Funksiyaga oid tabaqalashtirilgan individual masalalar yechish Isaqjonova Dildora 102-guruh Fizika matematika fakulteti Matematika va informatika yo'nalish Fan nomi: Matematika analiz Fan o'qtuvchisi: Rajabov Ulug'bek

Funksiyaga oid tabaqalashtirilgan individual masalalar yechish
Isaqjonova Dildora 102-guruh
Fizika matematika fakulteti
Matematika va informatika yo'nalish
Fan nomi:Matematika analiz
Fan o'qtuvchisi: Rajabov Ulug'bek

Funksiyaning differensiali.

1. 
   Elementar funksiyalarning differensiali jadvali.
   

2. Funksiyaning differensialining taqribiy hisoblashga tatbiqi.

f (x₀  x) 
f (x₀ ) 
f ^(x₀ )x
(3)

(3) formuladan funksiya qiymatini taqribiy hisoblashlarda foydalaniladi.

1-misol.
f (x)  3x ²  7
funksiyaning, argument 2 dan 2,001 gacha

o’zgargandagi orttirmasini taqriban toping.
Yechish. (3) formuladan foydalanamiz.


x₀  2 ,

x  0.001.

f ^(x)  6x,

 0.012.

f ^(x₀ )  6  2  12 ,
f (x₀ )  df (x₀ ) 
f ^(x₀ )

x  12  0.001

Funksiya orttirmasi o’rniga uning differensialini olib qancha xatoga yo’l qo’yilganini baholaymiz: buning uchun haqiqiy orttirmani topamiz,

f (x₀ ) 
f (x₀  x) 
f (x₀ )  3(x₀  x)²  7  (3x²  7) 

0
 3x²  6x x  3(x)²  7  3x²  7 

0 0 0

 6x₀x  3(x)²

 6  2  0.001  3  0.000001  0.012003.

Demak, absalyut xato

y  dy
Nisbiy xato

 0.012003  0.012

 0.000003.

 ^0.000003  0.00025

dy 0.012
yoki
0,025%.

Taqribiy hisoblash xatosi ancha kichik, bu esa yuqoridagi taqribiy tenglikdan taqribiy hisoblashlarda foydalanish mumkinligini ko’rsatadi.

3. Funksiyaning ikkinchi tartibli differensiali.

4. 
   Ferma teoremasi.
   

Ferma teoremasi. (1602-1665y. – atoqli fransuz matematigi).


f (x)

funksiya birorta X oraliqda aniqlangan va bu oraliqning ichki c
nuqtasida eng katta (eng kichik) qiymatga ega bo’lib, hamda bu nuqtada
chekli f ^(c) hosila mavjud bo’lsa,

f
tenglik o’rinli bo’lishi zarur.
^ ( c )  0

Ferma teoremasi sodda geometrik ma’noga ega. Teorema shartlari bajarilganda X oraliqda shunday c nuqta mavjud bo’ladiki, bu nuqtadan funksiya grafigiga o’tkazilgan urinma OX o’qiga parallel bo’ladi(21.1- chizma).


_y y
A B
^O _c x O ^{a c b x}
21.1-chizma 21.2-chizma

5. 
   Roll teoremasi.
   

6. Lagranj teoremasi.

7. 
   Teylor teoremasi .
   

8. Makloren formulasi.

“Funksiyaning differensiali va differensial hisobning asosiy teoremalari” mavzusi bo‘yicha test topshriqlari

I darajali testlar

1. 
   Funksiya orttirmasi uchun formulani toping.
   

A) y  y^x   x
D) y  y^x  x²
В) y  y^x   x
E) y  y^x²   x

2. 
   y 
   

f x funksiyaning differensialini toping.

1. 
   dy  y^dx
   

4. 
   dy  y^x  x²
   

В) dy  y^x   x

4. 
   dx  y^dy
   

3. 
   y 
   

f x funksiyaning 2-tartibli differensialini toping.

A) d ² y  d (dy)  d ( y^dx)  y ^dx² В) d ² y  y^dx²
D) d ² y  y^dx E) d ² y  y^dx

4. 
   Roll teoremasining shartlari quyidagilarning qaysilarida to’g’ri berilgan: 1)
   

f (x) funksiya a, b kesmada aniqlangan va uzluksiz; 2) aqalli a,b oraliqda
f ^(x) chekli hosila mavjud emas; 3) oraliqning chetki nuqtalarida funksiya

teng f (a) 
f (b)
qiymatlarni qabul qiladi

A) 1), 3) В) 1), 2) D) hammasi E) 2), 3)

5. 
   Lagranj teoremasining shartlari quyidagilarning qaysilarida to’g’ri berilgan: 1)
   

f (x) funksiya a, b kesmada aniqlangan va uzluksiz; 2) aqalli a,b ochiq

oraliqda chekli
f (x)
hosila mavjud; 3) oraliqning chetki nuqtalarida funksiya

teng f (a) 
f (b)
qiymatlarni qabul qiladi

A) 1), 2) В)1), 3) D) hammasi E) 2), 3)

6. 
   Lagranj formulasini toping.
   

A) f (b) 

4. 
   f (b) 
   

f (a) 
f (a) 
f ^(c)  (b  a)
f ^(c)  (b  a)
В) f (b) 

4. 
   f (b) 
   

f (a) 
f (a) 
f (c)  (b  a)
f ^(c)  (b  a)

7. 
   Teylor formulasini toping.
   

f (x)  f (a)  ^{f (a)} (x  a)  ^{f (a)} (x  a)²  

^A) n 
1! 2!
n 1_{ }

_ f (a) _{x  an } f a   (x  a) _{(x  a)n 1}
n! (n  1)!
f (x)  f (0)  ^{f (0)} (x  a)  ^{f (0)} (x  a)²  

^В) n 
1! 2!
n 1_{ }

 ^{f (0)} x  aⁿ  ^{f  (x)} (x  a)^n1
n! (n  1)!
f (x)  f (a)  ^{f (a)} x  ^{f (a)} x²  

D) _{ }
1! 2!
 

• 
  
  õ
  f ⁿ (a) _{n }
  

n!
f ^n1 a   (x  a) (n  1)!
_xn1

f (x) 
f (a)  ^{f (a)} (x  a)  ^{f (a)} (x  a)²  

^E) n 
1! 2!
n 1_{ }

_ f (a) _{x  an } f a   (x  a) _{(x  a)n1}
n! (n  1)!

8. 
   Makloren formulasini toping.
   

A) f (x) 

В
f (0) 

f ^(0) _{x }
1!
f ^(a)
f ^(0) _x2
2!
f ^(a) ₂
 ... 
f ^n(0)
xⁿ 
n!
f ^n(a) <sub>n

f</sub> n 1_(x) (n  1)!
_f n 1<sub>(x)


x</sub>n 1


n 1

) f (x) 
f (a)  x 
1!
x  ... 
2!
n! ^{x } (n  1)! ^x

4. 
   f (x) 
   

f (0) 
f (0) _{x }
1!
f (0) _x2
2!


 ... 
f ^n(0)
xⁿ 
n!
_f n 1 _(x) (n  1)!


_xn 1

4. 
   f (x) 
   

f (0) 
f ^(0) _{x }
1
f ^(0) _x2
2
 ... 
f ^n(0)
xⁿ 
n
_f n 1_(x) (n  1)
_xn 1