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CHAPTER 2. DERIVATIVE
Materials :
 Initial Concept of Derivative
 Gradient of Function at a Point
 Operation on Derivative
 Derivative Mind Map
Indicators of Achievement :
1. Students are able to explain the concept of derivative correctly.
2. Students are able to determine the average velocity, instantaneous
velocity, and gradient of a function at a point using the concept of
derivative.
3. Students are able to calculate the derivative and explain the operations
that apply to the derivative of a function.
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 1
SUB-CPMK020301 | Pertemuan 4
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 2
Tangent Line
Tangent line of the curve 𝑦 = 𝑓(π‘₯) at point 𝑃(π‘₯, 𝑓 𝑋 ) is a line that goes through 𝑃
with slope
π‘š = lim
β„Žβ†’0
𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯)
Ξ”π‘₯
= 2.5
2.1 Two Problems One Solution
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 3
Average speed vs instantaneous speed
Suppose π‘Ÿ(𝑑) is the position of an object at time 𝑑, hence average velocity at time
t1 to t2 be :
π‘£π‘Žπ‘£π‘” 𝑑1, 𝑑2 =
π‘Ÿ 𝑑2 βˆ’ π‘Ÿ(𝑑1)
𝑑2 βˆ’ 𝑑1
The smaller the difference between 𝑑1 and 𝑑2, we can get the "instantaneous
speed":
𝑣 𝑑 = lim
Δ𝑑→0
π‘£π‘Žπ‘£π‘” 𝑑, 𝑑 + Δ𝑑 = lim
Δ𝑑→0
π‘Ÿ 𝑑 + Δ𝑑 βˆ’ π‘Ÿ(𝑑)
Δ𝑑
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 4
Definition of Derivatives
A function 𝑓 is said to have a derivative at point π‘₯ if,
lim
Ξ”π‘₯β†’0
𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯)
Ξ”π‘₯
There is and is up.
If the above limit exists, then the derivative of f at point x i.e. fβ€²(x) is equal
to the limit value above. The form of the limit above is equivalent to,
𝑓′ 𝑐 = lim
x→𝑐
𝑓 π‘₯ βˆ’ 𝑓(𝑐)
x βˆ’ c
Here are some notations for derivatives of f at point x,
𝑓′
π‘₯ 𝐷π‘₯𝑓(π‘₯)
𝑑
𝑑π‘₯
𝑓(π‘₯)
𝑑(𝑓 π‘₯ )
𝑑π‘₯
2.2 Derivative
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 5
If f has a derivative at point x=c, then f is continuous at point x=c.
If f is not continuous at point x=c, then f has no derivative at point x=c.
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 6
Constant Function Rules
If 𝑓 π‘₯ = π‘˜, where k is a constant, then for any x,𝑓′
π‘₯ = 0.
Identity function rules
If f x = π‘₯, maka 𝑓′
π‘₯ = 1
Rank Rules
If 𝑓 π‘₯ = π‘₯𝑛
, with n rational numbers, then𝑓′
π‘₯ = 𝑛π‘₯π‘›βˆ’1
Constant Multiplier Rules
π‘˜π‘“ β€² π‘₯ = π‘˜ βˆ™ 𝑓′ π‘₯ , with k a constant
Addition and subtraction rules
𝑓 βˆ™ 𝑔 β€² π‘₯ = 𝑓 π‘₯ 𝑔′ π‘₯ + 𝑓′ π‘₯ 𝑔 π‘₯
Division Rules
𝑓
𝑔
β€²
π‘₯ =
𝑔 π‘₯ 𝑓′
π‘₯ βˆ’ 𝑓 π‘₯ 𝑔′(π‘₯)
𝑔2(π‘₯)
, 𝑔′
π‘₯ β‰  0
2.3 Rules for searching derivative
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 7
Trigonometric Derivatives
The sin x and cos x functions are differentiated across real numbers with
𝐷π‘₯ sin π‘₯ = cos π‘₯
𝐷π‘₯ cos π‘₯ = βˆ’ sin π‘₯
Advanced Trigonometric Derivatives
The derivative of the trigonometric function below can be obtained from the derivative
rules in the previous chapter and utilizes the basic derivative of trigonometry
𝐷π‘₯ tan π‘₯ = sec2
π‘₯
𝐷π‘₯ cot π‘₯ = βˆ’ csc2
π‘₯
𝐷π‘₯(sec π‘₯) = sec π‘₯ tan π‘₯
𝐷π‘₯ csc π‘₯ = βˆ’ csc π‘₯ cot π‘₯
2.4 Derivatives of Trigonometric Functions
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 8
Example, 𝑦 = 𝑓 𝑒 and 𝑒 = 𝑔 π‘₯ .
