UNIVERSIDAD DE CIENCIAS Y ADMINISTRACIÓN Ejercicio 9. 𝒚 = 𝟑𝒙𝟒 − 𝟐𝒙𝟐 + 𝟖 Solución Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑 (𝑐) = 0 𝑑𝑥 𝑑𝑦 𝑑 (3𝑥 4 − 2𝑥 2 + 8) = 𝑑𝑥 𝑑𝑥 Aplicando la deriva de cada elemento de la función 𝑑 𝑑 𝑑 (3𝑥 4 ) − (2𝑥 2 ) + (8) = 4 ∙ 3𝑥 4−1 − 2 ∙ 2𝑥 2−1 + 0 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑 (3𝑥 4 ) − (2𝑥 2 ) + (8) = 12𝑥 3 − 4𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 ∴ 𝑑 (3𝑥 4 − 2𝑥 2 + 8) = 12𝑥 3 − 4𝑥 𝑑𝑥 Ejercicio 10. 𝒚 = 𝟒 + 𝟑𝒙 − 𝟐𝒙𝟑 Solución Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑 (𝑐) = 0 𝑑𝑥 𝑑𝑦 𝑑 (4 + 3𝑥 − 2𝑥 3 ) = 𝑑𝑥 𝑑𝑥 Aplicando la deriva de cada elemento de la función 2 𝑑 𝑑 𝑑 (4) + (3𝑥) − (2𝑥 3 ) = 0 + 3 − 3 ∙ 2𝑥 3−1 𝑑𝑥 𝑑𝑥 𝑑𝑥 Aplicando la derivación de cada función 𝑑 𝑑 𝑑 (4) + (3𝑥) − (2𝑥 3 ) = 3 − 6𝑥 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 ∴ 𝑑 (4 + 3𝑥 − 2𝑥 3 ) = 3 − 6𝑥 2 𝑑𝑥 Ejercicio 11. 𝑠 = 𝑎𝑡 5 − 5𝑏𝑡 3 Solución 𝑑𝑠 𝑑 = (𝑎𝑡 5 − 5𝑏𝑡 3 ) 𝑑𝑡 𝑑𝑡 Formulas a utilizar: 𝑑 (𝑐𝑡 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑡 Aplicando la deriva de cada elemento de la función 𝑑 𝑑 (𝑎𝑡 5 ) − (5𝑏𝑡 3 ) = 5 ∙ 𝑎𝑡 5−1 − 3 ∙ 5𝑏𝑡 3−1 𝑑𝑡 𝑑𝑡 ∴ 𝑑 𝑑 𝑑 (𝑎𝑡 5 ) − (5𝑏𝑡 3 ) = (𝑎𝑡 5 − 5𝑏𝑡 3 ) = 5𝑎𝑡 4 − 15𝑏𝑡 2 𝑑𝑡 𝑑𝑡 𝑑𝑡 Ejercicio 12. 𝑦= 𝑧2 𝑧7 − 2 7 Solución 𝑑𝑦 𝑑 𝑧2 𝑧7 = ( − ) 𝑑𝑧 𝑑𝑧 2 7 Formulas a utilizar: 𝑑 (𝑐𝑧 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑧 𝑛−1 𝑑𝑧 Aplicando la deriva de cada elemento de la función 3 𝑑𝑦 𝑑 𝑧2 𝑧7 𝑑 𝑧2 𝑑 𝑧7 = ( − )= ( )− ( ) 𝑑𝑧 𝑑𝑧 2 7 𝑑𝑧 2 𝑑𝑧 7 𝑑 𝑧2 𝑑 𝑧7 𝑧 2−1 𝑧 7−1 −7∙ ( )− ( )= 2∙ 𝑑𝑧 2 𝑑𝑧 7 2 7 𝑑 𝑧2 𝑑 𝑧7 𝑧 𝑧6 − = 2 ∙ − 7 ∙ ( ) ( ) 𝑑𝑧 2 𝑑𝑧 7 2 7 𝑑 𝑧2 𝑑 𝑧7 ( ) − ( ) = 𝑧 − 𝑧6 𝑑𝑧 2 𝑑𝑧 7 ∴ 𝑑 𝑧2 𝑧7 ( − ) = 𝑧 − 𝑧6 𝑑𝑧 2 7 Ejercicio 13. 