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

СОДЕРЖАНИЕ: Задача 19 . Найти производную второго порядка от функции, заданной параметрически. 19.1. x= -2sin2t= -4sintcost y= 4sint/cos3t yxx= 4sint = -1 _ 16sin2tcos5t 4sintcos5t

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= e^t cost-e^t sint= e^t (cost-sint)

y= e^t sint+e^t cost= e^t (sint+cost)

y_xx = e^t (sint+cost) = sint+cost

e^2t (cost-sint)² e^t (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= 1/2t

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= 1/2t

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= -1/2t³

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= e^t

y= 1/(1-t² )

y_xx = e^t /(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 )

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