𝑓 ∘ 𝑔 β€² 𝑐 = 𝑓′ 𝑔 𝑐 βˆ™ 𝑔′ 𝑐
Or
𝑑𝑦
𝑑π‘₯
=
𝑑𝑦
𝑑𝑒
βˆ™
𝑑𝑒
𝑑π‘₯
Example:
Asked to look for derivatives of π‘₯2 + 3π‘₯ 3
𝑦 = 𝑒3, 𝑒 = π‘₯2 + 3π‘₯
So
𝑑𝑦
𝑑π‘₯
= 3 π‘₯2
+ 3π‘₯ 2
βˆ™ 2π‘₯
2.5 Chain Rules
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 9
Derivatives of Order 2 and 3
The second and third derivatives of the function f at point x are respectively
written 𝑓′′
π‘₯ and 𝑓′′′
(π‘₯) where 𝑓′′
(π‘₯) is obtained by deriving 𝑓′
(π‘₯) once and
𝑓′′′
(π‘₯) is obtained by decreasing 𝑓′′
(π‘₯) once again.
Derivatives of Order More than 3
The nth derivative of the function f at point x is written in the form 𝑓 𝑛
(π‘₯).
2.6 Higher Order derivatives
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 10
Implicit derivative y with respect to x.
For example, given the function of two variables f, g, and a curve equation
𝑓 π‘₯, 𝑦 = 𝑔 π‘₯, 𝑦
Can we determine the derivative of y to x?
We can proceed by considering y as a function of x (in other words, 𝑦(π‘₯)), and
deriving both segments of this equation to x. These are referred to as implicit
derivatives.
2.7 Implicit Derivatives
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 11
Example, Suppose we are asked to specify
𝑑𝑦
𝑑π‘₯
from the following equation:
π‘₯𝑦 = π‘₯2
+ 𝑦2
𝑑
𝑑π‘₯
π‘₯𝑦 =
𝑑
𝑑π‘₯
π‘₯2
+ 𝑦2
We apply the multiplication rule to the left field, and the addition rule to the right field
𝑑
𝑑π‘₯
π‘₯ βˆ™ 𝑦 + π‘₯ βˆ™
𝑑
𝑑π‘₯
𝑦 =
𝑑
𝑑π‘₯
π‘₯2 +
𝑑
𝑑π‘₯
𝑦2
Since we think of y as a function of x, we need to apply the chain rule to
𝑑
𝑑π‘₯
𝑦2 .
