Funksiyaning differensiali
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1. Funksiyaning differensiali
- Bu sahifa navigatsiya:
- 1-Misol. x tg y 2 4 funksiya differensialini toping. Yechish.
- 3-misol.
- Teylor formulasi
- Makloren formulasi
- Yuqori tartibli hosilalar.
Funksiyaning differensiali ) (x f y funksiyaning differensiali deb, uning orttirmasining erkli o’zgaruvchi x ning orttirmasiga nisbatan chiziqli bo’lgan bosh qismiga aytiladi. ) (x f y funksiyaning differensiali dy bilan belgilanadi. Funksiyaning differensiali uning hosilasi bilan erkli o’zgaruvchi orttirmasining ko’paytmasiga teng:
x x f dy ) ( ' yoki x y dy ' . Ravshanki, x dx . Shu sababli dx x f dy ) ( ' dx y dy ' . Differensial geometrik jixatdan ) (x f y funksiya grafigiga ) , ( y x M nuqtada o’tkazilgan o’rinma ordinatasining orttirmasiga teng (20- shakl). Funksiyaning differensiali dy uning y orttirmasidan x ga nisbatan yuqori tartibli cheksiz kichik miqdorga farq qiladi.
1- shakl Agar
) (x u u va ) (x v v funksiyalar differensiallanuvchi bo’lsa, u holda differensialning ta’rifi va differensiallash qoidalaridan bevosita differensialning asosiy xossalariga ega bo’lamiz: 1. ,
) ( C d bunda S — o’zgarmas. 2. .
( Cdu Cu d . 3. . ) ( dv du v u d . 4. .
( dv du v u d
5. , ) ( 2 v udv vdu v u d bunda 0 v . 6. .
( ' ' ) ( ' ) (
u f dx u u f u df .
x tg y 2 4 funksiya differensialini toping. Yechish. Oldin berilgan funksiyaning hosilasini topamiz: . 2 sec 2 8 2 cos
1 2 8 2 3 2 3 x x tg x x tg y U holda
xdx 2 xsec 2 tg 8 dy 2 3 . ) (x f y funksiyaning ikkinchi tartibli differensiali deb birinchi tartibli differensialdan olingan differensialga aytiladi va ) ( 2 dy d y d
kabi belgilanadi. ) (x f y
funksiyaning n - tartibli differensiali neb
) 1 ( n -tartibli differensialdan olingan differensialga aytiladi, ya’ni: ). ( 1 y d d y d n n ) (x f y funksiya berilgan bo’lib, bunda x — erkli o’zgaruvchi bo’lsa, u holda uning yuqori tartibli differensiallari ushbu formulalar bo’yicha hisoblanadi: . ,..., ' ' ' , " ) ( 3 3 2 2
n n dx y y d dx y y d dx y y d
) 1
x y funksiyaning ikkinchi tartibli differensialini toping. Yechish. Berilgan funksiyaning birinchi va ikkinchi tartibli hosilalarini topamiz: . 1
, ln 1 1 ln ' x y x x x x y
Demak, xdx dy ln , . 1 2 2
x y d
Funksiyaning dy differensiali uning y orttirmasidan dx x ga nisbatan yuqori tartibli cheksiz kichik miqdorga farq qiladi, shu sababli dy y yoki , ) ( ' ) ( ) ( x x f x f x x f bundan
, ) ( ' ) ( ) (
x f x f x x f formulaga ega bo’lamiz, bu formula funksiya qiymatlarini taqribiy hisoblashlarda ko’llaniladi.
15 , 0 arcsin
ning takribiy qiymatini hisoblang. Yechish. x y arcsin
funksiyani qaraymiz: 15 ,
x , 01 , 0 x deb olib va x x x x x )' (arcsin
arcsin ) arcsin( formuladan foydalanib topamiz: . 534 . 0 011 . 0 6 01 . 0 5 . 0 1 1 5 . 0 arcsin 51 . 0 arcsin 2
Shunday qilib, 534 , 0 15 , 0 arcsin radian. Teylor teoremasi (1685-1731y., ingliz matematigi).
) (x f y funksiya a x nuqtani o’z ichiga olgan biror oraliqda ) 1 ( n
tartibgacha barcha hosillarga ega bo’lsa, 1 1 2 ) ( )! 1 ( ) ( ! ) ( ....
