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grade 12 calculus revision
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- 12th Grade
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DIFFERENTIAL CALCULUSPART 1
(LIMIT, 1ST PRINCIPLE AND
DERIVATIVE USING RULES)
GRADE 12
PAST EXAM PAPERS
EMAILBY
ADDRESS: melulekishabalala@gmail.com
MR M. SHABALALA @NOMBUSO HIGH NUMBER: 0733318802 Page 1
CELLPHONE
, CALCULUS
It helps us to understand any change in different variables ,eg measure instantaneous changes in
rate of population growth, behavior of particles, economics profit and losses also determining the
spread of disease. It is basically the MATHEMATICS OF CHANGE. Wherever there is a
change in one variable produce changes in another, therefore calculus helps us to understand that
changes occur. LIMIT - The limit of the function is the value of 𝑦 to which the graph approaches
as the values of 𝑥 approach a certain value from both the left and right side of that point. It will be
explored more at University.
AVERAGE GRADIENT
The Average gradient (average rate of change) of a function, it is like the gradient of the line
between two points, change in 𝑦 divided by change in 𝑥.
EXAMPLE:
1.1 Find the limit of the following functions.
a) b)
1.2 If 𝑓(𝑥) = −2𝑥 3 find the average gradient between 𝑥 = −1 and 𝑥 = 2
1. “NEW MAIN TOPIC” FIRST PRINCIPLE
Reads as: Derivative (𝑓′(𝑥)) of 𝑓(𝑥) as ℎ approaches zero.
General formula to find average gradient(slope) of a function at any point(Derivative)
EXAMPLES OF DERIVATIVE(gradient) USING 1ST PRINCIPLE
1.3 Find 𝑓′(𝑥) from first principle.
1 5
a) 𝑓(𝑥) = 2𝑥 2 − 2 𝑥 b) 𝑓(𝑥) = c) 𝑓(𝑥) = 2𝑥 3 d) 𝑓(𝑥) = 5
𝑥
1.4 In each of the following case find the derivative of 𝑓(𝑥) by 1st principle at the point where 𝑥 = −1
1 2
a) 𝑓(𝑥) = 𝑏 − 𝑥 2 b) 𝑓(𝑥) = 𝑥 2 + 𝑥 − 2 c) 𝑓(𝑥) = − 𝑥 3 d) 𝑓(𝑥) =
2 5𝑥
BY MR M. SHABALALA @NOMBUSO HIGH Page 2
,DIFFERENTIATE THE FOLLOWING BY USING 1ST PRINCIPLE?
a) 𝑓(𝑥) = 𝑥 24 OR b) 𝑓(𝑥) = 𝑥 2 OR c) 𝑓(𝑥) = 3𝑥 2 + 𝑥 12
WOW! it seems as if it is a challenge now if exponent is too big!
So far we have dealt with average gradient (gradient between the two or more points)
Now we will be introduced to the concept of gradient at a point.
DIFFERENTIATION – Is used to find gradient at a specific point
TWO METHODS OF DETERMINING THE DERIVATIVE
( GRADIENT)
DETERMINE THE DERIVATIVE DETERMINE THE DERIVATIVE USING THE
RULES OF DIFFERENTIATION
FROM FIRST PRINCIPLE
1. Derivative of a constant
2. The power rule
′( )
𝒇(𝒙 + 𝒉) − 𝒇(𝒙) 3. Derivative of a function × a constant
𝒇 𝒙 = lim 4. The sum rule
𝒉→𝟎 𝒉 5. The difference rule
2. THE RULES FOR DIFFERENTIATION
THE RULES FOR DIFFERENTIATION - allow you to differentiate functions without
going through the process of differentiating from first principles.
