Контрольная работа по "Математике"
Автор: Никита Кузнецов • Февраль 27, 2019 • Контрольная работа • 2,552 Слов (11 Страниц) • 280 Просмотры
Задача 6. Найти производную.
6.1. [pic 1]
ex + 2e2x+ex
y' = 1- √(e2x+ex+1) = 2+ex+√(e2x+ex+1)-ex√(e2x+ex+1)-2e2x-ex =
2+ex+2√(e2x+ex+1) 2+ex+2√(e2x+ex+1)
= (2-ex)√(e2x+ex+1)+2+ex-2ex
2+ex+2√(e2x+ex+1)
6.2. [pic 2]
y' = 1/4*e2x(2-sin2x-cos2x)+1/8*e2x(-2cos2x+2sin2x)=1/8*e2x(4-2sin2x-2cos2x-2cos2x+2sin2x)=1/8*e2x(4-4cos2x)=e2x*sin2x
6.3. [pic 3]
y' = 1 * 1 * 2ex = ex .
2 1 + (ex-3)2 4 e2x-6ex+10
4
6.4. [pic 4]
y' = 1 * 1-2x * -2xln2(1+2x)-(1-2x)2xln2 = (2x-1)2xln4 = 2x(2x-1)
ln4 1+2x (1+2x)2 ln4(1+2x)3 (1+2x)3
6.5. [pic 5]
ex(√(ex+1)+1) _ ex(√(ex+1)-1)
y' = ex + √(ex+1)+1 * 2√(ex+1) 2√(ex+1) =
√(ex+1) √(ex+1)-1 (√(ex+1)+1)2
= ex + ex√(ex+1)+ex-ex√(ex+1)+ex = √(ex+1)
√(ex+1) 2ex√(ex+1)
6.6. [pic 6]
y' = 2/3*3/2*√(arctgex) * ex = ex√(arctgex)
1+ex 1+ex
6.7. [pic 7]
y' = 2ex - 2ex = ex
2(e2x+1) 1+e2x 1+e2x
6.8. [pic 8]
6.9. [pic 9]
y' = 2/ln2*((2xln2)/(2√(2x-1))-(2xln2)/(1+2x-1))=2x/√(2x-1)-2
6.10. [pic 10]
ex(√(1+ex)+1) _ ex(√(1+ex)-1)
y'= 2√(1+ex)+2ex(x-2) _ √(1+ex)+1 * 2√(1+ex) 2√(1+ex) =
2√(1+ex) √(1+ex)-1 (√(1+ex)+1)
= xex+2 _ 2ex√(1+ex)+2ex = xex .
√(1+ex) ex√(1+ex)( √(1+ex)+1) √(1+ex)
6.11. [pic 11]
y'= αeαx(αsinβx-βcosβx)+eαx(αβcosβx+β2sinβx) =
α2+β2
= eαx(α2sinβx+β2sinβx) = eαxsinβx
α2+β2
6.12. [pic 12]
y'= αeαx(βsinβx-αcosβx)+eαx(β2cosβx+αβsinβx) =
α2+β2
= eαx(β2cosβx+2αβsinβx-α2cosβx)
α2+β2
...