𝒚 = √𝒗 Solución Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑𝑦 𝑑 = (√𝒗) 𝑑𝑥 𝑑𝑥 1 𝑑 𝑑 (𝑣 2 ) (√𝒗) = 𝑑𝑥 𝑑𝑥 1 𝑑 1 1 𝑑𝑣 (𝑣 2 ) = 𝑥 2−1 𝑑𝑥 2 𝑑𝑥 1 𝑑 1 1 𝑑𝑣 (𝑣 2 ) = 𝑣 −2 𝑑𝑥 2 𝑑𝑥 𝑑𝑦 𝑑 1 1 𝑑𝑣 1 1 𝑑𝑣 = = (√𝒗) = 𝑣 −2 𝑑𝑥 𝑑𝑥 2 𝑑𝑥 2 2√𝑣 𝑑𝑥 ∴ 𝑑 1 1 𝑑𝑣 (√𝒗) = 2 𝑑𝑥 2 √𝑣 𝑑𝑥 Ejercicio 14. 4 𝑦= 2 3 − 𝑥 𝑥2 Solución 𝑑𝑦 𝑑 2 3 = ( − ) 𝑑𝑥 𝑑𝑧 𝑥 𝑥 2 Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 Aplicando la deriva de cada elemento de la función 𝑑𝑦 𝑑 2 3 𝑑 2 𝑑 3 = ( − 2) = ( )− ( ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 2 𝑑 2 𝑑 3 𝑑 𝑑 (2𝑥 −1 ) − (3𝑥 −2 ) ( )− ( 2) = 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑑𝑧 𝑑 2 𝑑 3 ( )− ( ) = (−1)(2)𝑥 −1−1 − (−2)(3)𝑥 −2−1 𝑑𝑥 𝑥 𝑑𝑥 𝑥 2 𝑑 2 𝑑 3 ( )− ( ) = −2𝑥 −2 + 6𝑥 −3 𝑑𝑥 𝑥 𝑑𝑥 𝑥 2 𝑑 2 𝑑 3 2 6 ( )− ( 2) = − 2 + 3 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑥 𝑥 ∴ 𝑑𝑦 𝑑 2 3 2 6 = ( − 2) = − 2 + 3 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑥 𝑥 Ejercicio 15. 4 2 𝑠 = 2𝑡 3 − 3𝑡 3 Solución 4 2 𝑑𝑠 𝑑 = (2𝑡 3 − 3𝑡 3 ) 𝑑𝑡 𝑑𝑡 Formulas a utilizar: 𝑑 (𝑐𝑡 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑡 Aplicando la deriva de cada elemento de la función 4 2 4 2 𝑑 𝑑 4 2 (2𝑡 3 ) − (3𝑡 3 ) = ∙ 2𝑡 3−1 − ∙ 3𝑡 3−1 𝑑𝑡 𝑑𝑡 3 3 4 2 1 𝑑 𝑑 8 1 (2𝑡 3 ) − (3𝑡 3 ) = ∙ 𝑡 3 − 2𝑡 −3 𝑑𝑡 𝑑𝑡 3 5 ∴ 4 2 4 2 𝑑𝑠 𝑑 𝑑 𝑑 83 2 = (2𝑡 3 − 3𝑡 3 ) = (2𝑡 3 ) − (3𝑡 3 ) = √𝑡 − 3 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 3 √𝑡 Ejercicio 16. 