20. Sodda qoidalar. f (x) va g(x) funksiyalar a,b intervalda aniqlan-gan bo’lib, ular xa,b da ntartibli f n(x), gn(x) hosilalarga ega bo’lsin. Buni quyidagicha tushinish lozim: f (x) va g(x) funksiyalar x nuqtani o’z ichiga olgan , a,b intervalda f (x), f (x),, f k1(x) hamda g(x),g(x),,gk1(x) hosilalarga ega bo’lib, x nuqtada esa
f k(x),gk(x) hosilaga ega. U holda
(cf (x))n cf n(x), c const;
( f (x) g(x))n f n(x) gn(x),
n f n(x)g(x) Cn1 f n1(x)g(x) Cn2 f n2(x)g(x)
( f (x)g(x))
Cnk f nk(x)gk(x) f (x)gn(x)
bo’ladi, bunda
Cnk n(n 1)(n k 1) , 1 k n. k!
6.9−misol. Ushbu
f (x) x2 2x5x3 6 x 2, x 3
funksiyaning ntartibli hosilasi topilsin.
◄ Berilgan funksiyani quyidagicha
7 9
f (x) x 2 x 3
yozib olamiz. So’ng
1 n (1)n n!
x (x a)n1
a
formuladan foydalanib topamiz:
n n1n! 9(1)nn!
2x 3 7(1)
x2 _5x 6 (x 2)n1 (x 3)n1 .►
30. Murakkab funksiyaning yuqori tartibli hosilalari. u f (x)
funksiya a,b intervalda, y F(u) funksiya esa c,d intervalda aniqlangan bo’lib, ular yordamida y F( f (x)) murakkab funksiya tuzilgan bo’lsin. u f (x) funksiya x(a,b) nuqtada ikkinchi tartibli f (x) hosilaga, y F(u)
funksiya esa mos u (u f (x)) nuqtada ikkinchi tartibli F(u) hosilaga ega bo’lsin. Ikkinchi tartibli hosila ta’rifiga ko’ra
y (F( f (x))) (F( f (x)))
bo’ladi. Murakkab funksiyaning hosilasini hisoblash formulasi (6.5) dan hamda ko’paytmaning hosilasini hisoblash formulasi (6.9) dan foydalanib topamiz:
(F( f (x))) F( f (x)) f (x) F( f (x)) f (x) F( f (x))( f (x)) F( f (x)) f (x) f (x) F( f (x)) f (x)
F( f (x)) f 2(x) F( f (x)) f (x).
Demak,
y F( f (x)) F( f (x)) f 2(x) F( f (x)) f (x).
Xuddi shunga o’xshash u f (x) funksiya x(a,b) nuqtada f (x) va y F(u) funksiya esa mos u (u f (x)) nuqtada F(u) hosilaga ega bo’lsa,
murakkab y F( f (x)) funksiya ham x(a,b) nuqtada 3-tartibli hosilaga ega bo’ladi. Bu hosila quyidagicha hisoblanadi:
y F( f (x)) ''' (F( f (x))) ' F( f (x)) f 2 (x) F( f (x)) f (x) '
F( f (x)) f 3 (x) F( f (x))2 f (x) f (x) F( f (x)) f (x) f (x) F( f (x)) f (x)
F( f (x)) (x) f 3 (x) 3F( f (x)) f (x) f (x) F( f (x)) f (x).
Shu yo’l bilan murakkab funksiya y F( f (x))ning istalgan tartibli hosilalari ham hisoblanishi mumkin.
40. Funksiyaning yuqori tartibli differensiallari.
Faraz qilaylik, f (x) funksiya xa,b nuqtada ikkinchi tartibli f (x) hosilaga ega bo’lsin.
6−ta’rif. f (x) funksiya differensiali dy ning xa,b nuqtadagi differensiali funksiyaning ikkinchi tartibli differensiali deb ataladi. Funksiyaning ikkinchi tartibli differensiali d 2 f (x) yoki d 2 y kabi belgilanadi: d 2 y d(dy) yoki d 2 f (x) d(df (x)).
Endi differensiallash qoidasidan foydalanib topamiz: d 2 y d(dy) d(ydx) dxd(y) dx ydx y(dx)2
Demak,
d 2 y ydx2 (6.19)
bunda dx2 dxdx (dx)2
Xuddi yuqoridagiga o’xshash, xa,b nuqtada funksiyaning 3–tartibli differensiali ta’riflanadi: d 3 y d(d 2 y) d(ydx2 ) dx2d(y) ydx3
bunda dx3 (dx)3. Umuman funksiyaning (n 1) tartibli differensiali d n1 y dan olingan differensial f (x) funksiya xa,b nuqtadagi ntartibli differensiali deb ataladi va u d n y yoki d n f (x) kabi belgilanadi: d n y d(d n1 y) yoki d n f (x) d(d n1 f (x)).
