asosiy elementar funksiyalarning hosilalari formulasini o‘rganish

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o‘zbekiston respublikasi oliy ta’lim, fan va innovatsiyalar vazirligi __universiteti kurs ishi mustaqil ish referat mavzu:________________ asosiy elementar funksiyalarning n - tartibli hosilalari formulasini keltirib chiqarish ta’rif. agar y=f(x) funksiyaning x=xo nuqtadagi orttirmasi u ning argument orttirmasi x ga nisbatining x nolga intilganda chekli limiti mavjud bo’lsa, bu limit f (x) funksiyaning x o nuqtadagi xosilasi deb ataladi va yo yoki yo(x) yoki f(xo) yoki yoki ko’rinishlarda belgilanadi. demak ta’rifga ko’ra f o(xo)= = . misollar. 1.y=f(х)=с=cоnst bo’lsin. y=f(х+х)-f(х)=с-с=0 yо==0 2.y=f(х)=х bo’lsin. ==1; y о==1 3.y=х2 funksiyaning х=3 nuqtadagi hosilasini toping: y+y=(3+х)2=9+6х+(х)2 yо===(6+х)=6; 4.y=y(х)=,(х>0) yо==== yig’indi, ko’paytma va bo’linmaning xosilasi. teorema. agar u(x) va v(x) funksiyalar x(a,b) nuqtada va xosilalarga ega bo’lsa, u xolda ularning algebraik yisindisi, ko’paytmasi va bo’linmasi shu x nuqtada xosilaga ega bo’lib, quyidagi formulalar bo’yicha topiladi: (u±v)o=uo±vo; (uv)o=uov+uvo () o = (v(x) 0) teskari funksiyaning xosilasi. teskari funksiyaning mavjudligi xaqidagi teoremani isbotsiz keltirib o’taylik. 1-teorema. agar y=f(x) …
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xam shu x nuqtada hosilaga ega bo’ladi. parametrik ko’rinishda berilgan funksiyaning xosilasi. agar tenglamamiz parametrik ko’rinishda berilgan bo’lib, (t), (t) funksiyalar differensiallanuvchi va o(t)0 bo’lsa yaoni formula o’rinli bo’ladi. asosiy elementar funksiyalarning hosilalari. 1. y=xn (x>0) darajali funksiyaning xosilasini topaylik. funksiya xosilasining ta’rifiga ko’ra у=(х+х) н-хн=х н[-1ё] , = =; =n ajoyib limitni e’tiborga olsak ==nхн-1. yо=(хн)о=nхн-1. 2. y=х (>0 , 1) ko’rsatkichli funksiyaning xosilasi. y=-х= х (-1); =, =ln ajoyib limitga ko’ra yо===х =х ln. demak, yо=(х)о=хln 3. y=logах (а>0, а1) logarifmik funksiyaning hosilasi ham yо=(logах)о= logае formula bilan topiladi. agar logае=; logеа=lnа ; logех=lnх ; logхе=. ekanligini e’tiborga olsak yо=(logах)о= kelib chiqadi. agar а=е desak lnа=lnе=1 bo’lib, y=lnх ; yо=(lnх)о= bo’ladi. 4. y=sinх funksiyanig xosilasini topish uchun x ga x orttirma bersak u xam у orttirma olib y=sin(х+х)-sinх=2sin()cоs[] , yо==[]=cоsх. yо=(sinх)о=cоsх xuddi shuningdek o’rta maktab dasturidan bizga mahlum bo’lgan boshqa trigonometrik funksiyalarning xosilalarini xisoblash mumkin: (cоsх)о=-sinх ; (tgгх) …
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ada differensiallanuvchi bo’lsa, bu funksiyaning x [a,b] nuqtadagi hosilasi fo(x)= (1) tenglik bilan aniqlanar edi. limitning ta’rifiga ko’ra da nisbat fo(x) ga intiladi. boshqacha aytganda ular orasidagi farq cheksiz kichik miqdor bo’ladi. shuning uchun =f о(х)+у=f о(х)-х+ х ( да ). bundan ko’rinadiki funksiya orttirmasi u ikkita qo’shiluvchidan iborat bo’lar ekan. shularning birinchisi fo(x) x ga funksiyaning differensiali deyiladi va dy orqali belgilanadi. dy=fо(х) х (2) desak (2) dan dх=хох dх=х ekanligini e’tiborga olsak, dу=f о(х)dх (3) yoki =f о(х) (4) (4) dan ko’rinadiki f о(х) xosilani funksiya differensialining argument differensialiga nisbati deb qarash mumkin ekan. differensialning asosiy hossalari. 1.dc=0(c-o’згармас) 2..d(cу)=cdy 3.d(u+v)=du+dv 4.d(uv)=vdu+udv 5. d() = (v(х) 0) differensialning taqribiy xisoblarga tatbiqi. agar y=fo(x)= chekli limit mavjud bo’lsa, y=fо(х) х+ х (х0 да chекsiz кichiк funksiya) yoki y=dy+ х =1+=1+ =1+ =(1+ )=1 y dy f(х+х)-f(х)f о(х) х yoki y(х+х)y(х)+y о(х) х misol. sin 310= sin (300+10)= sin (+); х=; …