3 1 𝑦 = 2𝑥 4 + 4𝑥 −4 Solución 3 1 𝑑𝑦 𝑑 = (2𝑥 4 + 4𝑥 −4 ) 𝑑𝑥 𝑑𝑥 Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 Aplicando la deriva de cada elemento de la función 3 1 3 1 𝑑 𝑑 3 1 (2𝑥 4 ) + (4𝑥 −4 ) = ∙ 2𝑥 4−1 − ∙ 4𝑥 −4−1 𝑑𝑥 𝑑𝑥 4 4 3 1 5 𝑑 𝑑 3 1 (2𝑥 4 ) + (4𝑥 −4 ) = 𝑥 −4 − 𝑥 −4 𝑑𝑥 𝑑𝑥 2 ∴ 3 1 3 1 𝑑𝑦 𝑑 𝑑 𝑑 3 1 = (2𝑥 4 + 4𝑥 −4 ) = (2𝑥 4 ) + (4𝑥 −4 ) = 4 − 4 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 2 √𝑥 √𝑥 5 Ejercicio 17. 2 2 𝑦 = 𝑥 3 − 𝑎3 Solución 2 2 𝑑𝑦 𝑑 = (𝑥 3 − 𝑎3 ) 𝑑𝑥 𝑑𝑥 Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑𝑣 𝑑𝑐 = =0 𝑑𝑥 𝑑𝑥 Aplicando la deriva de cada elemento de la función 6 2 2 𝑑 𝑑 2 2 (𝑥 3 ) − (𝑎3 ) = 𝑥 3−1 − 0 𝑑𝑥 𝑑𝑥 3 2 2 𝑑 𝑑 2 1 (𝑥 3 ) − (𝑎3 ) = 𝑥 −3 𝑑𝑥 𝑑𝑥 3 ∴ 2 2 2 2 𝑑𝑦 𝑑 𝑑 𝑑 2 = (𝑥 3 − 𝑎3 ) = (𝑥 3 ) − (𝑎3 ) = 3 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 3 √𝑥 Ejercicio 18 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 𝑦= 𝑥 Solución 𝑑𝑦 𝑑 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 = ( ) 𝑑𝑥 𝑑𝑥 𝑥 Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 Aplicando la deriva de cada elemento de la función 𝑑𝑦 𝑑 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 𝑑 𝑎 𝑑 𝑏𝑥 𝑑 𝑐𝑥 2 = ( )+ ( )+ ( )= ( ) 𝑑𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑 𝑎 𝑑 𝑏𝑥 𝑑 𝑐𝑥 2 𝑑 𝑑 (𝑏) + 𝑐 (𝑥) ( )+ ( )+ ( ) = −𝑎𝑥 −1−1 + 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑎 𝑑 𝑏𝑥 𝑑 𝑐𝑥 2 ( )+ ( )+ ( ) = −𝑎𝑥 −2 + 𝑐 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 ∴ 𝑑𝑦 𝑑 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 𝑑 𝑎 𝑑 𝑏𝑥 𝑑 𝑐𝑥 2 𝑎 = ( )+ ( )+ ( )= ( )=𝑐− 2 𝑑𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 𝑥 𝑥 Ejercicio 19. 