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Differentiation
(DIFFERENTIATION OF ALGEBRAIC EXPRESSION,
THEOREMS OF DIFFERENTIATION,
CHAIN RULE)
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.
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.
Ave speed= 100m/9.58s
=10.43 m/s
What is the speed of Bolt at 4 s?
Derivative comes in!
INTRODUCTION
• Consider the graph of a function 𝑓 and a point 𝑃 = (𝑎, 𝑓 𝑎 ) on the graph.
Tangent line
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
.
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
.
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
• 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“.
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 ∎
POWER RULE
POWER RULE
THEOREMS ON DIFFERENTIATION
THEOREMS ON DIFFERENTIATION
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 ∎
and
Differentiation

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Differentiation

  • 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!
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  • 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“.
  • 12.
  • 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 ∎
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  • 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 ∎
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  • 37. and