This explains the theorems of differentiation and Chain rule.

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