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# Tema 3 (Soluciones cálculo de derivadas)

## by jhbenito on Nov 06, 2008

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## Tema 3 (Soluciones cálculo de derivadas)Document Transcript

• Tema 3. Derivadas. “Soluciones cálculo de derivadas” 4 1 1. f ' ( x) = 9 x 2 + x −1+ 3 3 x2 3 3 18 2. f ' ( x) = x 3 + 3 x + − x2 x4 3 5 5 3. f ' ( x) = x− 3 + 2 2x x x 2 3 x 2 9 3 4. f ' ( x) = x− 4 10 10 x 5. ( ⎛ f ' ( x) = 2 x − x senx + ⎜ x 2 + ⎜ ) 1 ⎞ ⎟ cos x ⎟ ⎝ 2 x⎠ f ' ( x) = x 2 (1 + 3 ln x ) + tan x 1 6. − 2x x x cos 2 x 3 x cot 2 x + 2 cot x + 3 x 7. f ' ( x) = − − ex 3 2 3x x 8. f ' ( x) = 2e x cos x ⎛ 1 ⎞ 9. f ' ( x) = 4 x ⎜ ln 4 ⋅ arcsenx + ⎜ ⎟ ⎟ ⎝ 1− x2 ⎠ 1 1 10. f ' ( x) = arctan x + x 2 x 1+ x2 − 20 x 2 + 16 x − 5 11. f ' ( x) = (4 x 2 −1 )2 2e x ( x − 1) 12. f ' ( x) = (x − e )x 2 Fundamentos matemáticos en Arquitectura I Jesús Hernández Benito
• x2 arcsenx − ( x − arctan x ) 1 1+ x 2 1− x2 13. f ' ( x) = (arcsenx )2 1 (1 − arctan x ) + x 2 x 1+ x2 14. f ' ( x) = (1 − arctan x) 2 − 2 x + 1 − 3 ln x 15. f ' ( x) = x4 −2 16. f ' ( x) = 1 − sen 2 x 17. f ' ( x) = (2 + tan 2 ) x + cot 2 x x senx − (tan x − cot x )(senx + x cos x ) x 2 sen 2 x 2 x − 2 + ln x 18. f ' ( x) = x2 (2 x arctan x + 1) ln x − 1 + x 2 arctan x 19. f ' ( x) = x ln 2 x 20. f ' ( x) = e x [(1 + x ) senx + x cos x ] 21. f ' ( x) = (3x senx + x 2 3 ) cos x ln x − x 2 senx ln 2 x 1+ x 22. f ' ( x) = e x 2 2 x 23. ( f ' ( x) = 42 4 x 3 + 6 x − 2 ) (2 x 6 2 ) +1 2 x 3 − 3x 24. f ' ( x) = x 4 − 3x 2 + 6 − 2x 25. f ' ( x) = ( 3 3 x2 − 5 ) 4 Fundamentos matemáticos en Arquitectura I Jesús Hernández Benito
• f ' ( x) = 5(senx − cos x ) (cos x + senx ) 4 26. 2⎛ 3x ⎞ 27. f ' ( x) = (arctan x ) ⎜ arctan x + ⎟ ⎝ 1+ x2 ⎠ 28. ( f ' ( x) = 1 − x 2 ) (arcsenx ) (3 4 2 1 − x 2 − 10 xarcsenx ) 29. f ' ( x) = cot x log10 e 30. f ' ( x) = cos x 2 x sen x − x sen x − 2 cos x f ' ( x) = (x − cos x ) 31. 2 −1 32. f ' ( x) = 1− x2 1 + sen10 x 33. f ' ( x) = −30 (1 − sen10 x) 2 −1 34. f ' ( x) = x x2 −1 −1 35. f ' ( x) = 2 1− x2 − ex 36. f ' ( x) = 2e x − e 2 x 2 37. f ' ( x) = x 2x 2 − 1 ex 38. f ' ( x) = e2x − 1 1 arcsen − ln 8 ⋅ 8 x 39. f ' ( x) = x x2 −1 Fundamentos matemáticos en Arquitectura I Jesús Hernández Benito
• sen 2 x 40. f ' ( x) = 1 − cos 4 x −x 41. f ' ( x) = 2x − x 2 42. f ' ( x) = 2 a 2 − x 2 1 43. f ' ( x) = ln 3 2ax + x 2 x +1 44. f ' ( x) = x − a2 2 1 1 45. f ' ( x) = + 1 − x arcsenx 2 x 1 − ln 2 x x a + a + x2 46. f ' ( x) = a + x2 + ( x a + x2 1+ a + x2 ) arcsenx 47. f ' ( x) = (1 − x ) 2 3 1 48. f ' ( x) = 1+ x2 1 1 −2 49. f ' ( x) = ⎛ 1 − x ⎞ ln 1 − x 1 − x 2 ln⎜ ln ⎟ ⎝ 1+ x ⎠ 1+ x 50. f ' ( x) = 8sen (sen 2 (sen 2 x )) ⋅ cos(sen 2 (sen 2 x )) ⋅ sen (sen 2 x ) ⋅ cos(sen 2 x ) ⋅ senx ⋅ cos x ⎛ 1 ⎞ 1 51. f ' ( x) = −sen⎜ ⎜ arccos(senx ) ⎟ arccos 2 (senx ) ⎟ ⎝ ⎠ 5 1 1 52. f ' ( x) = 2 9 ⎛5 x 4⎞ 2 x 1 + ⎜ tan + ⎟ cos 2 ⎝3 2 3⎠ Fundamentos matemáticos en Arquitectura I Jesús Hernández Benito
• ⎛ 11 ⎛ 1 ⎞⎞ 53. f ' ( x) = ⎜1 + ⎜1 + ⎟⎟ ⎜ 2 x+ x ⎜ ⎟ 2 x+ x+ x ⎝ ⎝ 2 x ⎠⎟⎠ 2 tan x(1 + tan 2 x ) 54. f ' ( x) = 1 + tan 4 x 55. f ' ( x) = 3x 3 x (1 + ln x) 56. [ f ' ( x) = x x x x ln x(1 + ln x) + x x −1 x ] 1 x −1 ⎛ 1 ⎞ 57. f ' ( x) = x ⎜1 + ln x ⎟ 2 ⎝ 2 ⎠ f ' ( x) = x x +1 (1 + 2 ln x ) 2 58. 1 ⎛ 1 ⎞ senx ⎛ cos x 1 1 ⎞ 59. f ' ( x ) = −⎜ ⎟ ⎜ ln + ⎟ ⎝ x ⎠ ⎝ sen x x xsenx ⎠ 2 Fundamentos matemáticos en Arquitectura I Jesús Hernández Benito