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Section 2.5 Day 1 AP Calculus.

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Presentation on theme: "Section 2.5 Day 1 AP Calculus."β€” Presentation transcript:

1 Section 2.5 Day 1 AP Calculus

2 Learning Targets Define implicit and explicit functions
Evaluate derivatives using Implicit Differentiation

3 Explicit Form Everything is solved for one variable Ex: 𝑦=3π‘₯+1

4 Implicit Form There is a mix of variables on the same side Ex: π‘₯𝑦=1

5 Differentiating with Respect to x
Ex: 𝑑 𝑑π‘₯ ( π‘₯ 3 ) = 3 π‘₯ 2 𝑑π‘₯ 𝑑π‘₯ = 3 π‘₯ 2 Ex: 𝑑 𝑑π‘₯ ( 𝑦 3 ) = 3 𝑦 2 𝑑𝑦 𝑑π‘₯

6 Implicit Differentiation Steps
1. Take the derivative of each piece 2. Put 𝑑𝑦 𝑑π‘₯ after the derivative of β€œy” terms 3. Split 𝑑𝑦 𝑑π‘₯ terms away from other terms using the equal sign 4. Factor 𝑑𝑦 𝑑π‘₯ 5. Divide

7 Example 1 𝑑 𝑑π‘₯ [π‘₯ 𝑦 2 ] π‘₯ 2𝑦 𝑑𝑦 𝑑π‘₯ + 𝑦 2 2π‘₯𝑦 𝑑 𝑑π‘₯ + 𝑦 2

8 Example 2 Find the derivative of 𝑦 3 βˆ’ 𝑦 2 βˆ’5𝑦 βˆ’ π‘₯ 2 =βˆ’4
1. 3 𝑦 2 𝑑𝑦 𝑑π‘₯ βˆ’2𝑦 𝑑𝑦 𝑑π‘₯ βˆ’5 𝑑𝑦 𝑑π‘₯ βˆ’2π‘₯=0 2. 3 𝑦 2 𝑑𝑦 𝑑π‘₯ βˆ’2𝑦 𝑑𝑦 𝑑π‘₯ βˆ’5 𝑑𝑦 𝑑π‘₯ =2π‘₯ 3. 𝑑𝑦 𝑑π‘₯ 3 𝑦 2 βˆ’2π‘¦βˆ’5 =2π‘₯ 4. 𝑑𝑦 𝑑π‘₯ = 2π‘₯ 3 𝑦 2 βˆ’2π‘¦βˆ’5

9 Example 3 Find the derivative of ( π‘₯ 2 +4 𝑦 2 =4) 1. 2π‘₯+8𝑦 𝑑𝑦 𝑑π‘₯ =0
2. 8𝑦 𝑑𝑦 𝑑π‘₯ =βˆ’2π‘₯ 3. 𝑑𝑦 𝑑π‘₯ =βˆ’ 2π‘₯ 8𝑦 =βˆ’ π‘₯ 4𝑦

10 Example 4 𝑑𝑦 𝑑π‘₯ 3 π‘₯ 2 + 𝑦 =100𝑦 1. 6 π‘₯ 2 + 𝑦 2 2π‘₯+2𝑦 𝑑𝑦 𝑑π‘₯ =100 𝑑𝑦 𝑑π‘₯ 2. 6 π‘₯ 2 +6 𝑦 2 2π‘₯+2𝑦 𝑑𝑦 𝑑π‘₯ =100 𝑑𝑦 𝑑π‘₯ 3. 12 π‘₯ π‘₯ 2 𝑦 𝑑𝑦 𝑑π‘₯ +12 𝑦 2 π‘₯+12 𝑦 3 𝑑𝑦 𝑑π‘₯ =100 𝑑𝑦 𝑑π‘₯ 4. 12 π‘₯ 2 𝑦 𝑑𝑦 𝑑π‘₯ +12 𝑦 3 𝑑𝑦 𝑑π‘₯ βˆ’100 𝑑𝑦 𝑑π‘₯ =βˆ’12 π‘₯ 3 βˆ’12 𝑦 2 π‘₯ 5. 𝑑𝑦 𝑑π‘₯ = βˆ’12 π‘₯ 3 βˆ’12 𝑦 2 π‘₯ 12 π‘₯ 2 𝑦+12 𝑦 3 βˆ’100

11 Example 5 What is the slope of the line tangent to the curve 2 π‘₯ 2 βˆ’3 𝑦 2 =2π‘₯π‘¦βˆ’6 at the point (3, 2)?

12 Example 6 Find 𝑑 2 𝑦 𝑑 π‘₯ 2 (2 π‘₯ 3 βˆ’3 𝑦 2 =8)

13 Exit Ticket for Feedback
Find the derivative of 2π‘₯+𝑦 2 βˆ’π‘₯𝑦=10


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