Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.19

Задача 19. Найти производную второго порядка  от функции, заданной параметрически.

19.1. 

x'= -2sin2t= -4sintcost

y'= 4sint/cos3t

y''xx=      4sint     =       -1    _

         16sin2tcos5t   4sintcos5t

19.2. 

x'= -t/√(1-t2)

y'= -1/t2

y''xx= (1-t2)2             t4

19.3. 

x'= etcost-etsint= et(cost-sint)

y'= etsint+etcost= et(sint+cost)

y''xx= et(sint+cost) = sint+cost

         e2t(cost-sint)2   et(cost-sint)2

19.4. 

x'= 2shtcht

y'= -2sht/ch3t

y''xx=    -2sht   =  -1_

         4shtch4t    2ch4t

19.5. 

x'= 1+cost

y'= sint

y''xx= sint/(1+cost)2

19.6. 

x'= -1/t2

y'= -2t/(1+t2)2

y''xx=   -2t3 _

         (1+t2)2

19.7. 

x'= 1/2√t

y'= 1/√(1-t)3

y''xx=    4t _

         √(1-t)3

19.8. 

x'= cost

y'= sint/cos2t

y''xx= sint/cos4t

19.9. 

x'= 1/cos2t

y'= -2cos2t/sin22t

y''xx= -2cos2tcos4t

              sin22t

19.10. 

x'= 1/2√(t-1)

y'= (2-t)/(1-t)3/2

y''xx= 4(t-1)(2-t) = 2t-8

              (1-t)3/2    √(1-t)

19.11. 

x'= 1/2√t

y'= 1/3√(t-1)2

y''xx= 4t/3√(t-1)2

19.12. 

x'= -sint/(1+2cost)2

y'= (cost+2)/(1+2cost)2

y''xx= (cost+2)(1+2cost)4 = (cost+2)(1+2cost)2

            sin2t(1+2cost)2                  sin2t

19.13. 

x'= 3t2 / 2√(t3-1)

y'= 1/t

y''xx= 2(t3-1)

            3t5

19.14. 

x'= cht

y'= 2tht/ch2t

y''xx= 2tht/ch4t

19.15. 

x'= 1/2√(t-1)

y'= -1/2√t3

y''xx= -2t+2

           √t3

19.16. 

x'= -2cost sint

y'= 2sint/cos3t

y''xx=     2sint    = 1/2cos4t

         4cos4tsint

19.17. 

x'= 1/2√(t-3)

y'= 1/(t-2)

y''xx= 4(t-3)/(t-2)

19.18. 

x'= cost

y'= -sint/cost

y''xx= -sint/cos3t

19.19. 

x'= 1+cost

y'= -sint

y''xx= -sint/(1+cost)2

19.20. 

x'= 1-cost

y'= sint

y''xx= sint/(1-cost)2

19.21. 

x'= -sint

y'= cost/sint

y''xx= -cost/sin3t

19.22. 

x'= -sint+sint+tcost= tcost

y'= cost-cost+tsint= tsint

y''xx= sint/cos2t

19.23. 

x'= et

y'= 1/√(1-t2)

y''xx= et/√(1-t2)

19.24. 

x'= -sint

y'= 2sin3(t/2)cos(t/2)

y''xx= -2sin3(t/2)cos(t/2)/sint= -sin2(t/2)

19.25. 

x'= sht

y'= 2cht/33√sht

y''xx= 2cht/33√sh4t

19.26. 

x'= 1/(1+t2)

y'= t

y''xx= t(1+t2)2

19.27. 

x'= 2-2cost

y'= -4sint

y''xx= -2sint/(1-cost)

19.28. 

x'= cost-cost+tsint= tsint

y'= -sint+sint+tcost= tcost

y''xx= cost/sin2t

19.29. 

x'= -2/t3

y'= -2t/(t2+1)2

y''xx= -t7/2(t2+1)2

19.30. 

x'= cost-sint

y'= 2cos2t

y''xx= 2cos2t/( cost-sint)= 2cost+2sint

19.31. 

x'= 1/t

y'= 1/(1+t2)

y''xx= t2/(1+t2)