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T10-Derivada por Inc-Asqui
Boris Asqui
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ESCUELA SUPERIOR POLITÉCNICA DE CHIMBORAZO PRIMER SEMESTRE PARALELO ¨B¨ ANÁLISIS MATEMÁTICO ASQUI VACA BORIS JOSUE DERIVADAS 2020-2021 𝒚 = 𝒙𝒏 → 𝒚´ = 𝒏𝒙𝒏−𝟏 𝑦′ = lim ∆𝑥→0 (𝑥 + ∆𝑥)𝑛 − 𝑥𝑛 ∆𝑥 lim ∆𝑥→0 𝑥𝑛 + 𝑛𝑥𝑛−1∆𝑥 + ( 𝑛 2 ) 𝑥𝑛−2∆𝑥2 + ⋯ + ( 𝑛 𝑛 )∆𝑥𝑛 − 𝑥𝑛 ∆𝑥 lim ∆𝑥→0 ∆𝑥(𝑛𝑥𝑛−1 + ( 𝑛 2 ) 𝑥𝑛−2∆𝑥 + ⋯ + ( 𝑛 𝑛 ) ∆𝑥𝑛−1) ∆𝑥 = 𝒏𝒙𝒏−𝟏 𝒚 = 𝒔𝒆𝒏 𝒙 → 𝒚´ = 𝒄𝒐𝒔𝒙 𝑓′(𝑥) = 𝑙𝑖𝑚 𝛥𝑥→0 𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥) 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 sen(𝑥 + 𝛥𝑥) − 𝑠𝑒𝑛𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 sen 𝑥 ∗ 𝑐𝑜𝑠𝛥𝑥 + 𝑐𝑜𝑠𝑥 ∗ 𝑠𝑒𝑛𝛥𝑥 − 𝑠𝑒𝑛𝑥 𝛥𝑥 𝑙𝑖𝑚 𝛥𝑥→0 sen 𝑥(𝑐𝑜𝑠 𝛥𝑥 − 1) 𝛥𝑥 + 𝑙𝑖𝑚 𝛥𝑥→0 𝑐𝑜𝑠𝑥 ∗ 𝑠𝑒𝑛𝛥𝑥 𝛥𝑥 𝑙𝑖𝑚 𝛥𝑥→0 sen 𝑥(𝑐𝑜𝑠𝛥𝑥−1) 𝛥𝑥 + 𝑙𝑖𝑚 𝛥𝑥→0 𝑐𝑜𝑠𝑥∗𝑠𝑒𝑛𝛥𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 1−𝑐𝑜𝑠𝛥𝑥 𝛥𝑥 ( 𝑙𝑖𝑚 𝛥𝑥→0 sen𝑥) + ( 𝑙𝑖𝑚 𝛥𝑥→0 cos𝑥) 𝑙𝑖𝑚 𝛥𝑥→0 𝑠𝑒𝑛𝛥𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 1 − 𝑐𝑜𝑠𝛥𝑥 𝛥𝑥 − 0 ∧ 𝑙𝑖𝑚 𝛥𝑥→0 𝑠𝑒𝑛𝛥𝑥 𝛥𝑥 − 1 = 𝑓(𝑥) = 0 ∗ 𝑠𝑒𝑛𝑥 + 𝑐𝑜𝑠𝑥 ∗ 1 = 𝑪𝒐𝒔 𝒙 𝒚 = 𝒄𝒐𝒔 𝒙 → 𝒚´ = −𝒔𝒆𝒏𝒙 𝑓′(𝑥) = 𝑙𝑖𝑚 𝛥𝑥→0 𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥) 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 cos(𝑥 + 𝛥𝑥) − 𝑐𝑜𝑠𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 cos 𝑥 ∗ 𝑐𝑜𝑠𝛥𝑥 − 𝑠𝑒𝑛𝑥 ∗ 𝛥𝑥 − 𝑐𝑜𝑠𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 cos 𝑥 (𝑐𝑜𝑠𝛥𝑥 − 1) − 𝑠𝑒𝑛𝑥 ∗ 𝑠𝑒𝑛𝛥𝑥 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 cos 𝑥 (𝑐𝑜𝑠𝛥𝑥 − 1) 𝛥𝑥 − 𝑠𝑒𝑛𝑥 𝑠e𝑛𝛥𝑥 𝛥𝑥 = −𝒔𝒆𝒏 𝒙 𝒚 = 𝒌 ∗ 𝒇(𝒙) → 𝒚´ = 𝒌 ∗ 𝒇´(𝒙) 𝑓′(𝑥) = 𝑙𝑖𝑚 𝛥𝑥→0 𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥) 𝛥𝑥 = 𝑙𝑖𝑚 𝛥𝑥→0 𝑘 ∗ 𝑓(𝑥 + 𝛥𝑥) − 𝑘 ∗ 𝑓(𝑥) 𝛥𝑥 𝑘 ∗ 𝑙𝑖𝑚 𝛥𝑥→0 𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥) 𝛥𝑥 = 𝒌 ∗ 𝒇´(𝒙) 𝒚 = 𝑼(𝒙) + 𝑽(𝒙) + 𝑾(𝒙) 𝑦 + ∆𝑦 = 𝑈 + ∆𝑈 + 𝑉 + ∆𝑉 + 𝑊 + ∆𝑊 ∆𝑦 = 𝑈 + ∆𝑈 + 𝑉 + ∆𝑉 + 𝑊 + ∆𝑊 − 𝑈 − 𝑉 − 𝑊 = ∆𝑈 + ∆𝑉 + ∆𝑊 ∆𝑦 ∆𝑥 = ∆𝑈 + ∆𝑉 + ∆𝑊 ∆𝑥 lim ∆𝑥→𝑥 ∆𝑦 ∆𝑥 = lim ∆𝑥→𝑥 ( ∆𝑈 ∆𝑥 + ∆𝑉 ∆𝑥 + ∆𝑊 ∆𝑥 ) = 𝑼´𝑿 + 𝑽´𝑿 + 𝑾´𝑿 𝒚 = 𝑼. 𝑽 → 𝒚` = 𝑼,𝑽 +𝑽,𝑼 lim ℎ→0 𝑢(𝑥 + ℎ)𝑣(𝑥 + ℎ) − 𝑢(𝑥)𝑣(𝑥) ℎ lim ℎ→0 ( 𝑢(𝑥 + ℎ) − 𝑢(𝑥) ℎ )( 𝑣(𝑥 + ℎ) − 𝑣(𝑥) ℎ ) lim ℎ→0 𝑢(𝑥 + ℎ)𝑣(𝑥 + ℎ) − 𝑢(𝑥 + ℎ)𝑣(𝑥) − 𝑢(𝑥)𝑣(𝑥 + ℎ) + 𝑢(𝑥)𝑣(𝑥) ℎ lim ℎ→0 (𝑢(𝑥 + ℎ) − 𝑢(𝑥)) ∗ 𝑣(𝑥 + ℎ) ℎ +lim ℎ→0 (𝑣(𝑥 + ℎ) − 𝑣(𝑥)) ∗ 𝑢(𝑥) ℎ lim ℎ→0 (𝑢(𝑥 + ℎ) − 𝑢(𝑥)) ℎ ∗ lim ℎ→0 𝑣(𝑥 + ℎ) +lim ℎ→0 (𝑣(𝑥 + ℎ) − 𝑣(𝑥)) ℎ ∗ lim ℎ→0 𝑢(𝑥) 𝑈, ∗ 𝑣(𝑥 + 0) +𝑉 , ∗ 𝑢(𝑥) = 𝑼,𝑽 +𝑽,𝑼 𝒚 = 𝑼 𝑽 → 𝒚´ = 𝑼,𝑽 − 𝑽,𝑼 𝑽𝟐 lim ℎ→0 𝑈 𝑉 = lim ℎ→0 (𝑢,(𝑥) ∗ 1 𝑣(𝑥) ) → 𝒚, = 𝑼,𝑽 + 𝑽,𝑼 lim ℎ→0 𝑢,(𝑥) 1 𝑣(𝑥) + 𝑣(𝑥) 𝑣2(𝑥) 𝑢(𝑥) = 𝑢,(𝑥) 1 𝑣(𝑥) + (−𝑣 ,(𝑥)) 𝑣2(𝑥) 𝑢(𝑥) = 𝑢,(𝑥) 𝑣(𝑥) − 𝑣 ,(𝑥)𝑢(𝑥) 𝑣2(𝑥) = 𝑢,(𝑥)𝑣(𝑥)−𝑣 ,(𝑥)𝑢(𝑥) 𝑣2(𝑥) = 𝑼,𝑽 − 𝑽,𝑼 𝑽𝟐 𝒚 = 𝒇(𝒙) ± 𝒈(𝒙) ± 𝒘(𝒙) → 𝒚´ = 𝒇´(𝒙) ± 𝒈´(𝒙) ± 𝒘´(𝒙) *𝑌 = (𝑓 + 𝑔 + 𝑤)(𝑥) 𝑓(𝑥) + 𝑔(𝑥) + 𝑤(𝑥) = (𝑓 + 𝑔 + 𝑤)(𝑥) 𝑦´ = lim ∆𝑥→0 (𝑓 + 𝑔 + 𝑤)(𝑥 + ∆𝑥) − (𝑓 + 𝑔 + 𝑤)(𝑥) ∆𝑥 𝑦´ = lim ∆𝑥→0 𝑓(𝑥 + ∆𝑥) + 𝑔(𝑥 + ∆𝑥) + 𝑤(𝑥 + ∆𝑥) − 𝑓(𝑥) − 𝑔(𝑥) − 𝑤(𝑥) ∆𝑥 𝑦´ = lim ∆𝑥→0 [𝑓(𝑥 + ∆𝑥) + −𝑓(𝑥)] + [𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)] + [𝑤(𝑥 + ∆𝑥) − 𝑤(𝑥)] ∆𝑥 𝑦´ = lim ∆𝑥→0 [𝑓(𝑥 + ∆𝑥) + −𝑓(𝑥)] ∆𝑥 + lim ∆𝑥→0 [𝑔(𝑥 + ∆𝑥) + −𝑔(𝑥)] ∆𝑥 + lim ∆𝑥→0 [𝑤(𝑥 + ∆𝑥) + −𝑤(𝑥)] ∆𝑥 SABIENDO QUE: lim ∆𝑥→0 [𝑓(𝑥 + ∆𝑥) + −𝑓(𝑥)] ∆𝑥 = 𝑓´(𝑥) lim ∆𝑥→0 [𝑔(𝑥 + ∆𝑥) + −𝑔(𝑥)] ∆𝑥 = 𝑔´(𝑥) lim ∆𝑥→0 [𝑤(𝑥 + ∆𝑥) + −𝑤(𝑥)] ∆𝑥 = 𝑤´(𝑥) DEDUCIMOS: 𝑦´ = 𝑓´(𝑥) ± 𝑔´(𝑥) ± 𝑤´(𝑥) 𝑦 = 𝑓(𝑥) ± 𝑔(𝑥) ± 𝑤(𝑥) → 𝑦´ = 𝑓´(𝑥) ± 𝑔´(𝑥) ± 𝑤´(𝑥) 𝒚 = 𝟏 √𝒙 𝟑 + 𝟐 𝒄𝒐𝒔(𝒙) − √𝒙 𝟓 √𝒙 𝟒 𝑦´ = ( 1 √𝑥 3 ) ′ + (2 cos(𝑥))′ − ( √𝑥 5 √𝑥 4 )′ 𝑦′ = (𝑥− 1 3) ′ + 2 cos(𝑥) − (𝑥 1 20)′ 𝑦′ = 1 3 (𝑥− 4 3) − 2 sen(𝑥) + 1 