Toshkent 2023 Reja: Kirish



(7)

_ x  2 _

1 1

_ x  3 _

Endi
x  2 ^va

1

x  3
funksiyalarning n-tartibli hosilalarini topishimiz lozim. Buning

uchun u=


x  a
funksiyaning n-tartibli hosilasini bilish yyetarli. Bu funksiyani u=(x+a)^-1

ko‘rinishda yozib, ketma-ket hosilalarni hisoblaymiz. U holda

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u’=-(x+a)^-2, u’’=2(x+a)^-3, u’’’=-23(x+a)^-3=-6(x+a)^-4.
Matematik induksiya metodi bilan
u⁽ⁿ⁾=(-1)ⁿn!(x+a)^-n-1 (8) Shunday qilib, (8.7) va (8.8) tengliklardan foydalanib quyidagi
y⁽ⁿ⁾=-7(-1)ⁿn!(x-2)^-n-1+9(-1)ⁿn!(x-3)^-n-1=(-1)ⁿn! ^{ 9}  ^{7 }

natijaga erishamiz.


_ ( x  3 )ⁿ

2. Leybnits formulasi.


( x  2 )ⁿ _

Agar u(x) va v(x) funksiyalar n-tartibli hosilalarga ega bo‘lsa, u holda bu ikki funksiya ko‘paytmasining n -tartibli hosilasi uchun
( uv )^{( n )}  u^{( n )}v  C_n' u^{( n1)}v' C²u^{( n2 )}v'' ... C^ku^{( nk )}v^{( k )}  ...

n

n
_{+ C} ^n1_{u' v}( n1 ) _{ uv}( n )
n

(9)

n
formula o‘rinli bo‘ladi. Bunda C^k
n( n  1)...(n  k  1)
 .
k!

Isboti. Matematik induksiya usulini qo‘llaymiz. Ma’lumki,
(uv)’=u’v+uv’. Bu esa n=1 bo‘lganda (9) formulaning to‘g‘riligini ko‘rsatadi. Shuning uchun
(9) formulani ixtiyoriy n uchun o‘rinli deb olib, uning n+1 uchun ham to‘g‘riligini ko‘rsatamiz.
(9) ni differensiyalaymiz:

n

n
( uv )^n1  u^{( n1)}v  u^{( n )}v' C'_n u^{( n )}v' C_n' u^{( n1)}v' ' C ²u^{( n1)}v' ' C ²u^{( n2 )}v' ' ' 
_{ ... C}k_u( n k 1 )_v( k ) _{ C}k_u( n k )_v( k 1 ) _{ ... C}n 1_{u'' v}( n 1 ) _{ C}n 1_{u' v}( n ) _

n n
_{+ u' v}( n ) _{ uv}( n1 )
Ushbu
n n

(10)

1 C
'  1 n  C'
_{C ' C 2  n } n( n 1) _ ( n 1)n _{ C 2 ,}

n n 1, n n
_{2 2} n 1

_Ck 1 _{ C}k
_ n( n 1)...(n  2  k ) _ n( n 1)...(n  k 1) _


^{n n} ( k 1)! k!

 C
( n 1)n...(n 1 ( k 1))
=

k!

k n1

tengliklardan foydalanib, (10) ni quyidagicha yozamiz:

_{( uv )}n1 _{ u}( n1 )_{v  C}1
u^{( n )}v'C ²
u^{( n1)}v'' ... C^k
_un1k _v( k ) _{ ... uv}( n1 )

n1
n1
n1

Demak, (9) formula n+1 uchun ham o‘rinli ekan. Isbot etilgan (9) formula Leybnits formulasi deb ataladi.

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3. 
   Leybnits formulasi tatbiqlari.
   

Misol. y=x³e^x ning 20-tartibli hosilasi topilsin.
Yechish. u=e^x va v=x³ deb olsak, Leybnits formulasiga ko‘ra

_y( 20 ) _{ x}3_{( e}x ₎( 20 ) _{ C}1
( x³ )'( e^x )⁽¹⁹⁾  C ² ( x³ )'' ( e^x )⁽¹⁸⁾  C³ ( x³ )''' ( e^x )⁽¹⁷⁾ 

20 20 20

20
 C ⁴ ( x³ )^{( 4 )}( e^x )¹⁶  ... ( x³ )^{( 20)} e^x
bo‘ladi. (x³)’=3x², (x³)’’=6x, (x³)’’’=6, (x³)⁽⁴⁾=0

tengliklarni va y=x³ funksiyaning hamma keyingi hosilalarining 0 ga tengligini, shuningdek n
uchun (e^x)⁽ⁿ⁾=e^x ekanligini e’tiborga olsak,

y^{( 20)}  e^x( x³  3C¹ x²  6C ² x  6C³
) tenglik hosil bo‘ladi.

20 20 20
Endi koeffitsientlarni hisoblaymiz:

C¹  20,

C ²  ^{20 19}  190,

_{C3 } 20 19 18 _ 20 19 18 _{ 1140}

20

Demak,
20 ₂

²⁰ 3! 6

y^{( 20)}  e^x( x³  60x² 1140x  6840 ).