LESSON 7: CALCULUS
let's understand Basic Differentiation and Integration!
(a) DIFFERENTIATION
We know that the gradient of a curve changes at each and every point. We obtain this
gradient by getting the gradient of the tangent o the curve at the said point.
For Instant: (To be Illustrated during class)
Examples
a) Find the gradient of the given curves at the said points
i. 𝑦 = 𝑥 2 − 3𝑥, 𝑎𝑡 (2, −2).
ii. 𝑦 = 𝑥 5 − 7𝑥 + 3, 𝑎𝑡 (0, 3).
iii. 𝑦 = 𝑥 2 − 4, 𝑎𝑡 (−4, −12).
iv. 𝑦 = −𝑥 2 − 𝑥, 𝑎𝑡 (0, 0).
v. 𝑦 = 𝑥 2 − 𝑥, 𝑎𝑡 (0, 0).
b) At what point is the gradient of
i. 𝑦 = 𝑥 2 + 3𝑥 + 2 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 9?.
2
ii. 𝑦 = 3 𝑥 3 + 𝑥 2 , 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 4?
iii.
1 4 5
𝑦= 𝑥 − 2𝑥 3 + 𝑥 2 , 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 0
4 2
NOTE:
There are many techniques of getting this gradient function depending on the nature of
the curve.
Here, we shall discuss but a few:
1. POWER RULE
In this section we look at derivative of the function 𝒇(𝒙) = 𝒙𝒏 .
𝑑𝑓
Let 𝑓(𝑥) = 𝑥 𝑛 . The derivative of 𝑓(𝑥) with respect to 𝑥 denoted by 𝑑𝑥
𝑜𝑟 𝑓′(𝑥) is defined as
𝒅𝒇
= 𝒏𝒙𝒏−𝟏 .
𝒅𝒙
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Example
1. Find the derivative of 𝑓(𝑥) = 𝑥 2 .
2. Find the derivative of 𝑓(𝑥) = 2𝑥 3 .
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3. Find the derivative of 𝑓(𝑥) = 𝑥 2 .
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4. Find 𝑓 ′ (𝑥) given 𝑓 (𝑥 ) = 3
𝑥
1
5. If 𝑓(𝑥) = −𝑥 5 + 𝑥 3 , find 𝑓 ′ (𝑥).
Solution
𝑑𝑓
= 𝑛𝑥 𝑛−1
𝑑𝑥
2. Product Rule
If 𝒉(𝒙) = 𝒇(𝒙)𝒈(𝒙), then 𝒉′ (𝒙) = 𝒇′(𝒙) 𝒈(𝒙) + 𝒇(𝒙)𝒈′(𝒙) .
Example. Find the derivative of the function 𝑓(𝑥) = 𝑥 2 (2𝑥 + 1)
Solution
Let 𝑔(𝑥) = 𝑥 2 and ℎ(𝑥) = 2𝑥 + 1 = 2𝑥 + 1𝑥 0 ,
𝑔′ (𝑥) = 2𝑥 2−1 = 2𝑥 and ℎ′ (𝑥) = 2(1)𝑥 1−1 + 1(0)𝑥 0−1 = 2𝑥 0 + 0 = 2.
𝑑𝑓
= 𝑓 ′ (𝑥) = 𝑔′(𝑥) ℎ(𝑥) + 𝑔(𝑥)ℎ′(𝑥)
𝑑𝑥
= 2𝑥(2𝑥 + 1) + 𝑥 2 (2)
= 4𝑥 2 + 2𝑥 + 2𝑥 2
= 6𝑥 2 + 2𝑥
Example
Find the derivative of 𝑦 = 𝑥 4 (𝑥 2 + 2𝑥 + 3)
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3. Quotient Rule
𝒇(𝒙)
In this section, we look at derivatives of the form .
𝒈(𝒙)
𝑓(𝑥)
Quotient rule states that, if the function is of the form 𝑢(𝑥) = ,
𝑔(𝑥)
𝒇′ (𝒙)𝒈(𝒙)−𝒈′(𝒙)𝒇(𝒙)
the derivative of the function is 𝒖′ (𝒙) = .
[𝒈(𝒙)]𝟐
Example
𝟐𝒙
Find the derivative of the function 𝒖(𝒙) = .
𝒙𝟐
𝒙+𝟐
Find the derivative of the function .
𝒙−𝟑
4. Chain Rule
A function of a function is known as a composite function. In this lecture, we learn on how
to find derivatives of composite functions. The rule for differentiating a composite
function is known as the ‘’chain rule”.
A derivative gives the rate of change.
𝑑𝑦 𝑑𝑥
If 𝑑𝑥
= 2 and 𝑑𝑡
= 3. We may say ‘y changes 2 times as fast as x and x changes 3 times as
fast as t’. We conclude that ‘y changes 2 × 3 = 6 times as fast as t.’
