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UNIT 2 LESSON 8 CHAIN & QUOTIENT RULE 1. 2 Chain Rule and Quotient Rule Example 1 dy = (2x 2 – 1) 2 [4(1 + 3x) 3 (3)] – (1 + 3x) 4 [2(2x 2 – 1)(4x)] dx.

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Presentation on theme: "UNIT 2 LESSON 8 CHAIN & QUOTIENT RULE 1. 2 Chain Rule and Quotient Rule Example 1 dy = (2x 2 – 1) 2 [4(1 + 3x) 3 (3)] – (1 + 3x) 4 [2(2x 2 – 1)(4x)] dx."— Presentation transcript:

1 UNIT 2 LESSON 8 CHAIN & QUOTIENT RULE 1

2 2 Chain Rule and Quotient Rule Example 1 dy = (2x 2 – 1) 2 [4(1 + 3x) 3 (3)] – (1 + 3x) 4 [2(2x 2 – 1)(4x)] dx [(2x 2 – 1) 2 ] 2 dy = (2x 2 – 1) 2 [12(1 + 3x) 3 ] – (1 + 3x) 4 [8x(2x 2 – 1)] dx (2x 2 – 1) 4

3 3 Factor out common factors 4, (2x 2 – 1), and (1 + 3x) 3 dy = 4 (2x 2 – 1) (1 + 3x) 3 [3(2x 2 – 1) – 2x(1 + 3x)] dx(2x 2 – 1) 4 dy = 4 (2x 2 – 1) (1 + 3x) 3 [6x 2 – 3 – 2x – 6x 2 ] dx(2x 2 – 1) 4 dy = 4 (2x 2 – 1) (1 + 3x) 3 ( – 3 – 2x) dx (2x 2 – 1) 4 dy = (2x 2 – 1) 2 [12(1 + 3x) 3 ] – (1 + 3x) 4 [8x(2x 2 – 1)] dx (2x 2 – 1) 4 dy = 4 (1 + 3x) 3 ( – 3 – 2x) dx (2x 2 – 1) 3

4 4 Find the equation of the tangent line at x = 0. Slope = dy = 4(1 ) 3 ( – 3) = 12 dx ( – 1) 3 Equation of tangent line 1 = 12(0) + b b = 1 y = 12x + 1 dy = 4 (1 + 3x) 3 ( – 3 – 2x) dx (2x 2 – 1) 3 dy = 4 (1 + 3(0)) 3 ( – 3 – 2(0)) dx (2(0) 2 – 1) 3

5 5 Chain Rule and the Quotient Rule Example 2 Rewrite


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