1 βˆ™ 𝑦 + π‘₯ βˆ™
𝑑𝑦
𝑑π‘₯
= 2π‘₯ + 2𝑦 βˆ™
𝑑𝑦
𝑑π‘₯
π‘₯ βˆ™
𝑑𝑦
𝑑π‘₯
βˆ’ 2𝑦 βˆ™
𝑑𝑦
𝑑π‘₯
= 2π‘₯ βˆ’ 𝑦
𝑑𝑦
𝑑π‘₯
=
2π‘₯ βˆ’ 𝑦
π‘₯ βˆ’ 2𝑦
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 12
Rate-related problem-solving strategies,
In general, the steps that can be done:
1. Create simple modeling
2. Determine variables and the relationship between variables
3. Calculate the implicit derivative
4. Search according to what is asked
2.8 Related Rate
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 13
Approximation
Approximation with differentials is carried out by approaching Δ𝑦 with
𝑑𝑦 = 𝑓′ π‘₯ 𝑑π‘₯
Ξ”y β‰ˆ 𝑑𝑦
𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯) β‰ˆ 𝑓′ π‘₯ 𝑑π‘₯
𝑓 π‘₯ + Ξ”π‘₯ β‰ˆ 𝑓 π‘₯ + 𝑓′
π‘₯ Ξ”x
2.9 Approximation
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 14
Example: Use differential to approximate 4.5 !
Suppose 𝑓 π‘₯ = π‘₯ . We are asked to
approach 4.5 = 𝑓 4 + 0.5
𝑓 4 + 0.5 β‰ˆ 𝑓 4 + 𝑓′(4) βˆ™ 0.5
Then we need to find out first.𝑓′
4 .
𝑓 π‘₯ = π‘₯ = π‘₯
1
2 ⟹ 𝑓′
π‘₯ =
1
2 π‘₯
𝑓′
4 =
1
2.2
=
1
4
So 4.5 β‰ˆ 2 +
1
4
βˆ™ 0.5 = 2.215
We can use this technique to approach the
roots of other things. For example 5 with
𝑑π‘₯ = 1 . But of course, the approximate
results will be less accurate as the 𝑑π‘₯ gets
bigger.
9/26/23
Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 15
Specify the turnan of the following functions:
1. 2π‘₯3 + 3π‘₯
2. 4π‘₯3
+ π‘₯ βˆ’ 5 π‘₯2
+ 2π‘₯
3.
π‘₯2+1
2π‘₯βˆ’1

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kalkulus pertemuan ke 4 tentang derivative.pptx

  • 1. CHAPTER 2. DERIVATIVE Materials :  Initial Concept of Derivative  Gradient of Function at a Point  Operation on Derivative  Derivative Mind Map Indicators of Achievement : 1. Students are able to explain the concept of derivative correctly. 2. Students are able to determine the average velocity, instantaneous velocity, and gradient of a function at a point using the concept of derivative. 3. Students are able to calculate the derivative and explain the operations that apply to the derivative of a function. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 1 SUB-CPMK020301 | Pertemuan 4
  • 2. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 2 Tangent Line Tangent line of the curve 𝑦 = 𝑓(π‘₯) at point 𝑃(π‘₯, 𝑓 𝑋 ) is a line that goes through 𝑃 with slope π‘š = lim β„Žβ†’0 𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯) Ξ”π‘₯ = 2.5 2.1 Two Problems One Solution
  • 3. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 3 Average speed vs instantaneous speed Suppose π‘Ÿ(𝑑) is the position of an object at time 𝑑, hence average velocity at time t1 to t2 be : π‘£π‘Žπ‘£π‘” 𝑑1, 𝑑2 = π‘Ÿ 𝑑2 βˆ’ π‘Ÿ(𝑑1) 𝑑2 βˆ’ 𝑑1 The smaller the difference between 𝑑1 and 𝑑2, we can get the "instantaneous speed": 𝑣 𝑑 = lim Δ𝑑→0 π‘£π‘Žπ‘£π‘” 𝑑, 𝑑 + Δ𝑑 = lim Δ𝑑→0 π‘Ÿ 𝑑 + Δ𝑑 βˆ’ π‘Ÿ(𝑑) Δ𝑑
  • 4. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 4 Definition of Derivatives A function 𝑓 is said to have a derivative at point π‘₯ if, lim Ξ”π‘₯β†’0 𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯) Ξ”π‘₯ There is and is up. If the above limit exists, then the derivative of f at point x i.e. fβ€²(x) is equal to the limit value above. The form of the limit above is equivalent to, 𝑓′ 𝑐 = lim x→𝑐 𝑓 π‘₯ βˆ’ 𝑓(𝑐) x βˆ’ c Here are some notations for derivatives of f at point x, 𝑓′ π‘₯ 𝐷π‘₯𝑓(π‘₯) 𝑑 𝑑π‘₯ 𝑓(π‘₯) 𝑑(𝑓 π‘₯ ) 𝑑π‘₯ 2.2 Derivative
  • 5. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 5 If f has a derivative at point x=c, then f is continuous at point x=c. If f is not continuous at point x=c, then f has no derivative at point x=c.
