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/cos3
t

y''xx
= 4sint
= -1 _

16sin2
tcos5
t 4sintcos5
t

19.2.

x'= -t/√(1-t2
)

y'= -1/t2

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

19.3.

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

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

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

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

19.4.

x'= 2shtcht

y'= -2sht/ch3
t

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

4shtch4
t 2ch4
t

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
= 4
t _

√(1-t)3

19.8.

x'= cost

y'= sint/cos2
t

y''xx
= sint/cos4
t

19.9.

x'= 1/cos2
t

y'= -2cos2t/sin2
2t

y''xx
= -2cos2tcos4
t

sin2
2t

19.10.

x'= 1/2√(t-1)

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

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

sin2
t(1+2cost)2
sin2
t

19.13.

x'= 3t2
/ 2√(t3
-1)

y'= 1/t

y''xx
= 2
(
t
3

-1)

3t5

19.14.

x'= cht

y'= 2tht/ch2
t

y''xx
= 2tht/ch4
t

19.15.

x'= 1/2√(t-1)

y'= -1/2√t3

y''xx
= -
2t
+
2

√t3

19.16.

x'= -2cost sint

y'= 2sint/cos3
t

y''xx
= 2sint
= 1/2cos4
t

4cos4
tsint

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/cos3
t

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/sin3
t

19.22.

x'= -sint+sint+tcost= tcost

y'= cost-cost+tsint= tsint

y''xx
= sint/cos2
t

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
√sh4
t

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/sin2
t

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
)