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AP Calculus Chapter 3 Section 3

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Presentation on theme: "AP Calculus Chapter 3 Section 3"β€” Presentation transcript:

1 AP Calculus Chapter 3 Section 3
Rules for Differentiation

2 Rule 1: Derivative of a Constant
The first rule of differentiation is that the derivative of every constant function is the zero of the function. If f(x) = c, then Proof:

3 Rule 2: Power Rule for Positive Integer Powers of x
If n is a positive integer, then The proof is on page 116.

4 Rule 2: Examples

5 Rule 3: Constant Multiple Rule
If u is a differentiable function of x and c is a constant, then Example:

6 Rule 4: Sum & Difference Rule
If u and v are differentiable functions of x, then their sum and difference are differentiable at every point where u and v are differentiable. At such points, Example: This is a very powerful rule because it allows us to find the derivatives of polynomials.

7 Rule 5: Product Rule The product of 2 differentiable functions u and v is differentiable, and β€œthe first times the derivative of the second plus the second times the derivative of the first” Note: the order you do the differentiation does not matter because we are adding the two together.

8 Product Rule Example Find the derivative of 𝑓 π‘₯ =( π‘₯ 2 +1)( π‘₯ 3 +3)
Solution: Let 𝑒= π‘₯ & 𝑣= π‘₯ 3 +3 , then 𝒅 𝒅𝒙 ( 𝒙 𝟐 +𝟏)( 𝒙 πŸ‘ +πŸ‘) = 𝒙 𝟐 +𝟏 πŸ‘ 𝒙 𝟐 + 𝒙 πŸ‘ +πŸ‘ πŸπ’™ =πŸ‘ 𝒙 πŸ’ +πŸ‘ 𝒙 𝟐 +𝟐 𝒙 πŸ’ +πŸ”π’™ =πŸ“ 𝒙 πŸ’ +πŸ‘ 𝒙 𝟐 +πŸ”π’™

9 Rule 6: Quotient Rule At a point where v β‰  0, the quotient y = u/v of 2 differentiable functions is differentiable, and β€œthe bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared”

10 Quotient Rule Example 𝑓 β€² π‘₯ = π‘₯ 2 +1 2π‘₯ βˆ’( π‘₯ 2 βˆ’1)(2π‘₯) ( π‘₯ 2 +1) 2
Differentiate 𝑓 π‘₯ = π‘₯ 2 βˆ’1 π‘₯ 2 +1 Solution: Let 𝑒= π‘₯ 2 βˆ’1 & 𝑣= π‘₯ 2 +1, π‘‘β„Žπ‘’π‘› 𝑓 β€² π‘₯ = π‘₯ π‘₯ βˆ’( π‘₯ 2 βˆ’1)(2π‘₯) ( π‘₯ 2 +1) 2 = 2 π‘₯ 3 +2π‘₯ βˆ’(2 π‘₯ 3 βˆ’2π‘₯) ( π‘₯ 2 +1) 2 = 4π‘₯ ( π‘₯ 2 +1) 1

11 Rule 7: Power Rule for Negative Integer Powers of x
If n is a negative integer and x β‰  0, then We will use this rule after we discuss the Chain Rule. This rule can be used to solve for a derivative in place of the Quotient Rule because we can redefine the denominator as a negative integer power of x.

12 Second & Higher Order Derivatives
We can find the derivative of the derivative, which is called the second derivative. Denoted: We can also find the third, fourth, fifth, and nth derivatives. Denoted:

13 Example Find the first four derivatives of y = x3 – 5x2 + 2

14 Example Find the first four derivatives of y = x3 – 5x2 + 2

15 Finding Instantaneous Rate of Change
An orange farmer currently has 200 trees yielding an average of 15 bushels of oranges per tree. She is expanding her farm at the rate of 15 trees per year, while improved husbandry is improving her average annual yield by 1.2 bushels per tree. What is the current (instantaneous) rate of increase of her total annual production of oranges?

16 Solution Let the functions t and y be defined as follows:
𝑑 π‘₯ =π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘‘π‘Ÿπ‘’π‘’π‘  π‘₯ π‘¦π‘’π‘Žπ‘Ÿπ‘  π‘“π‘Ÿπ‘œπ‘š π‘›π‘œπ‘€. 𝑦 π‘₯ =𝑦𝑖𝑒𝑙𝑑 π‘π‘’π‘Ÿ π‘‘π‘Ÿπ‘’π‘’ π‘₯ π‘¦π‘’π‘Žπ‘Ÿπ‘  π‘“π‘Ÿπ‘œπ‘š π‘›π‘œπ‘€. Then, 𝑝 π‘₯ =𝑑 π‘₯ 𝑦 π‘₯ 𝑖𝑠 π‘‘β„Žπ‘’ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘œπ‘Ÿπ‘Žπ‘›π‘”π‘’π‘  𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿ π‘₯. We know: 𝑑 0 =200, 𝑑 β€² π‘₯ =15;𝑦 0 =15, 𝑦 β€² 0 =1.2 We need to find 𝑝 β€² 0 , π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑝=𝑑𝑦. Apply Product Rule: 𝑝 β€² π‘₯ =𝑑 0 𝑦 β€² 0 +𝑦(0) 𝑑 β€² (0) = (15)(15) =465 π‘π‘’π‘ β„Žπ‘’π‘™π‘  π‘π‘’π‘Ÿ π‘¦π‘’π‘Žπ‘Ÿ

17 Homework # 3 – 52 by multiples of 3 on pages


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