Bu holda funksiyaning ntartibli differensiali uning ntartibli hosilasi orqali quyidagi
d n y yndxn
ko’rinishda ifodalanadi. Uning to’g’riligini matematik induksiya usuli yordamida isbotlash mumkin.
f (x) va g(x) funksiyalar a,b intervalda aniqlangan bo’lib, ular
xa,b nuqtada ntartibli differensialga ega bo’lsin. U holda ushbu 1) d n (c f (x)) cd n f (x), c const
d n ( f (x) g(x)) d n f (x) d n g(x);
d n ( f (x)g(x)) d n(x)g(x) Cnd n1 f (x) dg(x) f (x)d ng(x)
formulalar o’rinli bo’ladi.
Endi murakkab funksiya y F( f (x)) ning differensialini hisoblaymiz. Ma’lumki, y F( f (x)) (x) funksiyaning differensiali dy (x)dx (F( f (x)))dx
(F( f (x))) F( f (x)) f (x)
bo’lib, u
dy d(F( f (x)) F( f (x)) f (x)dx F( f (x))df (x) (6.20)
ko’rinishga ega bo’ladi.
Demak, funksiya murakkab bo’lgan holda ham funksiya differensiali funksiya hosilasi F( f (x)) bilan (bu holda argument f (x) bo’ladi) argument f (x) ning differensiali df (x) ko’paytmasidan iborat ekanini ko’ramiz. Odatda bu xossani differensial formasining invariantligi deyiladi. Bunda (6.14) formuladagi dx argument x ning ixtiyori orttirmasi x ni
x dx bildiradi, (6.20) formuladagi df (x) esa x o’zgaruvchiga bog’liq bo’ladi.
Endi y F( f (x)) murakkab funksiyaning ikkinchi tartibli differensia- lini hisoblaymiz. Ta’rifga ko’ra d 2 y d 2 (F( f (x))) d(dF( f (x)))
bo’ladi. Differensiallash qoidasidan foydalanib topamiz:
d2y d(F( f (x))df (x)) d(F( f (x)))df (x) F( f (x))d(df (x))
2(x) F( f (x))d2 f (x),
F( f (x))df
bunda df 2 (x) df (x) df (x) (df (x))2.
Demak,
d 2 y d 2 (F( f (x)) F( f (x))df 2 (x) F( f (x))d 2 f (x) (6.21)
Bu (6.21) formula bilan (6.19) formulani taqqoslab, ikkinchi tartibli differensiallar differensial formasining invariantligi xossasiga ega emasligini ko’ramiz.
y F( f (x)) funksiyaning uchinchi va hokazo tartibli differensiallari yuqoridagidek birin-ketin hisoblanadi.
6–§. Differensial hisobning asosiy teoremalari
Ushbu paragrafda differensial hisobning asosiy teoremalarini keltiramiz. Bu teoremalar kelgusida, ayniqsa funksiyalarni tekshirishda muhum rol o’ynaydi.
4−teorema (Ferma teoremasi). f (x) funksiya biror X R oraliqda aniqlangan va bu oraliqning ichki c nuqtasida o’zining eng katta (eng kichik) qiymatiga erishsin. Agar bu nuqtada funksiya chekli f (c) hosilaga ega bo’lsa, u holda f (c) 0 bo’ladi.
◄ f (x) funksiya c nuqtada eng katta qiymatga ega, ya’ni xX da f (x) f (c) tengsizlik o’rinli, shu bilan birga bu c nuqtada chekli
f (c) hosila mavjud bo’lsin. Ravshanki
f (c) lim f (x) f (c) lim f (x) f (c) lim f (x) f (c)
xc x c xc0 x c xc0 x c
Ayni paytda (32-chizma) lim f (x) f (c) 0, lim f (x) f (c) 0.
xc0 x c xc0 x c
bo’ladi.
Yuqoridagi munosabatlardan
f (c) 0
ekani kelib chiqadi. ►
Shunga o’xshash, funksiya c nuqtada eng kichik qiymatga ega va bu nuqtada chekli f (c) hosilaga ega bo’lganda ham f (c) 0 bo’lishi ko’rsatiladi.
uzluksiz, f (a) f (b) bo’lsin. Agar bu funksiya a,b intervalda chekli
hosilaga ega bo’lsa, u holda shunday c a c b nuqta topiladiki, f (c) 0 bo’ladi.
◄ f (x) funksiya a,b segmentda uzluksiz. Demak, Veyershtrassning birinchi teoremasiga (5–bob, 7–§ ) ko’ra bu oraliqda funksiya o’zining eng katta qiymati M va eng kichik qiymati m ga erishadi.
m M bo’lsin. Bunda f (x) const, xa,b bo’ladi. Ravshanki, bu holda ca,b uchun f (c) 0 bo’ladi.
m M bol’sin. Bu holda f (a) f (b) bo’lgani uchun f (x) funksiya o’zining eng katta qiymati M , eng kichik qiymati m larning kamida bittasiga a,b segmentning ichki c a c b nuqtasida erishadi. Ferma teoremasiga asosan bu nuqtada
f (c) 0
bo’ladi. ►
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