This document summarizes differentiation and the concept of the derivative. It defines secant and tangent lines, and explains that the derivative at a point equals the slope of the tangent line at that point. It provides the definition of the derivative as the limit of the difference quotient. Examples are given of finding derivatives using this limit definition and applying power rule theorems of differentiation. The document also demonstrates using the product rule to take the derivative of a sum of terms.

1. Differentiation
(DIFFERENTIATION OF ALGEBRAIC EXPRESSION,
THEOREMS OF DIFFERENTIATION,
CHAIN RULE)

2. REVIEW
• Consider the graph below:
A secant line is a straight line
joining two points on a
function.
A tangent line is a straight
line that touches a function at
only one point.
Secant Line. It is also
equivalent to the average rate
of change, or simply the slope
between two points.

3. REVIEW
• .
• The tangent line represents the
instantaneous rate of change of the
function at that one point.
• The slope of the tangent line at a point
on the function is equal to the derivative
of the function at the same point.

4. Ave speed= 100m/9.58s
=10.43 m/s
What is the speed of Bolt at 4 s?
Derivative comes in!

7. INTRODUCTION
• Consider the graph of a function 𝑓 and a point 𝑃 = (𝑎, 𝑓 𝑎 ) on the graph.
Tangent line

8. INTRODUCTION
• Let 𝑡 be an arbitrary nonzero real number, and consider the point 𝑄 𝑡 = 𝑎 + 𝑡, 𝑓 𝑎 + 𝑡 ,
which, together with 𝑃 = 𝑎, 𝑓 𝑎 , lies on the graph of 𝑓.
The slope of the secant line 𝐿1 containing 𝑃
and 𝑄 𝑡 is equal to
𝑚 𝑃, 𝑄 𝑡 =
𝑓 𝑎 + 𝑡 − 𝑓(𝑎)
𝑎 + 𝑡 − 𝑎
,
𝑚 𝑃, 𝑄 𝑡 =
𝑓 𝑎 + 𝑡 − 𝑓(𝑎)
𝑡
.
We would like to define the tangent line 𝐿 to
be the limit, as 𝑡 approaches zero, of the
line 𝐿𝑡.
Hence we can express the limit of the slope
of 𝐿𝑡 which is
𝑚 𝑃, 𝑄 𝑡 = lim
𝑡→𝑎
𝑓 𝑎+𝑡 −𝑓 𝑎
𝑡
.

9. TANGENT LINE
• An arbitrary real-valued function 𝑓 of a real variable is differentiable at a number 𝑎 in its
domain if
lim
𝑡→𝑎
𝑓 𝑎 + 𝑡 − 𝑓(𝑎)
𝑡
exist (i.e.., is finite).
The derivative of 𝑓 at 𝑎, denoted 𝑓′
𝑎 , is this limit. Thus
𝒇′ 𝒂 = 𝒍𝒊𝒎
𝒕→𝒂
𝒇 𝒂+𝒕 −𝒇(𝒂)
𝒕
.

10. 𝑓′
𝑥 = lim
𝑡→0
2𝑥𝑡 + 𝑡2
𝑡
𝑓′ 𝑥 = lim
𝑡→0
𝑡(2𝑥 + 𝑡
𝑡
𝑓′ 𝑥 = lim
𝑡→0
2𝑥 + 𝑡
𝑓′ 𝑥 = 2𝑥 ∎
The derivative of 𝑓 𝑥 = 𝑥2 is 𝑓′ 𝑥 = 2𝑥 .
Example
1. Find the derivative of 𝑓 𝑥 = 𝑥2.
Solution:
Write the derivative formula
𝑓′ 𝑥 = lim
𝑡→0
𝑓(𝑥 + 𝑡)2
−𝑓(𝑥)
𝑡
Plug the function to the definition of
derivative. and do some algebra operations
𝑓′ 𝑥 = lim
𝑡→0
(𝑥 + 𝑡)2−𝑥2
𝑡
𝑓′ 𝑥 = lim
𝑡→0
𝑥2 + 2𝑥𝑡 + 𝑡2 − 𝑥2
𝑡

11. • Some other books are using Δ for the formula of slope (
𝛥𝑦
𝛥𝑥
).
Here we will be using
𝑑
𝑑𝑥
, which reads as “ the derivative of” and 𝑓′(𝑥) which reads as
“f prime of x”.
Example,
𝑑
𝑑𝑥
𝑥2 = 2𝑥.
"The derivative of x2 equals 2x"
or simply "d dx of x2 equals 2x“.

13. Solution:
Let h be the arbitrary nonzero number
𝑓′ 𝑥 = lim
ℎ→0
𝑓(𝑥 + ℎ)2
−𝑓(𝑥)
ℎ
𝑓′
𝑥 = lim
ℎ→0
2(𝑥 + ℎ)2
−16 𝑥 + ℎ + 35 − (2𝑥2
− 16𝑥 + 35)
ℎ
𝑓′
𝑥 = lim
ℎ→0
2(𝑥2
+2𝑥ℎ + ℎ2
) − 16 𝑥 + ℎ + 35 − (2𝑥2
− 16𝑥 + 35)
ℎ
𝑓′
𝑥 = lim
ℎ→0
2𝑥2 + 4𝑥ℎ + 2ℎ2 − 16𝑥 − 16ℎ + 35 − 2𝑥2 + 16𝑥 − 35)
ℎ
𝑓′ 𝑥 = lim
ℎ→0
2𝑥2
− 2𝑥2
+ 4𝑥ℎ + 2ℎ2
− 16𝑥 + 16𝑥 − 16ℎ + 35 − 35)
ℎ
𝑓′ 𝑥 = lim
ℎ→0
4𝑥ℎ + 2ℎ2
− 16ℎ
ℎ
𝑓′ 𝑥 = lim
ℎ→0
ℎ(4𝑥 + 2ℎ − 16)
ℎ
𝑓′ 𝑥 = lim
ℎ→0
4𝑥 + 2ℎ − 16
𝑓′
𝑥 = 4𝑥 − 16 ∎

15. POWER RULE

16. POWER RULE

22. THEOREMS ON DIFFERENTIATION

25. THEOREMS ON DIFFERENTIATION

28. Calculate
𝑑𝑦
𝑑𝑥
2x3 − 4x2 3x5 + x2 .
You can solve using power rule or using the product rule.
Using Product Rule:
=
𝑑
𝑑𝑥
2x3 − 4x2 3x5 + x2 + 2x3 − 4x2 [
𝑑
𝑑𝑥
3x5 + x2 ]
= (6𝑥2 − 8𝑥) 3x5 + x2 + 2x3 − 4x2 15x4 + 2x
= 18𝑥7
+ 6𝑥4
− 24𝑥6
− 8𝑥3
+ 30𝑥7
+ 4𝑥4
− 60𝑥6
− 8𝑥3
= 48𝑥7 − 84𝑥6 + 8𝑥4 − 16𝑥3 ∎
Using Power Rule:
=
𝑑
𝑑𝑥
(6𝑥8 + 2𝑥5 − 12𝑥7 − 4𝑥4
=
𝑑
𝑑𝑥
6𝑥8 +
𝑑
𝑑𝑥
2𝑥5 −
𝑑
𝑑𝑥
12𝑥7 −
𝑑
𝑑𝑥
4𝑥4
= 48𝑥7 − 84𝑥6 + 8𝑥4 − 16𝑥3 ∎

37. and