Grade 12
Mathematics
Topic
Differential
Calculus
Page 1 of 15
Lesson Description:
In this lesson we will deal with:
 Looking at the concept of a limit of a function.
 Calculating the derivative of a function from first principles.
 Calculating the derivative of a function by using the rules of
differentiation.
 Sketching cubic functions.
 Graph interpretation.
 Applications of calculus relating to maxima and minima problems and
rates of change.
Key Concepts:











Functional Notation
Average Gradient
Limits
Derivative from First Principles
Derivative from Rules
Tangents to a Curve
Remainder & Factor Theorems
Sketching a Cubic Function
Graph Interpretation
Maxima & Minima Problems
Rates of Change (Calculus of Motion)
What is Calculus?
Calculus is the study of rates of change and motion. It is one of the most
powerful mathematical tools ever invented and is used every day by
mathematicians, physicists, engineers, biologists and indirectly by many other
professions.
Page 2 of 15
Calculus has been developed over thousands of years. Many famous
Mathematicians have had some part in inventing and perfecting it. Some of the
more famous ones were:
 Archimedes (c. 287 – 212 BC)
He was an “Ancient Greek” Mathematician. He developed methods for
finding the areas of shapes that did not have straight edges.
 Bhaskara (1114 – c. 1185)
An Indian who developed methods of predicting eclipses which were
based on calculating planetary motion. The study of moving objects leads
naturally to calculus.
 Newton (1642 – 1727)
As a young Cambridge student, Newton found mathematical ways of
describing the exact motion of the moon, planets and comets. He was
reluctant to publish his discoveries in calculus.
 Leibniz (1646 – 1716)
A German philosopher who expressed the mathematics involved in
calculus in terms of “infinitely small” quantities. This approach initially
provided a more useful notation in calculus.
 Cauchy (1789 – 1857)
Cauchy, a German, contributed to the improvement of Newton’s
approach, through the concept of limits.
 Robinson
Robinson, a Mathematician, who as recently as the 1950s improved in
Leibniz’s mathematical approach to calculus through non-standard
analysis.
Calculus is divided into two parts, namely:
1. Differential Calculus – calculating rates of change (differentiation)
2. Integral Calculus – calculating area (integration)
We will only be dealing with differentiation in Grade 12.
NB!!! Calculus counts 30 – 40 marks in Paper 1 in the final examinations.
As a result, this is an important section to master properly!
Page 3 of 15
Average Gradient:
Average Gradient (m) =
=
=
=
"𝑟𝑖𝑠𝑒"
"𝑟𝑢𝑛"
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
𝑓(𝑥+ℎ)− 𝑓(𝑥)
(𝑥+ℎ)− 𝑥
𝑓(𝑥+ℎ)− 𝑓(𝑥)
ℎ
Limits:
The limit of a function is the value of the function as x approaches a particular
value.
lim
e.g.
𝑥→4
=
=
=
lim
𝑥→4
𝑥 2 − 2𝑥−8
𝑥−4
(𝑥−4)(𝑥−2)
(𝑥−4)
lim(𝑥 + 2)
𝑥→4
6
(Take note that there is NO value at x = 4)
Page 4 of 15
Definition of the Derivative (First Principles)
If we find an expression for the gradient/slope of a function f at each point
where this is possible, we obtain a new function called the derivative/gradient
function of f. We shall denote the derivative of f(x) with respect to x by f '(x).
By using first principles, we calculate the derivate using the following formula:
𝑓 (𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
′(
𝑓 𝑥 ) = lim
Rules for Differentiation:
RULE 1: The derivative of a constant is zero.
If
𝑓(𝑥 ) = 𝑘, k a constant,
Then
𝑓 ′ (𝑥) = 0.
RULE 2: The derivative of xn is nxn – 1.
If
Then
𝑦 = 𝑥𝑛,
𝑑𝑦
𝑑𝑥
= 𝑛𝑥 𝑛−1 .
Page 5 of 15
RULE 3: If f is differentiable (i.e. if f '(x) exists, then the derivative of f
multiplied by a constant is equal to the derivative of f multiplied by
the constant.
𝑑
𝑑𝑥
=
RULE 4: (a)
(b)
(𝑘. 𝑓(𝑥 )), where k is a constant
𝑘.
𝑑
𝑑𝑥
(𝑓(𝑥 ))
The derivative of a sum is the sum of the derivatives.
The derivative of a difference is the difference of the
derivatives.
𝐷𝑥 [𝑓(𝑥 ) ± 𝑔(𝑥)] = 𝐷𝑥 [𝑓(𝑥)] ± 𝐷𝑥 [𝑔(𝑥)]
Notation:
Mathematics is like a language. It is vitally important that you know how to
write the different notations and understand what they mean!