  • 6. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 6 Constant Function Rules If 𝑓 π‘₯ = π‘˜, where k is a constant, then for any x,𝑓′ π‘₯ = 0. Identity function rules If f x = π‘₯, maka 𝑓′ π‘₯ = 1 Rank Rules If 𝑓 π‘₯ = π‘₯𝑛 , with n rational numbers, then𝑓′ π‘₯ = 𝑛π‘₯π‘›βˆ’1 Constant Multiplier Rules π‘˜π‘“ β€² π‘₯ = π‘˜ βˆ™ 𝑓′ π‘₯ , with k a constant Addition and subtraction rules 𝑓 βˆ™ 𝑔 β€² π‘₯ = 𝑓 π‘₯ 𝑔′ π‘₯ + 𝑓′ π‘₯ 𝑔 π‘₯ Division Rules 𝑓 𝑔 β€² π‘₯ = 𝑔 π‘₯ 𝑓′ π‘₯ βˆ’ 𝑓 π‘₯ 𝑔′(π‘₯) 𝑔2(π‘₯) , 𝑔′ π‘₯ β‰  0 2.3 Rules for searching derivative
  • 7. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 7 Trigonometric Derivatives The sin x and cos x functions are differentiated across real numbers with 𝐷π‘₯ sin π‘₯ = cos π‘₯ 𝐷π‘₯ cos π‘₯ = βˆ’ sin π‘₯ Advanced Trigonometric Derivatives The derivative of the trigonometric function below can be obtained from the derivative rules in the previous chapter and utilizes the basic derivative of trigonometry 𝐷π‘₯ tan π‘₯ = sec2 π‘₯ 𝐷π‘₯ cot π‘₯ = βˆ’ csc2 π‘₯ 𝐷π‘₯(sec π‘₯) = sec π‘₯ tan π‘₯ 𝐷π‘₯ csc π‘₯ = βˆ’ csc π‘₯ cot π‘₯ 2.4 Derivatives of Trigonometric Functions
  • 8. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 8 Example, 𝑦 = 𝑓 𝑒 and 𝑒 = 𝑔 π‘₯ . 𝑓 ∘ 𝑔 β€² 𝑐 = 𝑓′ 𝑔 𝑐 βˆ™ 𝑔′ 𝑐 Or 𝑑𝑦 𝑑π‘₯ = 𝑑𝑦 𝑑𝑒 βˆ™ 𝑑𝑒 𝑑π‘₯ Example: Asked to look for derivatives of π‘₯2 + 3π‘₯ 3 𝑦 = 𝑒3, 𝑒 = π‘₯2 + 3π‘₯ So 𝑑𝑦 𝑑π‘₯ = 3 π‘₯2 + 3π‘₯ 2 βˆ™ 2π‘₯ 2.5 Chain Rules
  • 9. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 9 Derivatives of Order 2 and 3 The second and third derivatives of the function f at point x are respectively written 𝑓′′ π‘₯ and 𝑓′′′ (π‘₯) where 𝑓′′ (π‘₯) is obtained by deriving 𝑓′ (π‘₯) once and 𝑓′′′ (π‘₯) is obtained by decreasing 𝑓′′ (π‘₯) once again. Derivatives of Order More than 3 The nth derivative of the function f at point x is written in the form 𝑓 𝑛 (π‘₯). 2.6 Higher Order derivatives
  • 10. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 10 Implicit derivative y with respect to x. For example, given the function of two variables f, g, and a curve equation 𝑓 π‘₯, 𝑦 = 𝑔 π‘₯, 𝑦 Can we determine the derivative of y to x? We can proceed by considering y as a function of x (in other words, 𝑦(π‘₯)), and deriving both segments of this equation to x. These are referred to as implicit derivatives. 2.7 Implicit Derivatives
