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Реферат: Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.4
Задача 4 . Вычислить приближенно с помощью дифференциала.
4.1. 
x0 = 8
∆x= 7,76-8= -0,24
f(x0 )= 3 √8=2
f'= 1/(33 √х2 )
f'(x0 )= 1/(3*4)=1/12
f(x)= 2-0,24/12= 1,98
4.2. 
x0 = 1
∆x= 1,012-1= 0,012
f(x0 )= 3 √(1+7)= 2
f'= (3х2 +7)/(33 √(х3 +7х))
f'(x0 )= (3+7)/(3*2)= 5/3
f(x)= 2+0,06/3= 2,02
4.3. 
x0 = 1
∆x= 0,98-1= -0,02
f(x0 )= 1/2*(1+√4)=3/2
f'= 1/2-х/(2√(5-х2 ))
f'(x0 )= 1/2-1/4=1/4
f(x)= 3/2-0,02/4= 1,495
4.4. 
x0 = 27
∆x= 27,54-27= 0,54
f(x0 )= 3
f'= 1/(33 √х2 )
f'(x0 )= 1/27
f(x)= 3+0,54/27= 3,02
4.5. 
x0 = 0
∆x= 0,08
f(x0 )= arcsin0=0
f'= 1/√(1-х2 )
f'(x0 )= 1
f(x)= 0+0,08=0,08
4.6. 
x0 = 1
∆x= 0,97-1= -0,03
f(x0 )= 3 √8=2
f'= (2х+2)/33 √(х2 +2х+5)
f'(x0 )= 4/(3*2)=2/3
f(x)= 2-0,06/3= 1,98
4.7. 
x0 = 27
∆x= 26,46-27= -0,54
f(x0 )= 3 √27=3
f'= 1/(33 √х2 )
f'(x0 )= 1/27
f(x)= 3-0,54/27= 2,98
4.8. 
x0 = 2
∆x= 1,97-2= -0,03
f(x0 )= √(4+2+3)
f'= (2х+1)/2√(х2 +х+3)
f'(x0 )= 5/(2*3)= 5/6
f(x)= 3-0,15/6= 2,975
4.9. 
x0 = 1
∆x= 1,021-1= 0,021
f(x0 )= 1
f'= 11х10
f'(x0 )= 11
f(x)= 1+11*0,021= 1,231
4.10. 
x0 = 1
∆x= 1,21-1= 0,21
f(x0 )= 1
f'= 1/(33 √х2 )
f'(x0 )= 1/3
f(x)= 1+0,21/3= 1,07
4.11. 
x0 = 1
∆x= 0,998-1= -0,002
f(x0 )= 1
f'= 21х20
f'(x0 )= 21
f(x)= 1-0,002/21= 0,9999
4.12. 
x0 = 1
∆x= 1,03-1= 0,03
f(x0 )= 1
f'= 2/(33 √х)
f'(x0 )= 2/3
f(x)= 1+0,03*2/3= 1,02
4.13. 
x0 = 2
∆x= 2,01-2= 0,01
f(x0 )= 26 =64
f'= 6х5
f'(x0 )= 6*25 =192
f(x)= 64+0,01*192= 65,92
4.14. 
x0 = 8
∆x= 8,24-8= 0,24
f(x0 )= 3 √8= 2
f'= 1/(33 √х2 )
f'(x0 )= 1/12
f(x)= 2+0,24/12= 2,02
4.15. 
x0 = 2
∆x= 1,996-2= -0,004
f(x0 )= 27 =128
f'= 7х6
f'(x0 )= 7*26 = 448
f(x)= 128-0,004*448= 126,208
4.16. 
x0 = 8
∆x= 7,64-8= -0,36
f(x0 )= 3 √8= 2
f'= 1/(33 √х2 )
f'(x0 )= 1/12
f(x)= 2-0,36/12= 1,97
4.17. 
x0 = 2,5
∆x= 2,56-2,5= 0,06
f(x0 )= √(10-1)= 3
f'= 1/√(4х-1)
f'(x0 )= 1/√9= 1/3
f(x)= 3+0,06/3= 3,02
4.18. 
x0 = 1
∆x= 1,016-1= 0,016
f(x0 )= 1/√(2+1+1)= 1/2
f'= -(4х+1)/2√(2х2 +х+1)3
f'(x0 )= (-4-1)/2√(2+1+1)3 = -5/16
f(x)= 0,5-0,08/16= 0,495
4.19. 
x0 = 8
∆x= 8,36-8= 0,36
f(x0 )= 3 √8= 2
f'= 1/(33 √х2 )
f'(x0 )= 1/12
f(x)= 2+0,36/12= 2,03
4.20. 
x0 = 4
∆x= 4,16-4= 0,16
f(x0 )= 1/2
f'= -1/(2√х3 )
f'(x0 )= -1/16
f(x)= 0,5-0,16/16= 0,499
4.21. 
x0 = 2
∆x= 2,002-2= 0,002
f(x0 )= 27 =128
f'= 7х6
f'(x0 )= 7*26 = 448
f(x)= 128+0,002*448= 128,896
4.22. 
x0 = 1
∆x= 1,78-1= 0,78
f(x0 )= √(4-3)= 1
f'= 2/√(4х-3)
f'(x0 )= 2
f(x)= 1+0,78*2= 2,56
4.23. 
x0 = 1
∆x= 0,98-1= -0,02
f(x0 )= 1
f'= 3/(2√х)
f'(x0 )= 3/2
f(x)= 1-3*0,02/2= 0,97
4.24. 
x0 = 3
∆x= 2,997-3= -0,003
f(x0 )= 243
f'= 5х4
f'(x0 )= 5*81= 405
f(x)= 243-405*0,003= 241,785
4.25. 
x0 = 1
∆x= 1,03-1= 0,03
f(x0 )= 1
f'= 2/(55 √х3 )
f'(x0 )= 2/5
f(x)= 1+2*0,03/5= 1,012
4.26. 
x0 = 4
∆x= 3,998-4= -0,002
f(x0 )= 256
f'= 4х3
f'(x0 )= 4*64= 256
f(x)= 256-256*0,002= 255,488
4.27. 
x0 = 0
∆x= 0,01-0=0,01
f(x0 )= √(1+0+sin0)=1
f'= (1+cosx)/(2√(1+х+sinx))
f'(x0 )= (1+1)/2= 1
f(x)= 1+0,01= 1,01
4.28. 
x0 = 0
∆x= 0,01-0= 0,01
f(x0 )= 1
f'= (3-sinx)/(33 √3x+cosx)
f'(x0 )= 3/3= 1
f(x)= 1+0,01= 1,01
4.29. 
x0 = 1
∆x= 1,02-1= 0,02
f(x0 )= 4 √(2-1)= 1
f'= (2-π/2*cos(πx/2))/(44 √(2x-sin(πx/2)))
f'(x0 )= 2/4√1= 0.5
f(x)= 1+0,02*05= 1,01
4.30. 
x0 = 2
∆x= 1,97-2= -0,03
f(x0 )= √9=3
f'= х/√(х2 +5)
f'(x0 )= 2/3
f(x)= 3-0,03*2/3= 2,98
4.31. 
x0 = 1,5
∆x= 1,58-1,5= 0,08
f(x0 )= 1/2
f'= -1/(√(2х+1))
f'(x0 )= -1/2
f(x)= 0,5-0,5*0,08= 0,46