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Calculus I group study Mod 2

# Calculus I group study Mod 2 - – x 4 x 1 Quotient Rule ƒ...

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Calculus I Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0- 618-60624-6). Calculus SG2 Module 3 3a 3.1 18 ƒ(x) = 1 – x 2 Derivative ƒ 1 (x) = lim ƒ(x + ∆x) – ƒ(x) / ∆x ∆x >0 ƒ 1 (x) = lim 1 - (x + ∆x) 2 – 1 + x 2 / ∆x ∆x >0 ƒ 1 (x) = lim 1 – x 2 – (∆x) 2 – 2x∆x – 1 + x 2 / ∆x ∆x >0 ƒ 1 (x) = lim (- ∆x-2x) ∆x >0 using direct substitution ƒ 1 (x) = lim (- 0 - 2x) ∆x >0 ƒ 1 (x) = lim - 2x 3b 3.3 30 ƒ(x) = x 4 (1 - (2 / x + 1)) ƒ(x) = x 4 (x + 1 – 2/ x + 1)) = x 4 (x - 1) / (x + 1) ƒ(x) = x 5

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Unformatted text preview: – x 4 / x + 1 Quotient Rule ƒ 1 (x) = (x + 1) (d/dx) (x 5 – x 4 ) - (x 5 – x 4 ) (d/dx) (x + 1) (x + 1) 2 Constant rule, sum the difference rule and power rule ƒ 1 (x) = (x + 1) (d/dx) (5x 4 – 4x 3 ) - (x 5 – x 4 ) (1 + 0) (x + 1) 2 ƒ 1 (x) = 4x 5 + 2x 4-4x 3 (x + 1) 2 ƒ 1 (x) = 2x 3 (2x 2 + x – 2) (x + 1) 2 3c 3.4 56 y=sin πx Differentiating x dy/dx = (d/dx)(sin πx) y'=cos(πx) (d/dx)(πx) y'= cos(πx)π y'= πcos(πx) 3d 3.5 2 x 2 - y 2 = 81 (d/dx)(x 2 ) – (d/dx)(y 2 ) = (d/dx)(81) 2x – 2y(d/dx)(y) = 0 dy/dx = x/y...
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Calculus I group study Mod 2 - – x 4 x 1 Quotient Rule ƒ...

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