) ( ! 2 ) ( ) ( ! 1 ) ( ) ( ) ( n n n n a x n a x a f a x n a f a x a f a x a f a f x f
formula o’rinli bo’ladi, bunda 1 0 , bo’lgan son. Bu formulaga qoldiq hadi, Langranj formasida
1 1 ) ! 1 ) (
n n a x n a x a f x R
bo’lgan, Teylor formulasi deyiladi. Teylor formulasida 0 a bo’lsa,
1 1 2 )! 1 ( ) ( ! ) 0 ( ... ! 2 ) 0 ( ! 1 ) 0 ( ) 0 ( ) ( n n n n x n x f x n f x f x f f x f
formula hosil bo’ladi. Bunga Makloren formulasi deyiladi. Teylor va Makloren formulalari funksiyalrni x ning darajalari bo’yicha yoyishda va taqribiy hisoblashlarda katta ahamiyatga ega.
1 3 2
x f funksiya
; 1 intervalning ichki 0 x
nuqtasida o’zining eng kichik qiymatiga erishsa ham, bu funksiya uchun Ferma teoremasining xulosasi o’rinli emas. Shuni ko’rsating. Yechish. Berilgan funksiya 0 x nuqtada o’zining eng kichik qiymatiga erishadi. Biroq funksiya shu 0 x nuqtada chekli hosilaga ega emas. Bu ushbu
3 3 2 1 0 0 x x x x f x f x f nisbatning 0
x da chekli limitga ega emasligidan kelib chiqadi. Demak, Ferma teoremasining sharti bajarilmaydi. Binobarin, teoremaning xulosasi o’rinli emas. 2-misol. Ushbu
3 2 x x f funksiya [-1; 2] segmentda Lagranj teoremasining shartlarini qanoatlantiradimi?
2 ; 1
intervalda x x f 2 xosilaga ega. Demak,
3 3 x x f funksiya [-1; 2] segmentda Lagranj teoremasiga ko’ra shunday s nuqta (-1 < c < 2) topiladiki,
c c f f f 2 1 2 1 2 bo’ladi. Keyingi tenglikdan 2 1
c ekanini topamiz. Yuqori tartibli hosilalar. ) (x f y funksiyaning ikkinchi tartibli hosilasi deb, uning hosilasidan olingan hosilaga, ya’ni ) (
ga aytiladi. Ikkinchi tartibli hosila quyidagilarning biri bilan belgilanadi: 2 2
), ( , dx y d x f y . ) (x f y funksiyaning n -tartibli hosilasi deb uning ) 1
n tartibli hosilasidan olingan hosilaga aytiladi va quyidagilarning biri bilan belgilanadi ) (n y ,
) ( ) ( x f n ,
n dx y d / . Ta’rifga ko’ra
) 1 ( ) (
n y y .
1-misol. 3 2 ) 7 2 ( x y funksiyaning ikkinchi tartibli hosilasini toping. Yechish. 2 2 2 2 2 2 2 3 2 ) 7 2 ( 12 4 ) 7 2 ( 3 ) 7 2 ( ) 7 2 ( 3 ) 7 2 ( x x x x x x x y ; ). 7 10 )( 7 2 ( 12 ) 8 7 2 )( 7 2 ( 12 4 ) 7 2 ( 2 ) 7 2 ( 12 ) 7 2 ( ) 7 2 ( 12 ) 7 2 ( 12 ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 x x x x x x x x x x x x x x x y y
Demak, ) 7 10 )( 7 2 ( 12 2 2
x y .
n x y
funksiyaning
tartibli hosilasini toping. Yechish. 1
nx y , 2 ) 1 ( x n x n n y ,
3 ) 2 )( 1 (
n x n n n y ,
n n n n x n n n n n n y x n n n n y n n n n 2 ... ) 3 )( 2 )( 1 ( ) 2 ( ...
) 3 )( 2 )( 1 ( , .... , ) 3 )( 2 )( 1 ( ) 1 ( ) 1 ( 4 ) 4 (
! 1 2 3 ... ) 3 )( 2 ( 1 ( ) ( n n n n n y n
) ! (n 1 dan n gacha bo’lgan sonlar ko’paytmasining qisqa yozilishi). Download 178.99 Kb. Do'stlaringiz bilan baham: |
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