ALWAYS USE RULES TO DIFFERENTIATE UNLESS IF IT STATED TO USE 1ST
PRINCIPLE
RULE 1 RULE 2
DERIVATIVE OF CONSTANT IS 0 THE POWER RULE
f(x) = k → ∴ 𝑓 ′ (𝑥) = 0 𝑓(𝑥) = 𝑥 𝑛 →∴ 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1
Eg. f(x)= 𝑥 6 →∴ 𝑓 ′ (𝑥) = 6𝑥 6−1
EXAMPLE f(x) = 5 → ∴ 𝑓 ′ (𝑥) = 0 ∴ 𝑓 ′ (𝑥) = 6𝑥 5
RULE 3
DERIVATIVE OF FUNCTION MULTIPLIED BY A CONSTANT
𝑓(𝑥) = 𝑘. 𝑔(𝑥) →∴ 𝑓 ′ (𝑥) = 𝑘. 𝑔′ (𝑥)
EXAMPLE F(x) = 5 𝑥 3 →∴ 𝑓 ′ (𝑥) = 5 × 3𝑥 3−1 = 15𝑥 2
RULE 4 RULE 5
THE SUM RULE THE DIFFERENCE RULE
𝑓(𝑥) = 𝑔(𝑥) + ℎ(𝑥) 𝑓(𝑥) = 𝑔(𝑥) − ℎ(𝑥)
→∴ 𝑓 ′ (𝑥) = 𝑔′(𝑥) + ℎ′(𝑥) →∴ 𝑓 ′ (𝑥) = 𝑔(𝑥) − ℎ(𝑥)
EXAMPLE EXAMPLE
𝑓(𝑥) = 4𝑥 2 + 8𝑥 →∴ 𝑓 ′ (𝑥) = 4 × 2𝑥 2−1 + 8 𝑓(𝑥) = 3𝑥 2 − 5𝑥 →∴ 𝑓 ′ (𝑥) = 3 × 2𝑥 2−1 − 5
BY MR M. SHABALALA @NOMBUSO HIGH Page 3
, EXAMPLES OF DIFFERENTIATION USING THE RULES
Before differentiating remove the following:
Brackets
Fractions(no variable(e.g x) must be in denominator)
Surds(√𝑥 remove all radical signs, to exponential form)
Before differentiate,ensure format ( 𝑦 = 𝑎𝑥 𝑛 𝑜𝑟 𝑦 = 𝑎𝑥 −𝑛 ) NB: 𝑦 = 13 𝑥 2 − 2𝑦
𝑑𝑦
f ′(𝑥) , , 𝐷𝑡 it’s a notation one and same thing. 𝑓 ′′(𝑥) it is a second derivative
𝑑𝑥
1) Given that 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 + 5 determine 𝑓′(𝑥).
𝑑𝑦
2. Determine if 𝑦 = (2𝑥 + 1)(𝑥 − 3)
𝑑𝑥
𝑡 2 + 5𝑡 + 6
3) Determine 𝐷𝑡 = [ ]
𝑡+2
𝑥2 3
4) Given that 𝑓 (𝑥 ) = + find 𝑓′(𝑥)
3 𝑥3
𝑑 4
5) Determine [ − √𝑥 3 ]
𝑑𝑥 5𝑥 3
𝑑 3
6) Determine [ √8𝑥 2 + (2𝑥 )−4 ]
𝑑𝑥
𝑡 2 + 5𝑡 + 6 (𝒙 + 𝟐)𝟑
7) Determine 𝐷𝑡 = [ ] or Determine 𝑓′(𝑥) if 𝒇(𝒙) =
𝑡 √𝒙
8) Given that 𝑓 (𝜃 ) = 2𝜃 3 − 3𝜃 2 + 5 find 𝑓′′(𝜃)
𝑑
9) Determine [𝑦𝑥 − 𝑦 = 2𝑥 2 − 2𝑥 ]
𝑑𝑥
1
10) Given: 𝑦 = 𝑎𝑥 2 + 𝑎 or determine 𝑓 ′(−1) if 𝑓(𝑥) = 5𝑥 3 + 2𝑟 − 3𝑥3
Determine:
𝑑𝑦
10.1
𝑑𝑥
𝑑𝑦
10.2
𝑑𝑎
BY MR M. SHABALALA @NOMBUSO HIGH Page 4
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