Задача Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.10
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Задача 10. Найти производную.
10.1.
y'= 1 *2-√5thx*√5/ch2x*(2-√5thx)+ √5/ch2x*(2+√5thx) =
4√5 2+√5thx (2-√5thx)2
= 1 _
ch2x(2-√5thx)
10.2.
y'= ch5x-4ch3xsh2x + 3ch3x-6chxsh2x + 3chx = 1 + 3-3sh2x + 3 _
4ch8x 8ch4x 8(1+sh2x) 4ch5x 8ch3x 8chx
10.3.
1-√(thx) + 1+√(thx) _
y'= 1/2* 1-√(thx) * 2√(thx)ch2x 2√(thx)ch2x _ 1 =
1+√(thx) (1-√(thx))2 2√(thx)ch2x
= √thx _
(1-th2x)(ch2x)
10.4.
√2-thx + √2+thx 2-th2x + 2th2x
y'= 3 *√2-thx * ch2x ch2x _ ch2x ch2x =
8√2 √2+thx (√2-thx)2 4(2-th2x)2
= 1 _
2ch2x(2-th2x)2
10.5.
y'= 1 + 1-√2thx * √2(1-√2thx+1+√2thx) =
2ch2x 4√2(1+√2thx) ch2x(1-√2thx)2
= 1-th2x _
ch2x(1-√2thx)2
10.6.
y'= _ 1 _ sh3x-2shxch2x = 2ch3x+2chx-sh2x
4thxch2x 2sh4x 4sh3xchx
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)th2x)ch2x a2ch2x-(1+a2)sh2x a2-sh2x
10.8.
y'= 1-√2cthx * √2(-1+√2cthx+1+√2cthx) = √2cthx =
18√2(1+√2cthx) sh2x(1-√2cthx)2 9sh2x(1-√2cthx)
= -√2cthx _
9(1+ch2x)
10.9.
y'= 1 * ch2x/√(sh2x)-√(sh2x)(shx-chx) =
1+sh2x/(shx-chx)2 chx-shx
= (chx-shx)(ch2x-sh2x(shx-chx))
√(sh2x)(ch2x+sh2x)
10.10.
y'= 2+sh2x * -ch2x(2+sh2x)-ch2x(1-sh2x) = ch2x _
6(1-sh2x) (2+sh2x)2 12-6sh2x-sh22x
10.11.
y'= 4√(1-thx)3 * 1-thx+1+thx = 1 _
44√(1+thx)3 ch2x(1-thx) 4ch2x√(1+thx) 4√(1-th2x)
10.12.
y'= chx(1+chx)-sh2x = 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'= 16shxch3x*ln(chx)+8ch3xshx-16ch3xshxln(chx) = 4thx
2ch4x
10.16.
y'= 2shxchx(12sh2x+1)-24sh3xchx = 4chx
3sh4x 3sh3x
10.17.
y'= 2chxsh2x-ch3x + 3 = shx-1 + 3 _
2ch4x ch2x√(1-th2x) 2ch3x ch2x√(1-th2x)
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)√(ch2x-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(sh2x-3chx-ch2x) = 1+chx _
8ch2(x/2)th(x/2) 4sh2x(3+chx) shx(3+chx)
10.21.
y'= -(3+5chx)(3shx(3+5chx)-5shx(5+3chx)) = 1
4(3+5chx)2√(9+30chx+25ch2x-25-30chx-9ch2x)
10.22.
y'= -16ch5xshx-4ch3x(1-8ch2x) = -4ch2xshx-1+8ch2x
4ch8x ch5x
10.23.
y'= -2/sh2x+1/sh4x+ch3x-2chxsh2x+ 5chx = -2/sh2x+1/sh4x+1-sh2x+ 5_
2ch4x 2+2sh2x 2ch3x 2chx
10.24.
y'= -16 + sh4x+3sh2xch2x = 1-4sh2x
3sh22x 3ch2xsh6x ch2xsh4x
10.25.
y'= chx _ 1-sh2x = 1 _ 1-sh2x
2+2sh2x 2ch3x 2ch2x 2ch3x
10.26.
y'= 3 + shx – sh3x-2shxch2x = 1+sh2x
4ch2(x/2)th(x/2) 2sh4x shx
10.27.
y'= 2chxsh2x-ch3x + 2shxchx _ 3chx = sh2x-1 + 2chx _ 3 _
2ch4x sh4x 2+2sh2x 2ch3x sh3x 2chx
10.28.
y'= ch3x-2chxsh2x + chx = 1 _
2ch4x 2+2sh2x ch3x
10.29.
y'= ch3x-2chxsh2x + chx = 1 _
2ch4x 2+2sh2x ch3x
10.30.
y'= 2ch2xshx-sh3x _ 1 = 1/sh3x
2sh4x 4ch2(x/2)th(x/2)
10.31.
y'= -2 _ sh4x-3ch2xsh2x = 1/sh4x
3sh2x 3sh6x