𝒅𝒚 𝒅𝒚 𝒅𝒙
This implies = × .
𝒅𝒙 𝒅𝒙 𝒅𝒕
The Chain Rule
𝑑𝑦 𝑑𝑦 𝑑𝑢
Let 𝑦 = 𝑓(𝑢) and 𝑢 = 𝑔(𝑥), then 𝑑𝑥
= 𝑑𝑢 × 𝑑𝑥 .
1. Find the derivative of the function 𝑦 = (2𝑥 2 − 1)3 .
𝑑𝑦
2. Find if 𝑦 = (𝑥 2 + 1)100.
𝑑𝑥
3. Differentiate 𝑦 = √𝑥 3 + 2𝑥 + 1
𝑑𝑦
4. Find 𝑑𝑡
if 𝑦 = 𝑥 2 and 𝑥 = 3𝑡 2 + 𝑡 + 1.
Note: The derivatives of two mutually inverse functions are the reciprocals of each other.
𝒅𝒚 𝟏 𝒅𝒙 𝟏
= 𝒅𝒙 and = 𝒅𝒚 .
𝒅𝒙 𝒅𝒚
𝒅𝒚 𝒅𝒙
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Example
𝒅𝒚
5. Find if
𝒅𝒙
i. 𝒚 = 𝒕𝟐 − 𝟏 and 𝒙 = 𝟐𝒕 + 𝟑.
ii. 𝒙 = 𝟑𝒕 𝒂𝒏𝒅 𝒚 = 𝒕𝟑 + 𝟒𝒕𝟐 − 𝟑𝒕
iii. 𝒙 = 𝒕𝟐 𝒂𝒏𝒅 𝒚 = 𝟓 − 𝟐𝒕.
iv. 𝒙 = 𝒕𝟑 𝒂𝒏𝒅 𝒚 = 𝒕𝟐 .
v. 𝒙 = 𝟐𝒕𝟑 𝒂𝒏𝒅 𝒚 = 𝒕𝟐 − 𝟓𝒕.
𝟏 𝟏
vi. 𝒙 = 𝒕 + , 𝒂𝒏𝒅 𝒚 = 𝒕 −
𝒕 𝒕
More Examples and EXERCIZES
A. DIFFERENTIATION
I. Determine the derivative of the following polynomial functions with
respect to 𝒙: 𝒚 = (𝟒𝒙𝟐 − 𝟓𝒙𝟑 )(𝟔𝒙 + 𝟒)
II. Integrate each of the following polynomial functions with respect to 𝒙
(i) 𝒚 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 − 𝟗𝒙 + 𝟔
3 − 2x 2
(ii) y= (6 marks)
3 + 2x
III. Determine the derivative of polynomial functions with respect to 𝒙
(i) 𝒚 = (𝟒𝒙𝟐 − 𝟓𝒙𝟑 )(𝟔𝒙 + 𝟒) (4 marks)
𝒙𝟑 −𝟐𝒙−𝟐
(ii) y = (4 marks)
𝒙
IV. Find the derivative of the function using various rules. (8 Marks)
i. 𝒇(𝒙) = 𝒙𝟐 + 𝟐𝒙
ii. 𝒇(𝒙) = (−𝟑𝒙𝟑 + 𝒙𝟐 + 𝟒)
𝒙𝟐 +𝟑𝒙+𝟏
iii. 𝒇 (𝒙 ) =
𝒙+𝟏
V. Find the first derivatives of the following function:
i. 𝒇 (𝒙) = 𝒙−𝟐 + 𝟓𝒙 + 𝟏 (2 Marks)
ii. 𝒚 = (𝒙𝟑 + 𝟏)(𝒙𝟐 + 𝟐𝒙 − 𝟑) (3 Marks)
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VI. Determine the derivative of each of the following polynomial functions with
respect to 𝑥
(i) 𝒚 = (𝟒𝒙𝟐 − 𝟓𝒙𝟑 )(𝟔𝒙 + 𝟒) (3 Marks)
(ii) 𝒚 = 𝟓𝒙𝟒 − 𝟐𝒙−𝟑 + 𝟓 (3 marks)
(iii) 𝒚 = 𝒙(𝟓𝒙 − 𝟐) + 𝟑(𝒙 + 𝟐)𝟐 .
(𝒙−𝟓)𝟐
(iv) 𝒚=
𝟓𝒙
(v) 𝒚=𝒙 −𝟐 (𝒙
− 𝒙𝟑 ).
𝟔𝒙𝟑 +𝟏𝟒𝒙𝟐 −𝟏𝟐𝒙
(vi) 𝒚=
𝟑𝒙−𝟐
(vii) 𝒚 = (𝒙 − 𝟑)(𝒙 + 𝟏).
