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 )


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