If y = f(x) is a function with respect to x, we can then represent the derivative of
f with respect to x, using the following notations:
𝑑𝑓
𝑑 𝑓(𝑥)
𝑑𝑦
𝑑𝑥
𝑑𝑥
𝑑𝑥
Page 6 of 15
𝑓 ′ (𝑥)
𝐷𝑥 [𝑓(𝑥 )]
The Equation of a Tangent to a Curve:
DEFINITION: A tangent to a curve is a line drawn to a curve with only one
point of contact.
How do we find the equation of a tangent to a curve y = f(x) at the point x=a?
RECIPE:
1.
Find the derivative, f '(x).
2.
Substitute x=a into the derivative to find the gradient,
mT = f '(a).
3.
Using the co-ordinates of the point of contact (a; f(a)),
we can find the equation of the tangent using:
y = mx + c or y – y1 = m(x – x1)
The Remainder & Factor Theorems:
You don’t need to learn these theorems, but you need to be able to apply the
factor theorem to factorise cubic functions.
REMAINDER THEOREM:
𝑏
If a polynomial f(x) is divided by ax – b, the remainder is 𝑓 ( ).
𝑎
FACTOR THEOREM:
Suppose f(x) is a polynomial:
𝑏
1. If ax – b is a factor of f(x), then 𝑓 ( ) = 0.
𝑎
𝑏
2. If 𝑓 ( ) = 0, then ax – b is a factor of f(x).
𝑎
Page 7 of 15
Sketching a Cubic Function:
𝑦 = 𝑓(𝑥 ) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑
1. Shape:
(𝑎 ≠ 0)
a > 0 (+)
a < 0 (–)
2. Intercepts:
y-intercept:
x-intercept:
Put x = 0, solve for y.
Put y = 0, solve for x.
3. Stationary Points:
 Find f '(x).
 Put f '(x) = 0 and solve for x.
 Find the y-values of the stationary points by substituting the xvalues you got from solving f '(x) = 0 into y = f(x).
 Determine the nature of the stationary points using a table of signs.
NOTE:
f '(x) > 0
f '(x) < 0
function is increasing
function is decreasing
Page 8 of 15
Graph Interpretation:
1. Finding the Equation of a Cubic Function
If we are given the x-intercepts and one other point we use the following
equation:
𝑦 = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 )(𝑥 − 𝑥3 )
where x1; x2 and x3 are the values of the x-intercepts.
2. Distances
If A(xA; yA) and B(xB; yB), then the vertical distance AB = yB – yA.
If C(xc; yC) and D(xD; yD), then the horizontal distance CD = xD – xC.
3. Points of Intersection
To find the point of intersection between two functions, you must put the
two equations equal to each other and then solve for x. Then put the xvalue back into one of the functions to find the y-value of the point of
intersection.
4. Intercepts
Y-intercept(s):
X-intercept(s):
Put x = 0 and solve for y.
Put y = 0 and solve for x.
5. Maximum Distance
If you want the maximum or minimum distance between two functions
(say f(x) and g(x)), make a new function for the distance:
d(x) = f(x) – g(x).
Find d'(x) and use a table of signs to determine the maximum.
6. Domain & Range
Domain – all possible x-values (input).
Range – all possible y-values (output).
Page 9 of 15
7. Increasing/Decreasing
If y = f(x), then we find f '(x) first.
If f '(x) > 0, then the function is increasing on that interval.
If f '(x) < 0, then the function is decreasing on that interval.
Maxima & Minima Problems:
1. Read the question carefully – at least twice – and then extract the relevant
information.
2. Develop a function that describes the situation algebraically.
3. Find the derivative of the function you have made.
4. Find the x-values when the derivative is equal to zero.
5. Determine whether it is a maximum or minimum.
6. Answer the question that has been asked.
Rates of Change (Calculus of Motion):
We will be dealing with displacement (s), velocity (v) and time (t) of a moving
object or particle.
If s(t) = f(t) is the equation of motion, which gives the position of the
object/particle at time t, then:
𝑣 (𝑡 ) =
𝑑𝑠
𝑎 (𝑡 ) =
𝑑𝑣
𝑑𝑡
𝑑𝑡
𝑠 (𝑡 )
= 𝑓 ′ (𝑡) gives the velocity of the object/particle
= 𝑣 ′ (𝑡) gives the acceleration of the object/particle
𝑑𝑠
𝑑𝑡
𝑣 (𝑡 )
Page 10 of 15
𝑑𝑣
𝑑𝑡
𝑎(𝑡)
Past Exam Paper Questions:
These questions have been taken from past Department of
Basic Education final Grade 12 Examinations.
February/March 2012 Paper 1
Page 11 of 15
Page 12 of 15
Page 13 of 15
Page 14 of 15
All past Grade 12 examination papers and their memos can
be downloaded from www.education.gov.za
Page 15 of 15