(viii) 𝒚 = (𝒙𝟐 + 𝟏)(𝟐𝒙 − 𝟏)
VII. Determine the derivative of each of the following polynomial functions with
respect to 𝑥
𝟑−𝟐𝒙𝟐
𝒚= (4 marks)
𝟑+𝟐𝒙
(b) INTERGRATION
Definitions
Given a function, f (x), an anti-derivative of f (x) is any function F (x) such that F(x) = f ′(
x) If F (x) is any anti-derivative of f (x) then the most general anti-derivative of f (x) is
called an indefinite integral and denoted, ∫f (x) dx = F(x) + C, C is any constant In this
definition the ∫ is called the integral symbol, f (x) is called the integrand, x is called the
integration variable and the “c” is called the constant of integration
OR, to put it simply, Integration is the Reverse of Differentiation. It is the process of
getting a function from its gradient function.
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The Rule Follows That,
𝒅𝒚 𝒙𝒏+𝟏
Given = 𝒙𝒏 , 𝒕𝒉𝒆𝒏 𝒚 = + 𝑪, 𝒘𝒉𝒆𝒓𝒆 𝒏 ≠ −𝟏 𝒂𝒏𝒅 𝑪 𝒊𝒔 𝒂𝒏𝒚 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕.
𝒅𝒙 𝒏+𝟏
Examples and EXERCIZES
B. INTEGRATION
VIII. Integrate each of the following polynomial functions with respect to 𝑥
(iii) 𝑦 = 5𝑥 3 − 5𝑥 2 − 9𝑥 + 6
(iv) 𝑦 = 2𝑥 2 − 5𝑥 3 + 7𝑥 − 3 (6 marks)
IX. Integrate the following functions
i) ∫(3𝑥 2 + 2𝑥 + 2)𝑑𝑥 (3 Marks)
ii) ∫(𝑥2 + 1)(2𝑥 + 4)𝑑𝑥
X. Determine the area bounded by the curves
𝑦 = 𝑥 and 𝑦 = 𝑥2 for 0 ≤ 𝑥 ≤ 2.
XI. Evaluate the following integrals;
i. ∫(2𝑥 3 + 7𝑥 − 8)𝑑𝑥 (2 Marks)
3
ii. ∫−1(5𝑥 4 − 6𝑥 2 − 4𝑥)𝑑𝑥 (3 Marks)
XII. Evaluate the following integrals;
iii. ∫(𝒙𝟑 + 𝟐𝒙 − 𝟏)𝒅𝒙 (2 Marks)
𝟐
iv. ∫−𝟏(𝟐𝒙𝟒 − 𝒙𝟐 + 𝟓)𝒅𝒙 (3 Marks)
XIII. Integrate the following functions
iii) ∫(𝟔𝒙𝟐 + 𝟐𝒙 + 𝟐)𝒅𝒙 (3 Marks)
iv) ∫(𝟑𝒙𝟐 + 𝒙)(𝟐𝒙 + 𝟒)𝒅𝒙
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XIV. Integrate each of the following polynomial functions with respect to 𝒙
(v) 𝒚 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 − 𝟗𝒙 + 𝟔
(vi) ∫ 𝟑𝒙(𝒙𝟐 + 𝟐𝒙 − 𝟐)𝒅𝒙 (4 marks)
More for You
i.
𝟏
∫(𝟑 𝒙𝟑 − 𝒙 − 𝟖)𝒅𝒙 (2 Marks)
ii.
𝟐
∫(𝟑 𝒙𝟑 + 𝟐𝒙 − 𝟑𝒙𝟐 + 𝟐)𝒅𝒙
𝟏
𝟏
iii. ∫( 𝒙 + 𝟒𝒙 + 𝟕𝒙𝟐 − 𝟖)𝒅𝒙
𝟒 𝟐
𝟐
iv. ∫(𝟏 − 𝟒𝒙)(𝟏 + 𝟒𝒙)𝒅𝒙
v.
𝟑
∫−𝟏(𝟓𝒙𝟒 − 𝟔𝒙𝟐 − 𝟒𝒙)𝒅𝒙 (3 Marks)
vi.
𝟏
∫−𝟏(𝟑𝒙𝟒 − 𝟐𝒙𝟐 − 𝟒)𝒅𝒙
vii. ∫𝟎 (𝟓𝒙𝟑 − 𝟔𝒙𝟐 − 𝟒𝒙)𝒅𝒙
𝟐
viii. ∫−𝟏(𝒙𝟑 − 𝟔𝒙𝟐 − 𝒙)𝒅𝒙
𝟏
ix. ∫−𝟐(𝟓𝒕𝟒 − 𝟔𝒕𝟐 − 𝟐𝒕)𝒅𝒕
𝟎
x.
𝟓
∫𝟐 (𝟓𝒙𝟒 − 𝟑𝒙𝟐 − 𝟔𝒙)𝒅𝒙
xi. ∫−𝟏(𝟏𝟎𝒙𝟒 − 𝟔𝒙𝟐 − 𝟖𝒙)𝒅𝒙
𝟔
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