Yuqori tartibli hosila va differensiallar

Differensiallanuvchi funksiyalarning xossalari.

Differensiallanuvchi funksialar haqida teoremalar. Biz quyida amaliy masalarni yechishda zarur bo`ladigan differensiallanuvchi funksialar haqidagi ba`zi teoremalarni keltiramiz.

	ROLL
	TEOREMASI
^A^g^a^r^(^)^f^x^f^u^n^k^s^i^y^a[ , ]a b^k^e^s^m^a^d^a^u^z^l^u^k^s^i^z^,^(^,^)^a^b^o^r^a^li^q^d^a^e^s^a differensiallanuvchi bo`lib, ^(^)^(^)^f^b^a^o`lsa^f^,^u^b^^holda^hech^bo`lmaganda bittashunday ^(^,^)^cnuq^ata^bt^opiladiki bunda ^^(^)^^0^f. ^c

Bu teoremaning geometrik ma`nosini 5-rasmdan aniqlab olish mumkin.

Teoremadagi shartlar juda ham muhim ekanligi esa 6-,7-rasmlarda ko`rinib turibdi.


5-rasm 6-rasm


7-rasm 8-rasm

	LAGRANJ
	TEOREMASI
^A^g^a^r^(^)^f^f^x^u^n^ks^i^y^a[ , ]a ^kb^es^m^a^d^a^uz^lu^k^s^iz,^(^,^)^a^o^b^r^a^l^i^q^d^a^e^s^a ^differensiallanuvchi^bo`lib,^(^)^(^)^fb^ao`lsa^f, u^b^holda hech bo`lmaganda bitta shunday ^( ,^)^cnuq^ata^bt^opiladiki bunda ^(^)^^(^)^^^(^)^f ^b^^f^.^a^f^c^b^a

Agar bu teoremada ^{,a x b x} deb^xolsak,^u h^ol^da bo`lib chekli orttirmali Lagranj formulasi hosil bo`ladi:
^{f (x  x)  f (x)  f (x  x)x} .
   , 0  1c x

	KOSHI
	TEOREMASI
^A^gar^^(^)^,^^(^)^t^f^un^k^t^si^y^alar[ , ]a^kb^e^s^m^a^d^a^u^z^l^uk^siz,^(^,^)^a^o^b^r^a^l^i^q^da esa differensiallanuvchi bo`lib, ^^^(^)^^0^,^^(^,^)^tbo`ls^ta, ^au^bholda hech bo`lmaganda bitta shunday ^( ,^)^cnuq^ata^bt^opiladiki bunda (b)  (a)   ^(c) . (b)  (a) ^(c)

Teylor formulasi. Agar ^{( )y}fun^fks^xiy^a ^{x a}nuqtining qandaydir atrofida ^{( 1)n}tarti^bli hosilagacha bo`lgan barcha hosilalarga ega bo`lsa, u holda shu atrofdagi barcha ^x nuqtalar uchun quyidagi Teyloe formulasi o`rinli bo`ladi:

n
^f  ^x ^{ f} ^a ^{ f a}  ^{x  a} ^{ f a}  ^{x  a}2 ... f ^n _a_  ^{x  a}n ^{ R}
 ^x ^.

Bu yerda _{Rn  x }

deb ataladi.

1! 2!

^f ^{a    x  a}
^n1
 ^{x  a} n1 ^, ^{0    1}
^{n  1}^!
n!

- Lagranj qoldiq hadi

Agar bu formulada 0аdeb olsak, u holda Makleron formulasi hosil bo`ladi:

^f  ^x ^{ f} ⁰ ^{ f 0} _{x }
^{f }⁰ _x² _... ^{f n} ⁰ _xⁿ _
^{f n1} ^{ x}
^xn1, ^{0    1} ^.

^{1! 2! n!} ^{n 1}^!
Teylor formulasi yordamida funksiyalarni ko`phad ko`rinishda yozib olish mumkin. Bunga ba`zi misollarni keltiramiz:
^{ex 1  x  x}2 ^{ x}3 ^{... x}n ^{ R} ^x
_n ;

2! 3! n!