  • 11. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 11 Example, Suppose we are asked to specify 𝑑𝑦 𝑑π‘₯ from the following equation: π‘₯𝑦 = π‘₯2 + 𝑦2 𝑑 𝑑π‘₯ π‘₯𝑦 = 𝑑 𝑑π‘₯ π‘₯2 + 𝑦2 We apply the multiplication rule to the left field, and the addition rule to the right field 𝑑 𝑑π‘₯ π‘₯ βˆ™ 𝑦 + π‘₯ βˆ™ 𝑑 𝑑π‘₯ 𝑦 = 𝑑 𝑑π‘₯ π‘₯2 + 𝑑 𝑑π‘₯ 𝑦2 Since we think of y as a function of x, we need to apply the chain rule to 𝑑 𝑑π‘₯ 𝑦2 . 1 βˆ™ 𝑦 + π‘₯ βˆ™ 𝑑𝑦 𝑑π‘₯ = 2π‘₯ + 2𝑦 βˆ™ 𝑑𝑦 𝑑π‘₯ π‘₯ βˆ™ 𝑑𝑦 𝑑π‘₯ βˆ’ 2𝑦 βˆ™ 𝑑𝑦 𝑑π‘₯ = 2π‘₯ βˆ’ 𝑦 𝑑𝑦 𝑑π‘₯ = 2π‘₯ βˆ’ 𝑦 π‘₯ βˆ’ 2𝑦
  • 12. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 12 Rate-related problem-solving strategies, In general, the steps that can be done: 1. Create simple modeling 2. Determine variables and the relationship between variables 3. Calculate the implicit derivative 4. Search according to what is asked 2.8 Related Rate
  • 13. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 13 Approximation Approximation with differentials is carried out by approaching Δ𝑦 with 𝑑𝑦 = 𝑓′ π‘₯ 𝑑π‘₯ Ξ”y β‰ˆ 𝑑𝑦 𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓(π‘₯) β‰ˆ 𝑓′ π‘₯ 𝑑π‘₯ 𝑓 π‘₯ + Ξ”π‘₯ β‰ˆ 𝑓 π‘₯ + 𝑓′ π‘₯ Ξ”x 2.9 Approximation
  • 14. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 14 Example: Use differential to approximate 4.5 ! Suppose 𝑓 π‘₯ = π‘₯ . We are asked to approach 4.5 = 𝑓 4 + 0.5 𝑓 4 + 0.5 β‰ˆ 𝑓 4 + 𝑓′(4) βˆ™ 0.5 Then we need to find out first.𝑓′ 4 . 𝑓 π‘₯ = π‘₯ = π‘₯ 1 2 ⟹ 𝑓′ π‘₯ = 1 2 π‘₯ 𝑓′ 4 = 1 2.2 = 1 4 So 4.5 β‰ˆ 2 + 1 4 βˆ™ 0.5 = 2.215 We can use this technique to approach the roots of other things. For example 5 with 𝑑π‘₯ = 1 . But of course, the approximate results will be less accurate as the 𝑑π‘₯ gets bigger.
  • 15. 9/26/23 Informatics Enginnering | Calculus 1C | Rima Aulia Rahayu, S.Si,. M.Si. 15 Specify the turnan of the following functions: 1. 2π‘₯3 + 3π‘₯ 2. 4π‘₯3 + π‘₯ βˆ’ 5 π‘₯2 + 2π‘₯ 3. π‘₯2+1 2π‘₯βˆ’1