Exam (elaborations)
Tarea con Soluciones Derivadas Cálculo I
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Tarea con solucionario de derivadas de cálculo I
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CIPRIANO𝑑𝑦
𝑦′ = LA TAREA 1 DE DERIVADAS (𝑎2 𝑦)′ = 𝑎2 𝑦 ′
𝑑𝑥
𝑎 ′ 𝑎
( ) =− 2
RECOMENDACIÓN: Estimado estudiante los ejercicios pares debes resolverlo 𝑥 𝑥
de acuerdo a las reglas de derivación y tablas de derivadas.
𝑢 1 𝑢 ′
Regla de la cadena: 𝑦 = 𝑙𝑛 𝑣 → 𝑦′ = 𝑢 ∙ (𝑣 ) , como el argumento es un cociente al derivar se
𝑣
𝑣 𝑢 ′
invierte, es decir 𝑦 ′ = ∙ (𝑣 ) , ver ejercicios 5 y 7
𝑢
𝑦 = 𝑢𝑛 → 𝑦 ′ = 𝑛𝑢𝑛−1 ∙ 𝑢′ , ver ejercicio 9
Ejemplo: Derivar 𝑦 = 𝑐𝑜𝑠 3 𝑥 → 𝑦 ′ = 3𝑐𝑜𝑠 2 𝑥 ∙ (𝑐𝑜𝑠𝑥)′ = 3𝑐𝑜𝑠 2 𝑥(−𝑠𝑒𝑛𝑥) = −3𝑐𝑜𝑠 2 𝑥 ∙ 𝑠𝑒𝑛𝑥
Derivar y simplificar al máximo las siguientes funciones
1
1. 𝑦 = 𝑥 ∙ 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥 + √1 − 4𝑥 2 . Se aplicará las reglas del producto y de la cadena
2
SOLUCIÓN
1 1 1
𝑦 ′ = 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥 + 𝑥 ∙ √1−4𝑥 2 ∙ 2 + 2 ∙ ∙ (−8𝑥)
2√1−4𝑥2
2𝑥 2𝑥
𝑦 ′ = 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥 + √1−4𝑥2 − √1−4𝑥2 = 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥 ∴ 𝑦′ = 𝑎𝑟𝑐𝑠𝑒𝑛2𝑥
1
2. 𝑦 = 𝑥𝑎𝑟𝑐𝑐𝑜𝑠2𝑥 − 2 √1 − 4𝑥 2 r.- 𝑦 ′ = 𝑎𝑟𝑐𝑐𝑜𝑠2𝑥
𝑥
3. 𝑦 = 𝑥√𝑎2 − 𝑥 2 + 𝑎2 𝑎𝑟𝑐𝑠𝑒𝑛
𝑎
SOLUCIÓN
1 1 1
𝑦 ′ = √𝑎2 − 𝑥 2 + 𝑥 ∙ ∙ (−2𝑥) + 𝑎2 ∙ ∙𝑎
2√𝑎2 −𝑥2 2
√1−𝑥2
𝑎
𝑥2 𝑎 𝑎2 −𝑥2 𝑎2 −𝑥 2 +𝑎2 −𝑥 2
𝑦 ′ = √𝑎2 − 𝑥 2 − √𝑎2 2
+ 𝑎 ∙ √𝑎2 = √𝑎2 − 𝑥 2 + √𝑎2 =
−𝑥 −𝑥 2 −𝑥 2 √𝑎2 −𝑥 2
2𝑎2 −2𝑥 2 𝑎2 −𝑥 2
𝑦′ = √𝑎2 −𝑥 2
= 2 √𝑎2 = 2√𝑎2 − 𝑥 2 ∴ 𝑦 ′ = 2√𝑎2 − 𝑥 2
−𝑥 2
𝑥∙𝑎𝑟𝑐𝑐𝑜𝑠𝑥
4. 𝑦 = 𝑥 + √1 − 𝑥 2 ∙ 𝑎𝑟𝑐𝑐𝑜𝑠𝑥 r.- − √1−𝑥 2
1+𝑠𝑒𝑛𝑥
5. 𝑦 = 𝑙𝑛√
1−𝑠𝑒𝑛𝑥
1
SOLUCIÓN: Antes de derivar reescribiremos la función aplicando la propiedad: 𝑙𝑛√𝑓(𝑥) = 𝑙𝑛𝑓(𝑥)
2
1 1+𝑠𝑒𝑛𝑥 1 1−𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥∙(1−𝑠𝑒𝑛𝑥)−(1+𝑠𝑒𝑛𝑥)∙(−𝑐𝑜𝑠𝑥)
𝑦 = 2 ln 1−𝑠𝑒𝑛𝑥 → 𝑦 ′ = 2 ∙ 1+𝑠𝑒𝑛𝑥 ∙ [ (1−𝑠𝑒𝑛𝑥)2
]
1
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