asosiy elementar funksiyalarning n - tartibli hosilalari formulasini keltirib chiqarish

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powerpoint-präsentation 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 dy dx yoki df dx ko’rinishlarda belgilanadi. demak ta’rifga ko’ra f o(xo)= lim x0 y x = lim x0 x f(x0 + x) - f(x0 ) . misollar. 1.y=f(x)=s=const bo’lsin.  y=f(x+  x)-f(x)=s-s=0 yo= lim x0 x y =0 2.y=f(x)=x bo’lsin. y = (x  x) - x =1; y o= lim x x x0 x y =1 3.y=x2 funksiyaning x=3 nuqtadagi hosilasini toping: y+  y=(3+  x)2=9+6  x+(  x)2 yo= lim y = lim (6  x)x = lim (6+  x)=6; x x0 x0 x x0 4.y=y(x)= x ,(x>0) yo= …

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i x ning murakkab funksiyasi deyiladi. teorema. agar u= (x) funksiya o’zgaruvchi x nuqtada yxo= o(x) xosilaga, y=f(u) funksiya esa o’zgaruvchi u bo’yicha yo=f o(u) xosilaga ega bo’lsa, u xolda y=f( (x)) murakkab funksiya xam shu x nuqtada ux ' fu '(u) '(x) hosilaga ega bo’ladi. parametrik ko’rinishda berilgan funksiyaning xosilasi. agar tenglamamiz   y   (t) x   (t) parametrik ko’rinishda berilgan bo’lib,  (t),  (t) funksiyalar differensiallanuvchi va  o(t)  0 bo’lsa  't (t)  ' (t)  t y' x yaoni x't (t) y ' (t) y'  t x formula o’rinli bo’ladi. asosiy elementar funksiyalarning hosilalari. 1. y=xn (x>0) darajali funksiyaning xosilasini topaylik. funksiya xosilasining ta’rifiga ko’ra  u=(x+  x) n-xn=x n[ 1  x  -1yo] , p x      x u = x x  x p      x …

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h mumkin: (cosx)o=-sinx ; (tggx) o= sos 2 x sin 2 x 1 1 ; (ctgx) o=- . 5. endi y=arcsinx teskari trigonometrik funksiyaning xosilasini hisoblashni ko’raylik. y=arcsinx funksiya x=siny funksiyaga teskari funksiya bo’lgani uchun, teskari funksiyalarning hosilalariga ko’ra yo=(arcsinx) o= (sin y)' cos y 1 1 1 1 = = = 1  sin 2 y 1  x 2 (arcsinx) o= 1  x 2 1 , (-1<x<1). xuddi shuningdek (arccosx) o=- ; (arctgx) o= ; (arcctgx) o= - 1  x 2 1 1  x 2 1  x 2 1 1 . 6. y=lnx bo’lsa, yo= 1  x' = 1 ; agar y=lnu bo’lib y=f(x) bo’lsa, x x yo=(lnu) o= u' = f ( x)' ; u f ( x) agar y=uv(x)(x) bo’lsa, lny=vlnu – bundan xosila olsak u' =vo∙lnu+v∙ u' , yo=u v[vo∙lnu+v∙ u' ]. u u u differensiallash qoidalari: 5.y=f(u),u= (x),y=f[ (x)] …

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u ) = du  v  u  dv (v(x)  0) v v 2 differensialning taqribiy xisoblarga tatbiqi. agar y=fo(x)= lim x0 y x chekli limit mavjud bo’lsa,  y=fo(x)  x+  x (  x 0 da   0 cheksiz kichik funksiya) yoki  y=dy+  x  y =1+ x =1+ dy dy f ' ( x)x f ' ( x)  x =1+  lim y = lim (1+ x0 dy x0 f ' ( x) )=1   y  dy  f(x+  x)-f(x)  f o(x)  x yoki y(x+  x)  y(x)+y o(x)  x misol. sin 310= sin (300+10)= sin (  +  );  x=  6 180 180 ; x+  x=300+10 = 6  +  180 ; sin(300+10)= sin (  +  )  sin  + sin x' ∙ 6 …

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powerpoint-präsentation 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 dy dx yoki df dx ko’rinishlarda belgilanadi. demak ta’rifga ko’ra f o(xo)= lim x0 y x = lim x0 x f(x0 + x) - f(x0 ) . misollar. 1.y=f(x)=s=const bo’lsin.  y=f(x+  x)-f(x)=s-s=0 yo= lim x0 x y =0 2.y=f(x)=x bo’lsin. y = (x  x) - x =1; y o= lim x x x0 x y =1 3.y=x2 funksiyaning x=3 nuqtadagi hosilasini toping: y+ …

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