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6.bin image34.wmf oleobject77.bin oleobject78.bin oleobject79.bin image35.wmf image3.wmf oleobject80.bin oleobject81.bin image36.wmf oleobject82.bin image37.wmf oleobject83.bin image38.wmf oleobject84.bin oleobject85.bin oleobject86.bin oleobject5.bin oleobject87.bin oleobject88.bin oleobject89.bin oleobject90.bin oleobject91.bin image39.wmf oleobject92.bin oleobject93.bin oleobject94.bin oleobject95.bin image4.wmf oleobject96.bin oleobject97.bin oleobject98.bin image40.wmf oleobject99.bin image41.wmf oleobject100.bin image42.wmf oleobject101.bin image43.wmf oleobject6.bin oleobject102.bin image44.wmf oleobject103.bin oleobject104.bin oleobject105.bin oleobject106.bin oleobject107.bin image45.wmf oleobject108.bin image46.wmf image5.wmf oleobject109.bin oleobject110.bin oleobject111.bin oleobject112.bin image47.wmf oleobject113.bin image48.wmf oleobject114.bin image49.wmf oleobject115.bin oleobject7.bin image50.wmf oleobject116.bin image51.wmf oleobject117.bin image52.wmf oleobject118.bin image53.wmf oleobject119.bin oleobject120.bin oleobject121.bin oleobject8.bin image54.wmf oleobject122.bin oleobject123.bin image55.wmf oleobject124.bin image56.wmf oleobject125.bin image57.wmf oleobject126.bin image58.wmf image6.wmf oleobject127.bin image59.wmf oleobject128.bin oleobject129.bin oleobject130.bin oleobject131.bin oleobject132.bin oleobject133.bin oleobject134.bin image60.wmf oleobject9.bin oleobject135.bin image61.wmf oleobject136.bin image62.wmf oleobject137.bin image63.wmf oleobject138.bin image64.wmf oleobject139.bin image65.wmf oleobject10.bin oleobject140.bin oleobject141.bin oleobject142.bin oleobject143.bin image66.wmf oleobject144.bin oleobject145.bin oleobject146.bin image67.wmf oleobject147.bin oleobject11.bin oleobject148.bin image68.wmf oleobject149.bin oleobject150.bin oleobject151.bin image69.wmf oleobject152.bin oleobject153.bin oleobject154.bin oleobject155.bin oleobject12.bin oleobject156.bin image70.wmf oleobject157.bin oleobject158.bin oleobject159
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o‘zbekiston respublikasi oliy ta’lim, fan va innovatsiyalar vazirligi __universiteti kurs ishi mustaqil ish referat mavzu:________________ asosiy elementar funksiyalarning n - tartibli hosilalari formulasini keltirib chiqarish ta’rif. agar y=f(x) funksiyaning x=xo nuqtadagi orttirmasi u ning argument orttirmasi x ga nisbatining x nolga intilganda chekli limiti mavjud bo’lsa, bu limit f (x) funksiyaning x o nuqtadagi xosilasi deb ataladi va yo yoki yo(x) yoki f(xo) yoki yoki ko’rinishlarda belgilanadi. demak ta’rifga ko’ra f o(xo)= = . misollar. 1.y=f(х)=с=cоnst bo’lsin. y=f(х+х)-f(х)=с-с=0 yо==0 2.y=f(х)=х bo’lsin. ==1; y о==1 3.y=х2 funksiyaning х=3 nuqtadagi hosilasini toping: y+y=(3+х)2=9+6х+(х)2 yо===(6+х)=6; 4.y=y(х)=,(х>0) yо==== yig’indi, ko’paytma va bo’linmaning xosilasi. teorema....

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