𝒚= √𝒙 𝟐 − 𝟐 √𝒙 Solución 𝑑𝑦 𝑑 √𝑥 2 = ( − ) 𝑑𝑥 𝑑𝑥 2 √𝑥 Formulas a utilizar: 7 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 Aplicando la derivada de cada función 𝑑𝑦 𝑑 √𝑥 2 𝑑 √𝑥 𝑑 2 = ( ) ( − )= ( )− 𝑑𝑥 𝑑𝑥 2 𝑑𝑥 2 𝑑𝑥 √𝑥 √𝑥 1 𝑑 √𝑥 𝑑 2 1 1 1 1 ( ) = ( ) ( ) 𝑥 2−1 − (2) (− ) 𝑥 −2−1 ( )− 𝑑𝑥 2 𝑑𝑥 √𝑥 2 2 2 3 𝑑 √𝑥 𝑑 2 1 1 ( ) = 𝑥 −2 + 𝑥 −2 ( )− 𝑑𝑥 2 𝑑𝑥 √𝑥 4 𝑑 √𝑥 𝑑 2 1 1 1 1 1 1 ( )= + 3= +2 = + ( )− 𝑑𝑥 2 𝑑𝑥 √𝑥 4√𝑥 𝑥 2 4√𝑥 √𝑥 3 4√𝑥 𝑥 √𝑥 ∴ 𝑑𝑦 𝑑 √𝑥 2 1 1 = + ( − )= 𝑑𝑥 𝑑𝑥 2 4√𝑥 𝑥 √𝑥 √𝑥 Ejercicio 20 𝑠= 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 √𝑡 Solución 𝑑𝑠 𝑑 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 = ( ) 𝑑𝑡 𝑑𝑡 √𝑡 Formulas a utilizar: 𝑑 (𝑐𝑡 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑡 Aplicando la derivada de cada función 𝑑𝑠 𝑑 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 = ( ) = ( )+ ( )+ ( ) 𝑑𝑡 𝑑𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 √𝑡 1 1 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 𝑑 𝑑 𝑑 2 −1 ( ) + ( ) + ( ) = (𝑎𝑡 −2 ) + 𝑏 (𝑡 ∙ 𝑡 −2 ) + 𝑐 (𝑡 ∙ 𝑡 2 ) 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 𝑑𝑥 𝑑𝑥 1 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 𝑑 𝑑 1 𝑑 3 ( ) + ( ) + ( ) = (𝑎𝑡 −2 ) + 𝑏 (𝑡 2 ) + 𝑐 (𝑡 2 ) 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 𝑑𝑥 𝑑𝑥 1 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 1 1 1 3 3 ( ) + ( ) + ( ) = − 𝑎𝑡 −2−1 + 𝑏𝑡 2−1 + 𝑐𝑡 2−1 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 2 2 2 8 3 1 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 1 1 3 1 ( ) + ( ) + ( ) = − 𝑎𝑡 −2 + 𝑏𝑡 −2 + 𝑐𝑡 2 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 2 2 2 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 𝑎 𝑏 3𝑐 ( )+ ( )+ ( ) = − 2 + + √𝑡 3 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 