at 3 points on the graph.x y
The function is decreasing between A and B C A
The function is increasing between B and C
What is the function doing at B? B
If we look at the tangent to the curve at point ‘B’.
We see it is horizontal.
It will therefore have zero gradient. It has no rate of change.
A to B B B to C
The derivative of a function may be used to identify the shape of its curve.
negative
zero
positive

21. Let us first consider when the function is Increasing
(this is a smile parabola with roots at x = 0 and x = 2)
NB This is a graph of the GRADIENT
not the function.

22. The value of (x – 1)2 must always be positive for all values of x.
This means that the value of 3(x – 1)2 must always be positive which by default means that it can never be negative for all values of x.
Since the gradient is never negative the function is never decreasing.

23. Stationary points Let us remind our self of what we found earlier.x y C B A
A to B B B to C
negative
positive
zero
We saw previously that if f `(x) = 0 the function is neither increasing or decreasing.
The tangent is horizontal so the point (x, f(x)) is a stationary point.
Stationary points occur when f `(x) = 0.
The nature of the stationary point depends on the gradient either side of it.

24. Rising point of inflexion falling point of inflexiony
Maximum turning point x y
Minimum turning point x y
Rising point of inflexion x y
falling point of inflexion

25. We will now consider the gradient either side of the stationary points to determine their nature.

26. + + + - - + - + + + - +
slope
Maximum turning point at (1,5) and minimum turning point at (2,4).

27. We will now consider the gradient either side of the stationary points to determine their nature.

28. + + + + + - + + -
slope
Rising point of inflexion at (0,0) and maximum turning point at (3,27)

29. Curve sketching
When sketching curves, gather as much information as possible form the list below.
1. Intercepts.
Where does the curve cut the y axis? x = 0
Where does the curve cut the x axis (roots)? y = 0
2. Stationary Points.
Where is the curve stationary? f `(x) = 0
What are the nature of the stationary points?
3. Behaviour.
What happens for large + / - values of x?

31. - + + - - + + - +
slope
Maximum turning point (0,0) and minimum turning point at (2,-4)

32. Now lest put all this information onto a set of axis.y 3
(2,-4)

33. Now lest put all this information onto a set of axis.y 3
(2,-4)

34. Closed Intervals
In a closed interval the maximum and minimum values of a function are either at the stationary points or at the end points of the interval. x y
(-1,-2) 1
(3,18) x y
(-2,-12) 1
Maximum value = 0
Maximum value = 18
Minimum value = -12
Minimum value = -2

35. (4,48) x y 1
(-2,-12)
Maximum value = 48
Minimum value = -12

37. Maximum value = 21
Minimum value = -25/27

38. Graph of the derived function
We can sketch the graph of the derived function f `(x) by looking at the features of the graph of f(x).
This is a simple strategy but does require an understanding that you are drawing the graph of f `(x) from the graph of f(x).

39. x
f(x) x
f `(x)

40. x
f(x) x
f `(x)

41. x
f(x) x
f `(x)

42. x
f(x) x
f `(x)

43. x
f(x) x
f `(x)

44. x
f(x) x
f `(x)

45. x
f(x) x
f `(x)

46. x
f(x) x
f `(x)

47. x
f(x) x
f `(x)

48. Draw the graph of the derived function.x
f(x) x
f `(x)

49. Draw the graph of the derived function.x
f(x) x
f `(x)

50. Draw the graph of the derived function.x
f(x) x
f `(x)

51. Draw the graph of the derived function.x
f(x) x
f `(x)

52. Draw the graph of the derived function.x
f(x) x
f `(x)

53. Optimization: Maxima and minima
Without differentiation we would need to plot most functions to find their maximum and minimum values.
By finding stationary points we can easily identify and determine these values.
This allows us to examine all functions and find the functions maximum and / or minimum values. We can apply this to real life situations.

54. x
The glass front of a carriage clock has breadth x cm. The jeweller has used 16 cm of gold leaf to edge the perimeter of the glass.
Find, in terms of x, expressions for the height of the glass front and the area of the glass.
Find the dimensions of the rectangle that gives the maximum area.
Calculate the maximum area.
(b) Maximum and minimum area occurs when

55. We now need to determine whether this
gives a maximum or minimum area. + -
slope
Hence maximum area occurs when
x = 4
The dimensions that give the maximum area are breadth = 4cm, height = 4cm.

56. Further Applications
In many cases the solution to a problem cannot be found directly from the information provided.
Often an expression must be found for a value that is provided and this expression then used to obtain a function that can be differentiated.
A useful strategy:

58. The volume of the square based glass display case shown is 500cm3
The volume of the square based glass display case shown is 500cm3. The length of the base is x cm. The base of the case is not made of glass. x
(a) Show that the area of glass in the case is
(b) Find the dimensions of the case that minimise the use of glass and calculate this minimum area.
This formula has a ‘h’ in it that we do not need.
We need to find out what ‘h’ is in terms of x.
Is there any information in the question that will let us do this?
We are told that V = 500 so let us use that.
Let’s substitute this into the area formula.

59. We will now determine the nature of the s.p.

60. - +
slope
Hence minimum area occurs when x = 10
Hence the dimensions of the case are length =10cm, breadth = 10cm and height = 5cm.