hosila va differensialni hisoblash qoidalari

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1576494530.doc ) x ( v ) x ( u ) x ( v ) x ( ' v ) x ( u ) x ( v ) x ( ' u x ( v ) x ( u 2 × - × = ¢ ú û ù ê ë é ) x ( v ) x ( dv ) x ( u ) x ( v ) x ( du x ( v ) x ( u d 2 × - × = ú û ù ê ë é n x nx x n x cos 1 2 x sin 1 2 - 2 x 1 1 - a x 1 ln 2 x 1 1 - - 2 x 1 1 + x 1 2 x 1 1 + - arctgx x e 5 x y 2 x x 3 2 - + = x ln x 3 2 …

+ - + - = 0 ) x ( g lim ) x ( f lim a x a x = = ® ® ¥ = = ® ® ) x ( g lim ) x ( f lim a x a x ) x ( ' g ) x ( ' f lim a x ® ) x ( g ) x ( f lim a x ® ) x ( ' g ) x ( ' f lim ) x ( g ) x ( f lim a x a x ® ® = 0 0 ¥ ¥ ) x ( ' g ) x ( ' f 0 0 ¥ ¥ ) x ( ' g ) x ( ' f 0 0 ¥ ¥ 0 0 ¥ ¥ 7 4 x 1 1 x 2 lim )]' 4 x ( [ )' 12 x x ( …

nksiyalarning superpozitsiyasidan iborat y = f [g(x)] murakkab funksiya berilgan bo`lsin. agar u = g(x) funksiya x0 nuqtada differensiallanuvchi, o`z navbati-da y = f (u) funksiya u0 = g(x0) nuqtada differensiallanuvchi bo`lsa, u holda y = f [g(x)] murakkab funksiya ham x0 nuqtada differensiallanuv-chi bo`ladi va yoki y((x0) = f ((u0) · g((x0). murakkab funksiyaning erkli o`zgaruvchi bo`yicha hosilasi, shu funksiyani tashkil etgan (superpozitsiyalanuvchi) funksiya hosilalarining ko`paytmasiga teng. murakkab funksiya differensiali uchun dy = y((x0) · dx = f ((u0) · du tengliklar o`rinli, bu yerda du = g((x0) · dx. murakkab funksiya birinchi tartibli differensialini hisoblash uchun uning biror o`zgaruvchi bo`yicha hosilasini shu o`zgaruvchining differensialiga ko`paytirish yetarli. bun-da differensialni hisoblash shakli o`zgarishsiz qolib, o`zgaruvchilarning tanlanilishiga yoki ularning erkli yoki erksizligiga bog`liq emas.ushbu xossa birinchi tartibli differensial shaklining invariantlik xossasi deyiladi. misol. 1. funksiyaning birinchi tartibli hosilasi va differensialini hisoblaymiz: 2. y = xsin x (x > 0) funksiya hosilasini …

(sin2x)( = 2sin x(sinx)( = 2sin x cos x = sin2x, y(( = (sin 2x)(= = 2cos2x, y((( = (2cos2x)( = - 4sin2x va hokazo. quyida keltirilgan ba`zi funksiyalarning yuqori n – tartibli hosila-lari uchun tegishli formulalarni olish va ularni jadval holida yig`ish mumkin: f (x) f (n)(x) xp p(p-1)(p-2)…(p-n+1)xp-n ex ex ekx knekx lnx sin kx kn sin(kx+ ) cos kx kn sin(kx+ ) y = f (x) funksiyaning yuqori tartibli differensiallari ham ketma – ket ravishda, mos hosilalari kabi aniqlanadi: d2y = d(dy) – ikkinchi tartibli differensial; d3y = d(d2y) – uchinchi tartibli differensial; . . . . . . . . . . . . . . . . . . . . . . . . . . . . dny = d(dn -1y) - n-tartibli differensial. agar y = f (u) funksiya berilgan bo`lib, u erkli o`zgaruvchi yoki x ning chiziqli u = kx + …

ervalda differensiallanuvchi bo`lib, f (a) = f (b) tenglik o`rinli bo`lsa, u holda (a; b) intervalga tegishli hech bo`l-maganda bitta shunday bir s nuqta topiladiki, f ((c) = 0 bo`ladi. teoremani geometrik izohlaydigan bo`lsak, teorema shartlari bajarilganda, y = f (x) funksiya grafigi ab yoyga tegishli hech bo`lmagan-da bitta (1-rasmda ikkita d va e) nuqta topiladiki, chiziqning shu nuq-tasiga o`tkazilgan urinma 0x abssissalar o`qiga parallel bo`ladi. teo-remaning har bir sharti ahamiyatlidir, chunki ulardan biri bajarilmasa, (a; b) intervalda f ((c) = 0 tenglikni qanoatlantiruvchi s nuqta topilmasli-gi mumkin. masalan, 2-rasmda grafigi keltirilgan funksiya uchun uzluk-sizlik sharti bajarilmagan, a1 nuqta uning uzilish nuqtasi. 3-rasmda tasvirlangan funksiya uchun esa uning differensiallanuv-chanlik sharti bajarilmagan, a2 nuqtada funksiya hosilaga ega emas. egri chiziqlarga tegishli va (a; b) interval doirasida urinmalari 0x o`qiga pa-rallel bo`ladigan biror-bir nuqta mavjud emas. lagranj teoremasi: y = f (x) funksiya [a; b] kesmada aniqlangan va uzluksiz bo`lib, (a; b) …

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1576494530.doc ) x ( v ) x ( u ) x ( v ) x ( ' v ) x ( u ) x ( v ) x ( ' u x ( v ) x ( u 2 × - × = ¢ ú û ù ê ë é ) x ( v ) x ( dv ) x ( u ) x ( v ) x ( du x ( v ) x ( u d 2 × - × = ú û ù ê ë é n x nx x n x cos 1 2 x sin 1 2 - 2 x 1 1 - a x 1 ln 2 x 1 1 - - 2 x …

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