2√𝑡 2 2√𝑡 Otra forma de expresarlo 𝑑 𝑎 𝑑 𝑏𝑡 𝑑 𝑐𝑡 2 𝑎 𝑏 3𝑐√𝑡 ( )+ ( )+ ( ) = − + + 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 𝑑𝑡 √𝑡 2 2𝑡√𝑡 2√𝑡 ∴ 𝑑𝑠 𝑑 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 𝑎 𝑏 3𝑐√𝑡 = ( + + )=− 𝑑𝑡 𝑑𝑡 2 2𝑡√𝑡 2√𝑡 √𝑡 Ejercicio 21 𝒚 = √𝒂𝒙 + 𝒂 √𝒂𝒙 Solución 𝑑𝑦 𝑑 𝑎 = (√𝑎𝑥 + ) 𝑑𝑥 𝑑𝑥 √𝑎𝑥 Formulas a utilizar: 𝑑 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 Aplicando la derivada de cada función 𝑑𝑦 𝑑 𝑎 𝑑 𝑑 𝑎 = (√𝑎𝑥 + )= ( ) (√𝑎𝑥) + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 √𝑎𝑥 √𝑎𝑥 1 1 1 𝑑 𝑑 𝑎 1 1 1 ( ) = (𝑎2 ) ( ) 𝑥 2−1 + (𝑎) (𝑎−2 ) (− ) 𝑥 −2−1 (√𝑎𝑥) + 𝑑𝑥 𝑑𝑥 √𝑎𝑥 2 2 1 1 𝑑 𝑑 𝑎 𝑎2 1 𝑎2 3 ( ) = 𝑥 −2 − 𝑥 −2 (√𝑎𝑥) + 𝑑𝑥 𝑑𝑥 √𝑎𝑥 2 2 𝑑 𝑑 𝑎 √𝑎 √𝑎 ( )= − (√𝑎𝑥) + 𝑑𝑥 𝑑𝑥 √𝑎𝑥 2√𝑥 2𝑥√𝑥 1 − √𝑎 = 𝑎 ∙ 𝑎 2 = 𝑎 √𝑎 Expresado de otra forma: 9 𝑑 𝑑 𝑎 𝑎 𝑎 𝑎 𝑎 ( )= − = − (√𝑎𝑥) + 𝑑𝑥 𝑑𝑥 √𝑎𝑥 2√𝑎√𝑥 2𝑥 √𝑎√𝑥 2√𝑎𝑥 2𝑥 √𝑎𝑥 ∴ 𝑑𝑦 𝑑 𝑎 𝑎 𝑎 = (√𝑎𝑥 + )= − 𝑑𝑥 𝑑𝑥 2√𝑎𝑥 2𝑥√𝑎𝑥 √𝑎𝑥 Ejercicio 22 𝒓 = √𝟏 − 𝟐𝜽 𝑑𝑟 𝑑 = (√1 − 2𝜃) 𝑑𝜃 𝑑𝜃 Formulas a utilizar: 𝑑 𝑑𝑟 (𝑐𝜃 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝜃 𝑛−1 𝑑𝜃 𝑑𝜃 1 𝑑𝑟 𝑑 1 = (√1 − 2𝜃) = (1 − 2𝜃)2−1 (−2) 𝑑𝜃 𝑑𝜃 2 1 𝑑𝑟 𝑑 = (√1 − 2𝜃) = −(1 − 2𝜃)−2 𝑑𝜃 𝑑𝜃 𝑑𝑟 𝑑 = (√1 − 2𝜃) = 𝑑𝜃 𝑑𝜃 ∴ −1 1 (1 − 2𝜃)2 𝑑𝑟 𝑑 1 = (√1 − 2𝜃) = − 𝑑𝜃 𝑑𝜃 √(1 − 2𝜃) Ejercicio 23 𝟑 𝒇(𝒕) = (𝟐 − 𝟑𝒕𝟐 ) 𝑑𝑠 𝑑 𝟑 = [(𝟐 − 𝟑𝒕𝟐 ) ] 𝑑𝑡 𝑑𝑡 Formulas a utilizar: 10 𝑑 𝑑𝑠 (𝑐𝑡 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑡 𝑑𝑡 𝑑𝑠 𝑑 = [(2 − 3𝑡 2 )3 ] = 3(2 − 3𝑡 2 )3−1 (−6𝑡) 𝑑𝑡 𝑑𝑡 𝑑𝑠 𝑑 [(2 − 3𝑡 2 )3 ] = −18𝑡(2 − 3𝑡 2 )2 = 𝑑𝑡 𝑑𝜃 ∴ 𝑑𝑠 𝑑 [(2 − 3𝑡 2 )3 ] = −18𝑡(2 − 3𝑡 2 )2 = 𝑑𝑡 𝑑𝜃 Ejercicio 24 3 𝑓(𝑥) = √4 − 9𝑥 𝑑𝑦 𝑑 3 = [ √4 − 9𝑥 ] 𝑑𝑥 𝑑𝑥 Formulas a utilizar: 𝑑 𝑑𝑦 