Funksiyaning hosilasi va differensiali

2⁰. Sodda qoidalar. f (x) va g(x) funksiyalar a,b intervalda aniqlan-gan bo’lib, ular xa,b da ntartibli f ^n(x), g^n(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 ^k1(x) hamda g(x),g(x),,g^k1(x) hosilalarga ega bo’lib, x nuqtada esa
f ^k(x),g^k(x) hosilaga ega. U holda

1. 
   (cf (x))^n  cf ^n(x), c  const;
   
2. 
   ( f (x)  g(x))^n  f ^n(x)  g^n(x),
   

^n f ^n(x)g(x)  C_n¹ f ^n1(x)g(x)  C_n² f ^n2(x)g(x) 

3. 
   ( f (x)g(x)) 
   

 C_n^k f ^nk(x)g^k(x)  f (x)g^n(x)
bo’ladi, bunda

Cnk  n(n 1)(n  k 1) , 1 k  n. k!
6.9−misol. Ushbu
f (x)  x₂ 2x5^x³_ 6 x  2, x  3
_
funksiyaning ntartibli 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)n1
 a
formuladan foydalanib topamiz:
n ^n1n! 9(1)ⁿn!
 2x  3  7(1)

_^ x² _5x  6^_  (x  2)n1  (x  3)n1 .►
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 xa,b nuqtada ikkinchi tartibli f (x) hosilaga ega bo’lsin.
6−ta’rif. f (x) funksiya differensiali dy ning xa,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(ydx)  dxd(y)  dx ydx  y(dx)²
Demak,
d ² y  ydx² (6.19)
bunda dx²  dxdx  (dx)²
Xuddi yuqoridagiga o’xshash, xa,b nuqtada funksiyaning 3–tartibli differensiali ta’riflanadi: d ³ y  d(d ² y)  d(ydx² )  dx²d(y)  ydx³
bunda dx³  (dx)³. Umuman funksiyaning (n 1) tartibli differensiali d ^n1 y dan olingan differensial f (x) funksiya xa,b nuqtadagi ntartibli differensiali deb ataladi va u d ⁿ y yoki d ^n f (x) kabi belgilanadi: d ^n y  d(d ^n1 y) yoki d ⁿ f (x)  d(d ^n1 f (x)).
Bu holda funksiyaning ntartibli differensiali uning ntartibli hosilasi orqali quyidagi
d n y  yndxn
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
xa,b nuqtada ntartibli differensialga ega bo’lsin. U holda ushbu 1) d ⁿ (c  f (x))  cd ⁿ f (x), c  const

2. 
   d ⁿ ( f (x)  g(x))  d ⁿ f (x)  d ⁿ g(x);
   
3. 
   d ⁿ ( f (x)g(x))  d ^n(x)g(x)  C_nd ^n1 f (x)  dg(x)  f (x)d ^ng(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 xX 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)
xc x  c xc0 x  c xc0 x  c
Ayni paytda (32-chizma) lim f (x)  f (c)  0, lim f (x)  f (c)  0.
xc0 x  c xc0 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.

1. 
   m  M bo’lsin. Bunda f (x)  const, xa,b bo’ladi. Ravshanki, bu holda ca,b uchun f (c)  0 bo’ladi.
   
2. 
   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. ►