Differentiate: 4 27. 2 1. (π₯ + 2)(2π₯ − 3π₯ + 1) 2. π₯π 3 π₯ 28. 3. 4. 2 3π₯ sin π₯ π₯ ln π₯ 29. 5. 2π (1 + cos π₯) 30. 3 6. 2π₯ (sin π₯ + 2 π₯) 7. ( 8. (π₯ + π₯) sin π₯ 9. )( 1 π₯ 4 2 3 2 2π₯ +1 11. 12. π₯π (π₯ + 1) ln π₯ 13. (2 π₯ − 2) cos π₯ 16. 3 π₯ π₯ 3 π₯ 17. sin π₯ cos π₯ 18. ( 19. 20. 21. 1 2 π₯ + 2) cos π₯ π₯ 2 3π (π₯ + ) 36. 23. (1 + 1 4 π₯ 3 sinπ₯ 2 π₯ +1 3 25. (2π₯ − 1)(1 + ln π₯) 26. 2π₯ cos π₯ +2 π₯ π₯ −3 sin π₯ cosπ₯ π₯ π₯ lnπ₯ π₯ π cos π₯ 38. π₯ −4π₯ +1 6 2 2 π₯ −1 π₯ 2 3π₯ 3 π₯ −2 sinπ₯ sinπ₯+cosπ₯ π₯ π 3 42. (π₯ − 2)(2π₯ + 1) 43. 3π₯ π 47. )(4π₯ + 3) 3/2 π₯ 37. 46. π₯ 3π₯−1 π₯ 45. − 2π₯π tan π₯ ln π₯ 34. 44. 2 ln π₯ π₯ tanπ₯+1 π₯ 22. 24. 1 π₯ 4 π₯ π₯−2 2π₯ sin π₯ + π 41. π π₯ tan π₯ 33. 40. 4 ln π₯ sin π₯ π₯ 3 3π ln π₯ 39. 3 2π₯ −5 π₯ π 5 32. 35. π₯ (2π₯ − 1) sin π₯ 15. ) + 1 3π₯ − 2 10. 14. 31. π₯ π₯ 2 2 π₯ 3 π₯ cosπ₯ 2 π₯ +2 π₯ sin π₯ cos π₯ π₯ π π₯ cosπ₯ sinπ₯+1 48. π₯ ln π₯ − π₯ 49. π tan π₯ 50. π₯ ln π₯ 51. 3 52. π₯ 2 π₯ π sinπ₯ 4 (2π₯ − 3π₯) sin π₯ Solutions (OK to have different format due to expansion/factoring): 3 2 4 1. 4π₯ (2π₯ − 3π₯ + 1) + (π₯ + 2)(4π₯ − 3) 2. (π₯ + 3π₯ )π 3. 4. 6π₯ sin π₯ + 3π₯ cos π₯ ln π₯ + 1 5. 2π (1 + cos π₯ − sin π₯) 6. 6π₯ (sin π₯ + 2 π₯) + 2π₯ (cos π₯ + 1/ π₯) = 2π₯ (3 sin π₯ + π₯ cos π₯ + 7 π₯) 7. −π₯ 8. (π₯ + π₯) cos π₯ + (4π₯ + 1) sin π₯ 9. 3 2 π₯ 2 π₯ 2 3 (3π₯2 − 2) + 6π₯( 1π₯ −2 4 2 ) 2 2 π₯ (2π₯ +3) 2 2 (2π₯ +1) (2π₯ − 1) cos π₯ + 2 sin π₯ 11. π₯ π + 3π₯ π = (π₯ + 3π₯ )π 12. ln π₯ + (π₯ + 1)/π₯ = 3 π₯ 13. cosπ₯ 14. 5 2 π₯ 3 ( 1 π₯ 1 π₯ )π 17. cos π₯ − sin π₯ 19. 20. + 1 + ln π₯ + 4 ln π₯ cos π₯ 3 − 2 π₯ 2 −2 π₯ 1 π₯ π₯ + 4π₯ 2 π₯ 4sinπ₯ π₯ 3 2 − (2 π₯ − 2) sin π₯ π₯ 16. 18. + 6π₯ + 3 3 10. 15. −2 + 1 = 2π₯ 2 cos π₯ − ( π₯ 2 3π (π₯ + 1 π₯ 1 2 π₯ + 2) sin π₯ + 2π₯ − 1 2 π₯ ) 2(1−lnπ₯) 2 π₯ tanπ₯+1 2 sec π₯+1 π₯ 21. − 22. − 2π (π₯ + 1) 23. (1 + π₯ )12π₯ − 4π₯ (4π₯ + 3) = 4(3π₯ − π₯ 2 π₯ π₯ −4 + 2 −5 3 2 −2 −5 − 3π₯ ) 24. 2 (π₯ +1)cosπ₯−2π₯ sinπ₯ 2 (π₯2+1) 3 2 2 2 25. (2π₯ − 1)/π₯ + 6π₯ (1 + ln π₯) = 8π₯ − 1/π₯ + 6π₯ ln π₯ 26. − 6π₯ cos π₯ − 2π₯ sin π₯ 27. π 28. 3π₯ tan π₯ + π₯ sec π₯ 2 ( π₯ 2 3 π₯ ) 10 − 5 6 π₯ 2 3 4 2 3 29. 3π₯ −8π₯ 30. tanπ₯ π₯ 31. 3 π₯ 2 (π₯−2) 2 + sec π₯ ln π₯ 3/2 ( ) 1/2/2)(3π₯−1) = 2 (π₯3/2+2) π₯ 1 π₯ 32. 3π (ln π₯ + 33. 2 sin π₯ + 2π₯ cos π₯ + π 34. π₯ −3 −4 36. lnπ₯−1 39. 40. 1/2 +π₯ 3/2 2(π₯ +4) 2 +2) ) cos π₯ − 3π₯ sinπ₯ π₯ 38. 3(−π₯ π₯ 35. 37. 3/2 +2 −(3π₯ − sin π₯ cosπ₯ 2 π₯ 2 ln π₯ π₯ π (cos π₯ − sin π₯) 7 5 4π₯ −6π₯ +6π₯ 2 π₯ −1 π₯ 2 3 (π₯ ln 3 + 2π₯) 2 3 3π₯ sinπ₯−cosπ₯(π₯ −2) 2 sin π₯ π₯ π₯ π (cosπ₯−sinπ₯)−π (sinπ₯+cosπ₯) 41. 2π₯ π 2 −π₯ = 2π 3 sin π₯ 3 2 42. 3π₯ (2π₯ + 1) + 2(π₯ − 2) = 8π₯ + 3π₯ − 2 43. 3π (π₯ + 2π₯) 44. (π₯2+2)(3π₯2cosπ₯−π₯3sinπ₯)−2π₯4cosπ₯ 2 (π₯2+2) 45. − π₯ sin π₯ + π₯cos π₯ + sin π₯ cos π₯ 46. π π₯ 2 2 ( π₯ 1 π₯ 2 − 2 1 2 π₯ ) 47. 2 −(sinπ₯+1)sinπ₯−cos π₯ 2 (sinπ₯+1) 48. ln π₯ 49. 50. π (tan π₯ + sec π₯) π₯ + 2π₯ ln π₯ 51. 52. π₯ 2 π₯ π₯ π sinπ₯−π cosπ₯ 2 sin π₯ 4 π₯ = π csc π₯(1 − cot π₯) 3 (2π₯ − 3π₯) cos π₯ + (8π₯ − 3) sin π₯