vektor funksiya. vektor funksiya hosilasi va differensiali

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1576309835.doc i r j r k r r r i r j r k r r r r r r r r r r r r r 0000 lim()lim()lim()lim() tttttttt rtxtiytjztk ®®®® =++ r rr r 0 0 lim()() tt rtrt ® = rr r r 0 () rt r r r 0 () rtt +d r 000 ()()() xttiyttjzttk +d++d++d r rr 00000000 ()()(()()()()()() rttrtxttxtyttytzttzt r ijk ttttt +d-+d-+d-+d- d ==++ ddddd rr r r rr r t d d r r r r r 0 () drt dt r 0 0 '()lim t r rt t d® d = d r r 0 '() rt r r r 000 '()'()'() xtiytjztk ++ r rr 1212 (()())''()'()) rtrtrtrt +=+ rrrr (()) dartdr a dtdt = rr 1212 21 () drrdrdr rr dtdtdt × =+ rrrr rr r r r r (()()) dftrtdfdr rf dtdtdt =+ rr r r r …

a (a;b) intervalning har bir x nuqtasida chekli hosilaga ega. bu shart, xususan c nuqta uchun ham o`rinli. demak, ferma teoremasi shartlari bajariladi. bundan f`(c)=0 ekanligi kelib chiqadi. f(c)=m bo`lgan holda teorema yuqoridagi kabi isbotlanadi. roll teoremasiga quyidagicha geometrik talqin berish mumkin (20-rasm). agar [a,b] kesmada uzluksiz, (a,b) intervalda differensiallanuvchi f(x) funksiya kesma uchlarida teng qiymatlar qabul qilsa, u holda f(x) funksiya grafigida abssissasi x=c bo`lgan shunday c nuqta topiladiki, shu nuqtada funksiya grafigiga o`tkazilgan urinma abssissalar o`qiga parallel bo`ladi. eslatma. roll teoremasining shartlari yetarli bo`lib, zaruriy shart emas. masalan, 20-rasm 1) f(x)=x3, x([-1:1] funksiya uchun teoremaning 3-sharti bajarilmaydi. (f(-1)=-1(1=f(1)), lekin f`(0)=0 bo`ladi. 2) funksiya uchun roll teoremasining barcha shartlari bajarilmaydi, lekin (-1;0) ning ixtiyoriy nuqtasida f`(x)=0 bo`ladi. 3. lagranj teoremasi, koshi teoremasi teorema (lagranj teoremasi). agar f(x) funksiya [a,b] kesmada uzluksiz va (a,b) da chekli f`(x) hosila mavjud bo`lsa, u holda (a,b) da kamida bitta shunday c nuqta mavjud …

ksiya lagranj teoremasining shartlarini qanoatlantirsin deylik (21-rasm). funksiya grafigining a(a;f(a)), b(b;f(b)) nuqtalar orqali kesuvchi o`tkazamiz, uning burchak koeffitsienti bo`ladi. hosilaning geometrik ma`nosiga binoan f`(c) - bu f(x) funksiya grafigiga uning (s;f(s)) nuqtasida o`tkazilgan urinmaning burchak koeffitsienti: tg(=f`(c) demak, (1.1) formula (a,b) intervalda kamida bitta shunday c nuqta mavjudligini ko`rsatadiki, f(x) funksiya grafigiga (c;f(c)) nuqtada o`tkazilgan urinma ab kesuvchiga paralell bo`ladi. isbot qilingan (1.1) formulani boshqacha ko`rinishda ham yozish mumkin. buning uchun a<c<b tengsizliklarni e`tiborga olib, belgilash kiritamiz, u holda c=a+(b-a)(, 0<(<1 bo`lishi ravshan. natijada (1) formula ushbu f(b) - f(a) = f`(a+((b-a))(b-a) ko`rinishga keladi. agar (1) formulada a=x0; b=x0+(x almashtirishlar bajarsak, u f(x0+(x)-f(x0)=f`(c)(x (1.3) bu yerda x0 <c<x0+(x, ko`rinishga keladi. bu formula argument orttirmasi bilan funksiya orttirmasini bog`laydi, shu sababli (1.3) formula chekli orttirmalar formulasi deb ataladi. agar (1.1) lagranj formulasida f(a)=f(b) deb olsak, roll teoremasi kelib chiqadi, ya`ni roll teoremasi lagranj teoremasining xususiy holi ekan. misol. ushbu [0,2] kesmada …

o`ladi. isbot. ravshanki, (1.4) tenglik ma`noga ega bo`lishi uchun g(b)(g(a) bo`lishi kerak. bu esa teoremadagi g`(x)(0, x((a;b) shartdan kelib chiqadi. haqiqatdan ham, agar g(a)=g(b) bo`lsa, u holda g(x) funksiya roll teoremasining barcha shartlarini qanoatlantirib, biror c((a;b) nuqtada g`(c)=0 bo`lar edi. bu esa (x((a;b) da g`(x)(0 shartga ziddir. demak, g(b)(g(a). endi yordamchi funksiyani tuzaylik. shartga ko`ra f(x) va g(x) funksiyalar [a,b] da uzluksiz va (a,b) intervalda differensialanuvchi bo`lgani uchun f(x) birinchidan [a,b] kesmada uzluksiz funksiyalarning chiziqli kombinatsiyasi sifatida uzluksiz, ikkinchidan (a,b) intervalda hosilaga ega. so`ngra f(x) funksiyaning x=a va x=b nuqtalardagi qiymatlarini hisoblaymiz: f(a)(f(b)(0. demak, f(x) funksiya [a,b] kesmada roll teoremasiinng barcha shartlarini qanoailantiradi. shuning uchun hech bo`lmaganda bitta shunday c (a<c<b) nuqta topiladiki, f`(c)(0 bo`ladi. shunday qilib, va bundan (1.4) tenglikning o`rinli ekani kelib chiqadi. isbot tugadi. isbotlangan (1.4) tenglik koshi formulasi deb ham ataladi. endi koshi teoremasining geometrik ma`nosini aniqlaymiz. aytaylik x=((t), y=f(t), a(t(b tekislikdagi chiziqning parametrik tenglamasi bo`lsin. …

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1576309835.doc i r j r k r r r i r j r k r r r r r r r r r r r r r 0000 lim()lim()lim()lim() tttttttt rtxtiytjztk ®®®® =++ r rr r 0 0 lim()() tt rtrt ® = rr r r 0 () rt r r r 0 () rtt +d r 000 ()()() xttiyttjzttk +d++d++d r rr 00000000 ()()(()()()()()() rttrtxttxtyttytzttzt r ijk ttttt +d-+d-+d-+d- d ==++ ddddd rr r r rr r t d d r r r r r 0 () drt dt r 0 0 '()lim t r rt t d® d = d r r 0 '() rt r r r 000 '()'()'() xtiytjztk ++ r rr 1212 (()())''()'()) rtrtrtrt +=+ …

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