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑥 𝑑𝑥 1 𝑑𝑦 𝑑 3 1 = [ √4 − 9𝑥 ] = (4 − 9𝑥)3−1 (−9) 𝑑𝑥 𝑑𝑥 3 2 𝑑𝑦 𝑑 3 = [ √4 − 9𝑥 ] = −3(4 − 9𝑥)−3 𝑑𝑥 𝑑𝑥 ∴ 𝑑𝑦 𝑑 3 = [ √4 − 9𝑥 ] = − 𝑑𝑥 𝑑𝑥 3 2 (4 − 9𝑥)3 Ejercicio 25 𝑦= 1 √𝑎2 − 𝑥 2 11 𝑑𝑦 𝑑 1 = [ ] 2 𝑑𝑥 𝑑𝑥 √𝑎 − 𝑥 2 Formulas a utilizar: 𝑑 𝑑𝑦 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑥 𝑑𝑥 1 𝑑𝑦 𝑑 1 1 = [ ] = − (𝑎2 − 𝑥 2 )−2−1 (−2𝑥) 𝑑𝑥 𝑑𝑥 √𝑎2 − 𝑥 2 2 3 𝑑𝑦 𝑑 1 = [ ] = −𝑥(𝑎2 − 𝑥 2 )−2 𝑑𝑥 𝑑𝑥 √𝑎2 − 𝑥 2 ∴ 𝑑𝑦 𝑑 1 = [ ]= 2 𝑑𝑥 𝑑𝑥 √𝑎 − 𝑥 2 𝑥 3 (𝑎2 − 𝑥 2 )2 Ejercicio 26 𝟑 𝒇(𝜽) = (𝟐 − 𝟓𝜽)𝟓 3 𝑑𝑟 𝑑 = ((2 − 5𝜃)5 ) 𝑑𝜃 𝑑𝜃 Formulas a utilizar: 𝑑 𝑑𝑟 (𝑐𝜃 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝜃 𝑛−1 𝑑𝜃 𝑑𝜃 3 3 𝑑𝑟 𝑑 3 = ((2 − 5𝜃)5 ) = (2 − 5𝜃)5−1 (−5) 𝑑𝜃 𝑑𝜃 5 3 2 𝑑𝑟 𝑑 = ((2 − 5𝜃)5 ) = −3(1 − 2𝜃)−5 𝑑𝜃 𝑑𝜃 ∴ 3 𝑑𝑟 𝑑 = ((2 − 5𝜃)5 ) = − 𝑑𝜃 𝑑𝜃 3 2 (2 − 5𝜃)5 Ejercicio 27 12 𝑏 2 𝑦 = (𝑎 − ) 𝑥 Solución 𝑑𝑦 𝑑 𝑏 2 = [(𝑎 − ) ] 𝑑𝑥 𝑑𝑥 𝑥 Formulas a utilizar: 𝑑 𝑑𝑦 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑𝑥 Aplicando la deriva de cada elemento de la función 𝑑𝑦 𝑑 𝑏 2 𝑏 2−1 (−1)(−1)(𝑏𝑥 −1−1 ) = [(𝑎 − ) ] = 2 (𝑎 − ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑑𝑦 𝑑 𝑏 2 𝑏 = [(𝑎 − ) ] = 2𝑏𝑥 −2 (𝑎 − ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑑𝑦 𝑑 𝑏 2 2𝑏 𝑏 ∴ = [(𝑎 − ) ] = 2 (𝑎 − ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑥 Ejercicio 28 𝑏 3 𝑦 = (𝑎 + 2 ) 𝑥 Solución 13 𝑑𝑦 𝑑 𝑏 3 = [(𝑎 + 2 ) ] 𝑑𝑥 𝑑𝑥 𝑥 Formulas a utilizar: 𝑑 𝑑𝑦 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑𝑥 Aplicando la deriva 𝑑𝑦 𝑑 𝑏 3 𝑏 3−1 (−2𝑏𝑥 −2−1 ) = [(𝑎 + 2 ) ] = 3 (𝑎 + 2 ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑑𝑦 𝑑 𝑏 3 𝑏 2 = [(𝑎 + 2 ) ] = −6𝑏𝑥 −3 (𝑎 + 2 ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 ∴ 𝑑𝑦 𝑑 𝑏 3 6𝑏 𝑏 2 = [(𝑎 + 2 ) ] = − 3 (𝑎 + 2 ) 𝑑𝑥 𝑑𝑥 𝑥 𝑥 𝑥 Ejercicio 29 𝑦 = 𝑥√𝑎 + 𝑏𝑥 𝑑𝑦 𝑑 = [𝑥√𝑎 + 𝑏𝑥 ] 𝑑𝑥 𝑑𝑥 Formulas a utilizar: 14 𝑑 𝑑𝑦 (𝑐𝑥 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑥 𝑛−1 𝑑𝑥 𝑑𝑥 𝑑 𝑑𝑣 𝑑𝑢 (𝑢𝑣) = 𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 1 𝑑𝑦 𝑑 𝑑 = [𝑥√𝑎 + 𝑏𝑥 ] = [𝑥(𝑎 + 𝑏𝑥)2 ] 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑢=𝑥 1 𝑣 = √𝑎 + 𝑏𝑥 = (𝑎 + 𝑏𝑥)2 1 1 1 𝑑 𝑑 𝑑 𝑥 [𝑥(𝑎 + 𝑏𝑥)2 ] = (𝑥) [(𝑎 + 𝑏𝑥)2 ] + (𝑎 + 𝑏𝑥)2 𝑑𝑥 𝑑𝑥 𝑑𝑥 1 1 1 𝑑 1 [𝑥(𝑎 + 𝑏𝑥)2 ] = (𝑥) ( ) (𝑏)(𝑎 + 𝑏𝑥)2−1 + (𝑎 + 𝑏𝑥)2 𝑑𝑥 2 1 1 1 𝑑 𝑏𝑥 [𝑥(𝑎 + 𝑏𝑥)2 ] = ( ) (𝑎 + 𝑏𝑥)−2 + (𝑎 + 𝑏𝑥)2 𝑑𝑥 2 1 𝑑 𝑏𝑥 + √𝑎 + 𝑏𝑥 [𝑥(𝑎 + 𝑏𝑥)2 ] = 𝑑𝑥 2√𝑎 + 𝑏𝑥 1 𝑑 𝑏𝑥 √𝑎 + 𝑏𝑥 𝑏𝑥 + 2𝑎 + 2𝑏𝑥 + = [𝑥(𝑎 + 𝑏𝑥)2 ] = 𝑑𝑥 1 2√𝑎 + 𝑏𝑥 2√𝑎 + 𝑏𝑥 1 𝑑 𝑏𝑥 √𝑎 + 𝑏𝑥 𝑏𝑥 + 2𝑎 + 2𝑏𝑥 2𝑎 + 3𝑏𝑥 + = = [𝑥(𝑎 + 𝑏𝑥)2 ] = 𝑑𝑥 1 2√𝑎 + 𝑏𝑥 2√𝑎 + 𝑏𝑥 2√𝑎 + 𝑏𝑥 ∴ 1 𝑑 2𝑎 + 3𝑏𝑥 [𝑥(𝑎 + 𝑏𝑥)2 ] = 𝑑𝑥 2√𝑎 + 𝑏𝑥 Ejercicio 30 𝑠 = 𝑡√𝑎2 + 𝑡 2 𝑑𝑠 𝑑 = [𝑡√𝑎2 + 𝑡 2 ] 𝑑𝑡 𝑑𝑡 15 Formulas a utilizar: 𝑑 𝑑𝑠 (𝑐𝑡 𝑛 ) = 𝑐 ∙ 𝑛 ∙ 𝑡 𝑛−1 𝑑𝑡 𝑑𝑡 𝑑 𝑑𝑣 𝑑𝑢 (𝑢𝑣) = 𝑢 +𝑣 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑠 𝑑 = [𝑡√𝑎2 + 𝑡 2 ] 𝑑𝑡 𝑑𝑡 1 𝑑𝑠 𝑑 𝑑 = [𝑡√𝑎2 + 𝑡 2 ] = [𝑡(𝑎2 + 𝑡 2 )2 ] 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑢=𝑡 1 𝑣 = √𝑎2 + 𝑡 2 = (𝑎2 + 𝑡 2 )2 1 1 1 𝑑 𝑑 𝑑 (𝑡) [𝑡(𝑎2 + 𝑡 2 )2 ] = (𝑡) [(𝑎2 + 𝑡 2 )2 ] + (𝑎2 + 𝑡 2 )2 𝑑𝑡 𝑑𝑡 𝑑𝑥 1 1 1 𝑑 1 −1 [𝑡(𝑎2 + 𝑡 2 )2 ] = ( ) (𝑡)(𝑎2 + 𝑡 2 )2 (2𝑡) + (𝑎2 + 𝑡 2 )2 𝑑𝑡 2 1 1 1 𝑑 [𝑡(𝑎2 + 𝑡 2 )2 ] = (𝑎2 + 𝑡 2 )−2 + (𝑎2 + 𝑡 2 )2 𝑑𝑡 1 𝑑 1 + √𝑎2 + 𝑡 2 [𝑡(𝑎2 + 𝑡 2 )2 ] = 𝑑𝑡 √𝑎2 + 𝑡 2 ∴ 1 𝑑 1 √𝑎2 + 𝑡 2 1 + 𝑎2 + 𝑡 2 + = [𝑡(𝑎2 + 𝑡 2 )2 ] = 𝑑𝑡 1 √𝑎2 + 𝑡 2 √𝑎2 + 𝑡 2 16