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Escuela Preparatoria Uno UNIVERSIDAD AUTÓNOMA DE YUCATÁN Página 48 de 81 𝑦 = 𝑓(𝑥) = 𝑥3 + 2𝑥, (1, 3). 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 𝑓(𝑥+∆𝑥)−𝑓(𝑥) ∆𝑥 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 (𝑥+∆𝑥)3+2(𝑥+∆𝑥)−(𝑥3+2𝑥) ∆𝑥 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 𝑥3+3𝑥2(∆𝑥)+3𝑥(∆𝑥)2+(∆𝑥)3+2𝑥+2(∆𝑥)−𝑥3−2𝑥 ∆𝑥 𝒇′(𝒙) = 𝒍𝒊𝒎 ∆𝒙→𝟎 𝟑𝒙𝟐(∆𝒙)+𝟑𝒙(∆𝒙)𝟐+(∆𝒙)𝟑+𝟐(∆𝒙) ∆𝒙 𝑓′(𝑥) = 3𝑥2(0)+3𝑥(0)2+(0)3+2(0) 0 = 𝟎 𝟎 =? 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 ∆𝑥[3𝑥2+3𝑥(∆𝑥)+(∆𝑥)2+2] ∆𝑥 𝒇′(𝒙) = 𝒍𝒊𝒎 ∆𝒙→𝟎 [𝟑𝒙𝟐 + 𝟑𝒙(∆𝒙)+(∆𝒙)𝟐 + 𝟐] 𝑓′(𝑥) = 3𝑥2 + 3𝑥(0)+(0)2 + 2 𝒇′(𝒙) = 𝟑𝒙𝟐 + 𝟐 𝒎𝒕𝒂𝒏(𝟏, 𝟑) = 𝒇 ′(𝟏) = 𝟑(𝟏)𝟐 + 𝟐 = 𝟓 𝑓(𝑥) = √𝑥 , (1, 1) y (4, 2) 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 𝑓(𝑥+∆𝑥)−𝑓(𝑥) ∆𝑥 𝒇′(𝒙) = 𝒍𝒊𝒎 ∆𝒙→𝟎 √𝒙+∆𝒙−√𝒙 ∆𝒙 𝑓′(𝑥) = √𝑥+0−√𝑥 0 = 𝟎 𝟎 =? 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 √𝑥+∆𝑥−√𝑥 ∆𝑥 ∙ √𝑥+∆𝑥+√𝑥 √𝑥+∆𝑥+√𝑥 𝑓′(𝑥) = 𝑙𝑖𝑚 ∆𝑥→0 𝑥+∆𝑥−𝑥 ∆𝑥(√𝑥+∆𝑥+√𝑥) 𝒇′(𝒙) = 𝒍𝒊𝒎 ∆𝒙→𝟎 𝟏 √𝒙+∆𝒙+√𝒙 𝑓′(𝑥) = 1 √𝑥+0+√𝑥 𝒇′(𝒙) = 𝟏 𝟐√𝒙 𝒎𝒕𝒂𝒏(𝟏, 𝟏) = 𝒇 ′(𝟏) = 𝟏 𝟐√𝟏 = 𝟏 𝟐 𝒎𝒕𝒂𝒏(𝟒, 𝟐) = 𝒇 ′(𝟒) = 𝟏 𝟐√𝟒 = 𝟏 𝟒