20 (𝑥 21 20)′ 𝑦′ = (𝑥− 1 3) ′ + 2 cos(𝑥) − (𝑥 1 20)′ 𝑦′ = − 1 3√𝑥4 3 − 2𝑠𝑒𝑛(𝑥) + 1 20 √𝑥21 20 𝒚′ = − 𝟏 𝟑 √𝒙𝟒 𝟑 − 𝟐𝒔𝒆𝒏(𝒙) + 𝟏 𝟐𝟎𝒙 √𝒙 𝟐𝟎 y´=𝒆𝒙 → 𝒚´ = 𝒆𝒙 𝑦 = log𝑎 𝑥 → 𝑦´ = 1 𝑥 log𝑎 𝑒 𝑦´ = 1 𝑥 𝑙𝑛𝑒 𝑙𝑛𝑎 = 𝑦´ = 1 𝑥𝑙𝑛𝑎 𝒚 = 𝐥𝐧(𝒙) 𝑦´ = 1 𝑥 𝒚 = 𝟐𝒙𝟑 + 𝟑 𝑦′ = 2(𝑥3) + 0 𝑦′ = 2(𝑥2) + 0 𝒚′ = 𝟔𝒙𝟐 𝒚 = 𝒙𝟒 + 𝒙𝟐 𝑦′ = 4𝑥3 + 2𝑥 𝒚′ = 𝟐𝒙(𝟐𝒙𝟐 + 𝟏) 𝒚 = √𝒙 𝟑 𝑦′ = 𝑥 1 3 𝑦′ = 1 3 𝑥 1 3−1 𝑦′ = 1 3 𝑥− 2 3 𝒚′ = 𝟏 𝟑𝒙 𝟐 𝟑 𝒚 = 𝒙√𝒙 √𝒙 𝟒 → 𝑦 ′ = 𝑥√𝑥 𝑥 1 4 → 𝑦′ = 𝑥. 𝑥 1 2. 𝑥− 1 4 𝑦´ = 5 4 𝑥− 1 4 → 𝑦´ = 5 4 √𝑥 4 𝒚 = √𝒙 𝒂 √𝒙 𝒃 → 𝑦′ = ( √𝑥 𝑎 ) √𝑥 𝑏 + ( √𝑥 𝑏 ) √𝑥 𝑎 → 𝑦′ = 𝑥 1 𝑎 √𝑥 𝑏 + 𝑥 1 𝑏 √𝑥 𝑎 𝑦′ = 1 𝑎 𝑥 1 𝑎−1 √𝑥 𝑏 + 1 𝑏 𝑥 1 𝑏−1 → 𝑦′ = 𝑥 1−𝑎 𝑎 𝑎 √𝑥 𝑏 + 𝑥 1−𝑏 𝑏 𝑏 √𝑥 𝑎 → 𝑦′ = 𝑥 −𝑎+1 𝑎 √𝑥 𝑏 𝑎 + 𝑥 −𝑏+1 𝑏 √𝑥 𝑎 𝑏 𝑦′ = 𝑥 1−𝑎 𝑎 √𝑥 𝑏 𝑎 + 𝑥 1−𝑏 𝑏 √𝑥 𝑎 𝑏 𝒚 = √𝒙√𝒙√𝒙 ⋅ 𝟏 𝒙 𝑦′ = √ 𝑥√𝑥. 𝑥 1 2. 𝑥−1 → 𝑦′ = √𝑥 7 4. 𝑥−1 → 𝑦´ = 3 8 𝑥 3 8−1 𝑦´ = 3 8 𝑥− 5 8 → 𝑦´ = 3 8 1 √𝑥5 8 → 𝑦´ = 3 8 √𝑥5 8 𝒚 = 𝝅𝐜𝐨𝐬 (𝒙) 𝑦′ = 𝜋(−𝑠𝑒𝑛(𝑥)) 𝑦′ = − 𝜋 𝑠𝑒𝑛 (𝑥) 𝒚 = √𝒂 + 𝒃 . 𝒙𝟓 𝑦′ = √𝑎 + 𝑏 .5 𝑥5−1 𝑦′ = 5 𝑥4√𝑎 + 𝑏 𝒚 = 𝒆. 𝒔𝒆𝒏(𝒙) 𝑦′ = 𝑒. 𝑐𝑜𝑠(𝑥) 𝒚 = 𝟏 𝟐 𝒙𝟏𝟎𝟎 𝑦′ = 1 2 . 100𝑥100−1 𝑦′ = 1 2 . 100𝑥99 𝑦′ = 100 . 1 2 𝑥99 𝑦′ = 1 . 100 2 𝑥99 𝑦′ = 100 2 𝑥99 𝑦′ = 50 𝑥99
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