 
_x2 _x4 _x6 _{n x}2n

cos x  1 ... 1

^{2! 4! 6!} ²ⁿ<sup>!


 R</sup>2n1 ^{x ;}

 
_x3 _x5 _x7 _{n1 x}2n1

sin x  x ... 1

^{3! 5! 7!} ^{2n  1}^!

• 
  R_2n
  

^x ^;

^{1  x}m ^{ 1  mx  mm  1 x2 ... mm 1 m  n  1 xn  R} _x_;

2!

x x x
2 3 4
^ln^{1  x} ^{ x ...} ^1

2 3 4

n1 ^xn

n

n! ⁿ

• 
  ^Rn ^x _.
  

Differensiallanuvchi ratsional kasr ifodalardagi ⁰ , ^ _{ko`rinishdagi}

0 

noaniqliklarning aniq qiymatlarini Lopital qoidasidan:
_lim^  ^x _{ lim}^{ } ^x
^{xa }  ^x ^{xa } ^x

^{foydalanib hisoblash mumkin. Bu yerda lim}  ^x ^{ lim}  ^x ^{ 0}
x^a x^a

 
lim _ x_  lim _ x_   ^{. Bu qoidani misollarda ko`rib chiqamiz:}
x a x a
yoki

5. 
   misol. lim ^lnx2 ^{ 3}
   

0
⁰lim
2x


^{x2  3} _ ⁴ ; 16) lim



lim
3x  1_^ 3

 0 ^{. ►}

^x2 x²  7x  18 ^x2 2x  7 11 ^x x²  5 ^x2 2x
Bu jarayon noaniqlikdan qutimaguncha chekli marta davom ettirilishi mumkin. Agar noaniqlik ^{0  , 1 , 0 , 00,   } ko`rinishda bo`lsa, u holda ma`lum bir
elementar almashtirishlar bajarib uni <sup>0</sup> , <sup></sup> <sub>ko`rinishdagi noaniqlikka keltiriladi</sub> 0 
so`ngra Lopital qoidasi qo`llaniladi.

5. 
   ^{misol. lim x  4tg  x 0}^{ lim x  4}
   

 ⁰ 
 

; ►
^^0lim ¹ _{ } ⁸

5. misol.

^x2 ₈ ^x4 _ctg  x ^x4 _  1 _
^{8 8} sin^{2  x}

^{ }  
8

lim sin x ^tgx  1^ . Bu kabi misollar quyidagicha yechiladi:
x /2


lim _sin x _^tgx 
x /2
^{lim }_^{tg xln}^{sin x}^_ ^ ^{ 0} ^

x /2
 lim ^{lnsin x} _ ^{ 0 } _ lim ^{ctg x} _ ^{lim sin x cos x  0. ►}

 
x /2 _{ctg x} ^ ₀ ^ x /2 ₁ x /2

O‘z-o‘zini tekshirish uchun savollar

1. 
   Funksiya hоsilаsini tа’rifi.
   
2. 
   Hоsilаning iqtisоdiy mа’nоsi.
   
3. 
   Yig‘indi va ayirmaning hosilasi qanday topiladi?
   
4. 
   Hosilaga ega bo‘lmagan funksiyalar yig‘indisining hosilasi mavjud bo‘lishi mumkinmi, misollar keltiring.
   
5. 
   Hosilaga ega bo‘lmagan va hosilaga ega bo‘lgan funksiyalar yig‘indisining hosilasi mavjud bo‘lishi mumkinmi, javobingizni asoslang.
   
6. 
   Ko‘paytmaning hosilasi qanday topiladi?
   
7. 
   Hosilaga ega bo‘lmagan funksiyalar ko‘paytmasining hosilasi mavjud bo‘lishi mumkinmi, misollar keltiring.
   
8. 
   Bo‘linmaning hosilasi qanday topiladi?
   
9. 
   Murakkab funksiyaning differensiallanuvchi bo‘lishligi haqidagi teoremani ayting.
   
10. 
    Teskari funksiyaning hosilasi qanday topiladi?