Yuqori tartibli hosilalar

-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)ⁿn!(x+a)-ⁿ~¹ (8)

Shunday qilib, (8.7) va (8.8) tengliklardan foydalanib quyidagi

y
(x - 3 )ⁿ (x - 2)

n
J
ⁱⁿ=-7-(-1)ⁿ -n!(x-2)~ⁿ+9-(-1)ⁿ -n!(x-3)~ⁿ=(-1)ⁿ n! natijaga erishamiz

2. 
   .Leybnits formulasi.
   

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'+C²_nu^{(n-2 )}v''+... + C^k_nu^{(n - k)}v^(k) +... + + C_n^n-1u'v^(n-1) + uv⁽ⁿ⁾ (9) _k n(n -1 )...(n - k +1)
formula o‘rinli bo‘ladi. Bunda C
n
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:
(uv)ⁿ +¹ = u⁽ⁿ +¹⁾v + u⁽ⁿ⁾v'+C'_nu⁽ⁿ⁾v'+C_n' u^(n-1)v''+Clu^(n-1)v''+C^u^{(n - 2 )}v'''+ +... + C^k_nu^(n-k+^1}v^(k) + C^k_nu^(n-k)v^(k +¹⁾ +... + Cⁿ_n ^-1u"v^(n-1) + Cⁿ_n ^-1u’v⁽ⁿ⁾ +
+ u'v⁽ⁿ⁾ + uv⁽ⁿ+¹⁾ (10)
Ushbu
1 + Cn' = 1 + n = Cn+1, Cn'= n + ^ ^
rk-1 . nk _ ^{n(n - 1)}...⁽ⁿ + ^{2 - k) n(n - 1 )}...^{(n - k} +¹⁾ _ ^{n n} = (k -1)! k! ~
_(n +1 )n...(n +1 - (k -1)) k k! ^ tengliklardan foydalanib, (10) ni quyidagicha yozamiz:
(uv )ⁿ+¹ = u⁽ⁿ+^{1 )}v + C^l_n+_lu⁽ⁿ⁾v'+C¹_n+_lu^{(n-1 )}v''+...+Ck+₁u⁽ⁱⁿ+^1-k\^(k) +... + uv⁽ⁿ+¹⁾
Demak, (9) formula n+1 uchun ham o‘rinli ekan. Isbot etilgan (9) formula Leybnits formulasi deb ataladi

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
/²⁰⁾ = x³(e^x}⁽²⁰⁾ + C¹₂₀(x³ )'(e^x )⁽¹⁹⁾ + C2₀(x³ }"(e^x}⁽¹⁸⁾ + C³₂₀(x³ }'"(e^x}⁽¹¹⁾ +
+ C4(x³)⁽⁴⁾(e^x/⁶ +...+ (x³)⁽²⁰⁾e^x bo‘ladi. (x³)’=3x², (x³)’’=6x, (x³)’’’=6, (x³)⁽⁴⁾=0
tengliklarni va y=x funksiyaning hamma keyingi hosilalarining 0 ga tengligini, shuningdek Vn uchun (e^x)⁽ⁿ=e^x ekanligini e’tiborga olsak,
y(²⁰) = e^x (x³ + 3C^₀x² + 6C2₀x + 6C2₀ ) tenglik hosil bo‘ladi.
Endi koeffitsientlarni hisoblaymiz:
CL = 20, C22₀ =^=190, c³0 = ^ ^=1140
Demak,
y^{( 20)} = e^x (x³ + 60x² +1140x + 6840).

MUNDARIJA.

d ² y d ² f(x) 5
dx dx 6
d³ y d³ f(x ) 7
dx dx 7
dⁿy dⁿf(x) 8
_dx ⁿ _dx ⁿ 9
1 15
^ 1 Л^(п) 1 (_-1 )ⁿ ■ _n! = (-1)(-2)... (-n)x -¹-ⁿ = ⁽ • (2) 16
v x J xⁿ⁺¹ 16
y^(ny = (У) 18
xⁿ 18
v x J 18
y' = cos x = sin( x + ^) 19
y” = (cosx)' = - sinx = sin(x + 2 ■—), 7Г y''' = (- sinx)' = - cosx = sin( x + 3 ■—), y^(IV)) = (- cos x)' = sinx = sin( x + 4 ⁷) 20
(cosx)⁽ⁿ = cos( x + n^) (5) 23
(cosx)⁽¹¹⁵ = cos( x +115 ~) = cos( x + = sinx. 25
2.Ikkinchi tartibli hosilaning mexanik ma’nosi. 25
2 x + 3 34
x — 5x + 6 35
2x + 3 A B 36
(x — 2)(x — 3) x — 2 x — 3 37
A + B = 2, [— 3 A — 2 B = 3 38
1 1 38
x — 2 x — 3 39
1 , 39
x + a 40
(uv )⁽ⁿ⁾ = u⁽ⁿ⁾v + C_n' u^(n-1)v'+C²_nu^{(n-2 )}v''+... + C^k_nu^{(n - k)}v^(k) +... + + C_n^n-1u'v^(n-1) + uv⁽ⁿ⁾ (9) _k n(n -1 )...(n - k +1) 3
k! 4
(uv)ⁿ +¹ = u⁽ⁿ +¹⁾v + u⁽ⁿ⁾v'+C'_nu⁽ⁿ⁾v'+C_n' u^(n-1)v''+Clu^(n-1)v''+C^u^{(n - 2 )}v'''+ +... + C^k_nu^(n-k+^1}v^(k) + C^k_nu^(n-k)v^(k +¹⁾ +... + Cⁿ_n ^-1u"v^(n-1) + Cⁿ_n ^-1u’v⁽ⁿ⁾ + 5
CL = 20, C22₀ =^=190, c³0 = ^ ^=1140 11

1. 
   Azlarov. T., Mansurov. X., Matematik analiz. T.: «O‘zbekiston». 1 t: 2005, 2 t . 1995
   
2. 
   Fixtengols G. M. „Kurs differensialnogo i integralnogo ischeleniya“ M.: 1970.
   
3. 
   Sa’dullayev A. va boshqalar. Matematik analiz kursi misol va masalalar to'plami. T., «O‘zbekiston». 1-q. 1993., 2-q. 1995.
   
4. 
   Demidovich B. P. “Sbornik zadach i uprajneni po matematicheskomu analizu” T.: 1972.
   
5. 
   Ilin V. A., Poznyak E. G. “Maematik analiz asoslari” I qism, T.: 1981.