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Section 2.5 Day 1 AP Calculus
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Learning Targets Define implicit and explicit functions
Evaluate derivatives using Implicit Differentiation
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Explicit Form Everything is solved for one variable Ex: π¦=3π₯+1
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Implicit Form There is a mix of variables on the same side Ex: π₯π¦=1
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Differentiating with Respect to x
Ex: π ππ₯ ( π₯ 3 ) = 3 π₯ 2 ππ₯ ππ₯ = 3 π₯ 2 Ex: π ππ₯ ( π¦ 3 ) = 3 π¦ 2 ππ¦ ππ₯
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Implicit Differentiation Steps
1. Take the derivative of each piece 2. Put ππ¦ ππ₯ after the derivative of βyβ terms 3. Split ππ¦ ππ₯ terms away from other terms using the equal sign 4. Factor ππ¦ ππ₯ 5. Divide
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Example 1 π ππ₯ [π₯ π¦ 2 ] π₯ 2π¦ ππ¦ ππ₯ + π¦ 2 2π₯π¦ π ππ₯ + π¦ 2
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Example 2 Find the derivative of π¦ 3 β π¦ 2 β5π¦ β π₯ 2 =β4
1. 3 π¦ 2 ππ¦ ππ₯ β2π¦ ππ¦ ππ₯ β5 ππ¦ ππ₯ β2π₯=0 2. 3 π¦ 2 ππ¦ ππ₯ β2π¦ ππ¦ ππ₯ β5 ππ¦ ππ₯ =2π₯ 3. ππ¦ ππ₯ 3 π¦ 2 β2π¦β5 =2π₯ 4. ππ¦ ππ₯ = 2π₯ 3 π¦ 2 β2π¦β5
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Example 3 Find the derivative of ( π₯ 2 +4 π¦ 2 =4) 1. 2π₯+8π¦ ππ¦ ππ₯ =0
2. 8π¦ ππ¦ ππ₯ =β2π₯ 3. ππ¦ ππ₯ =β 2π₯ 8π¦ =β π₯ 4π¦
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Example 4 ππ¦ ππ₯ 3 π₯ 2 + π¦ =100π¦ 1. 6 π₯ 2 + π¦ 2 2π₯+2π¦ ππ¦ ππ₯ =100 ππ¦ ππ₯ 2. 6 π₯ 2 +6 π¦ 2 2π₯+2π¦ ππ¦ ππ₯ =100 ππ¦ ππ₯ 3. 12 π₯ π₯ 2 π¦ ππ¦ ππ₯ +12 π¦ 2 π₯+12 π¦ 3 ππ¦ ππ₯ =100 ππ¦ ππ₯ 4. 12 π₯ 2 π¦ ππ¦ ππ₯ +12 π¦ 3 ππ¦ ππ₯ β100 ππ¦ ππ₯ =β12 π₯ 3 β12 π¦ 2 π₯ 5. ππ¦ ππ₯ = β12 π₯ 3 β12 π¦ 2 π₯ 12 π₯ 2 π¦+12 π¦ 3 β100
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Example 5 What is the slope of the line tangent to the curve 2 π₯ 2 β3 π¦ 2 =2π₯π¦β6 at the point (3, 2)?
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Example 6 Find π 2 π¦ π π₯ 2 (2 π₯ 3 β3 π¦ 2 =8)
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Exit Ticket for Feedback
Find the derivative of 2π₯+π¦ 2 βπ₯π¦=10
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