Реферат: Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.10
Задача 10. Найти производную.
10.1.
y'= 1 *2-√5 thx *√5/ ch 2 x *(2-√5 thx )+ √5/ ch 2 x *(2+√5 thx ) =
4√5 2+√5thx (2-√5thx)2
= 1 _
ch2 x(2-√5thx)
10.2.
y'= ch5 x-4ch3x sh2 x + 3ch3 x-6chxsh2 x + 3chx = 1 + 3-3sh2 x + 3 _
4ch8 x 8ch4 x 8(1+sh2 x) 4ch5 x 8ch3 x 8chx
10.3.
1-√(thx) + 1+√(thx) _
y'= 1/2* 1-√(thx) * 2√(thx)ch2 x 2√(thx)ch2 x _ 1 =
1+√(thx) (1-√(thx))2 2√(thx)ch2 x
= √thx _
(1-th2 x)(ch2 x)
10.4.
√2-thx + √2+thx 2-th2 x + 2th2 x
y'= 3 *√2-thx * ch2 x ch2 x _ ch2 x ch2 x =
8√2 √2+thx (√2-thx)2 4(2-th2 x)2
= 1 _
2ch2 x(2-th2 x)2
10.5.
y'= 1 + 1-√2 thx * √2(1-√2thx+1+√2thx) =
2ch2 x 4√2(1+√2thx) ch2 x(1-√2thx)2
= 1-th2 x _
ch2 x(1-√2thx)2
10.6.
y'= _ 1 _ sh3 x-2shxch2 x = 2ch3 x+2chx-sh2 x
4thxch2 x 2sh4 x 4sh3 xchx
10.7.
y'= a-√(1+a2 )thx * √(1+a2 )thx(a-√(1+a2 )thx+a+√(1+a2 )thx) =
2a√(1+a2 )(a+√(1+a2 )thx) (a-√(1+a2 )thx)2
= thx = thx = thx _
(a2 -(1+a2 )th2 x)ch2 x a2 ch2 x-(1+a2 )sh2 x a2 -sh2 x
10.8.
y'= 1-√2cthx * √2(-1+√2cthx+1+√2cthx) = √2cthx =
18√2(1+√2cthx) sh2 x(1-√2cthx)2 9sh2 x(1-√2cthx)
= -√2cthx _
9(1+ch2 x)
10.9.
y'= 1 * ch2x/√(sh2x)-√(sh2x)(shx-chx) =
1+sh2x/(shx-chx)2 chx-shx
= (chx-shx)(ch2x-sh2x(shx-chx))
√(sh2x)(ch2 x+sh2 x)
10.10.
y'= 2+sh2x * -ch2x(2+sh2x)-ch2x(1-sh2x) = ch2x _
6(1-sh2x) (2+sh2x)2 12-6sh2x-sh2 2x
10.11.
y'= 4 √(1-thx)3 * 1-thx+1+thx = 1 _
44 √(1+thx)3 ch2 x(1-thx) 4ch2 x√(1+thx) 4 √(1-th2 x)
10.12.
y'= chx(1+chx)-sh2 x = 1 _
(1+chx)2 1+chx
10.13.
y'= shx√(sh2x)-chxch2x/√(sh2x) = shx-chxcth2x
sh2x
10.14.
y'= 3ch3x√(ch6x)-3sh6xsh3x/√(ch6x) = 3sh3x-3th6xsh3x
ch6x
10.15.
y'= 16shxch3 x*ln(chx)+8ch3 xshx-16ch3 xshxln(chx) = 4thx
2ch4 x
10.16.
y'= 2shxchx(12sh2 x+1)-24sh3 xchx = 4chx
3sh4 x 3sh3 x
10.17.
y'= 2chxsh2 x-ch3 x + 3 = shx-1 + 3 _
2ch4 x ch2 x√(1-th2 x) 2ch3 x ch2 x√(1-th2 x)
10.18.
y'= 1 * shx(1+3chx)-3shx(3+chx) =
√8√(1-(3+chx)2 /(1+3chx)2 ) (1+3chx)2
= -8shx = -1 _
8(1+3chx)√(ch2 x-1) 1+3chx
10.19.
y'= 4-√8th(x/2) * √8(4-√8th(x/2)+4+√8th(x/2)) =
√8(4+√8th(x/2)) 2ch2 (x/2)(4-√8th(x/2))2
= 1 _
ch2 (x/2)(4-√8th(x/2))
10.20.
y'= 1 _ shx(sh2 x-3chx-ch2 x) = 1+chx _
8ch2 (x/2)th(x/2) 4sh2 x(3+chx) shx(3+chx)
10.21.
y'= -(3+5chx)(3shx(3+5chx)-5shx(5+3chx)) = 1
4(3+5chx)2 √(9+30chx+25ch2 x-25-30chx-9ch2 x)
10.22.
y'= -16ch5 xshx-4ch3 x(1-8ch2 x) = -4ch2 xshx-1+8ch2 x
4ch8 x ch5 x
10.23.
y'= -2/sh2 x+1/sh4 x+ch3 x-2chxsh2 x +5chx = -2/sh2 x+1/sh4 x+1-sh2 x +5_
2ch4 x 2+2sh2 x 2ch3 x 2chx
10.24.
y'= -16 + sh4 x+3sh2 xch2 x = 1-4sh2 x
3sh2 2x 3ch2 xsh6 x ch2 xsh4 x
10.25.
y'= chx _ 1-sh2 x = 1 _ 1-sh2 x
2+2sh2 x 2ch3 x 2ch2 x 2ch3 x
10.26.
y'= 3 + shx – sh3 x-2shxch2 x = 1+sh2 x
4ch2 (x/2)th(x/2) 2sh4 x shx
10.27.
y'= 2chxsh2 x-ch3 x + 2shxchx _ 3chx = sh2 x-1 + 2chx _ 3 _
2ch4 x sh4 x 2+2sh2 x 2ch3 x sh3 x 2chx
10.28.
y'= ch3 x-2chxsh2 x + chx = 1 _
2ch4 x 2+2sh2 x ch3 x
10.29.
y'= ch3 x-2chxsh2 x + chx = 1 _
2ch4 x 2+2sh2 x ch3 x
10.30.
y'= 2ch2 xshx-sh3 x _ 1 = 1/sh3 x
2sh4 x 4ch2 (x/2)th(x/2)
10.31.
y'= -2 _ sh4 x-3ch2 xsh2 x = 1/sh4 x
3sh2 x 3sh6 x