Hosilani hisoblash

HOSILANI HISOBLASH QOIDALARI

1. Ikki funksiya yigʻindisi, ayirmasi, koʻpaytmasi va nisbatining hosilasi. Aytaylik f (x) va g (x) funksiyalari (a, b)$\subset$ R da

berilgan boʻlib, $x_0$$\in$(a, b) nuqtada $f^{\prime }(x_0)$ va $g^{\prime }(x_0)$ hosilalarga ega boʻlsin.

Hosila taʼrifiga koʻra

$\underset{n\to \mathrm{\infty }}{\lim }$$\frac{f(x)-f(x_0)}{x-x_0}=f^{\prime }(x_0)$, (1)

$\underset{n\to \mathrm{\infty }}{\lim }$ $\frac{g(x)-g(x_0)}{x-x_0}=g^{\prime }(x_0)$ (2)

boʻladi.

1) $f(x)\pm g(x)$ funksiya $x_0$ nuqtada hosilaga ega boʻlib,

$(f(x)\pm g(x)){\text{'}}_x=f^{\prime }(x_0)\pm g^{\prime }(x_0)$

$(f(x)\pm g(x)){\text{'}}_x=f^{\prime }(x_0)\pm g^{\prime }(x_0)$

boʻladi.

$◂F(x)=f(x)\pm g(x)$ deb topamiz:

$\frac{F(x)-F(x_0)}{x-x_0}=\frac{f(x)-f(x_0)}{x-x_0}\pm \frac{g(x)-g(x_0)}{x-x_0}.$

Bu tenglikda $x\to x_0$ da limitga oʻtib, yuqoridagi munosabatlarni eʼtiborga olsak. Unda

$\underset{n\to x_0}{\lim }\frac{F(x)-F(x_0)}{x-x_0}=\underset{n\to x_0}{\lim }=\frac{f(x)+f(x_0)}{x-x_0}\pm$

$\pm \underset{n\to x_0}{\lim }\frac{g(x)-g(x_0)}{x-x_0}=f^{\prime }(x_0)\pm g^{\prime }(x_0)$

boʻlishi kelib chiqadi. Demak,

$F^{\prime }(x_0)=(f(x)\pm g(x)){\text{'}}_x=f^{\prime }(x_0)\pm g^{\prime }(x_0).▸$

2) $f(x)\cdot g(x)$ funksiya $x_0$ nuqtada hosilaga ega boʻlib,

$(f(x)\cdot g(x)){\text{'}}_x=f^{\prime }(x_0)\cdot g(x_0)\pm f(x_0)\cdot g^{\prime }(x_0)$

boʻladi.

$◂$ F$(x)=f(x)\cdot g(x)$ deb

$\frac{F^{\prime }(x)-F(x_0)}{x-x_0}$

nisbatini quyidagicha

$\frac{F(x)-F(x_0)}{x-x_0}=\frac{f(x)-f(x_0)}{x-x_0}\cdot g(x_0)+\frac{g(x)-g(x_0)}{x-x_0}\cdot f(x)$

yozib olamiz. Soʻng $x⟶x_0$ da limitga oʻtib topamiz.

$\underset{n\to x_0}{\lim }\frac{F(x)-F(x_0)}{x-x_0}=g(x_0)\cdot \underset{n\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}+\underset{n\to x_0}{\lim }\frac{g(x)-g(x_0)}{x-x_0}\cdot f(x)=$

$=f^{\prime }(x_0)\cdot g(x_0)+f(x_0)\cdot g^{\prime }(x_0).$

Demak,

$F^{\prime }(x_0)=(f(x)\cdot g(x)){\text{'}}_{x0}=f^{\prime }(x_0)\cdot g(x_0)+f(x_0)\cdot g^{\prime }(x_0)▸$

3) $\frac{f(x)}{g(x)}$ funksiya $(g(x_0)\ne 0)$ $x_0$ nuqtada ho silaga ega boʻlib,

$\left(\frac{f(x)}{g(x)}\right){\text{'}}_{x0}=\frac{f^{\prime }(x_0)\cdot g(x_0)-f(x_0)\cdot g^{\prime }(x_0)}{g^2(x_0)}$

boʻladi.

$◂$ Modomiki, $g(x_0)\ne 0$ ekan, unda $x_0$ nuqtaning biror atrofidagi $x$ larda $g(x)\ne 0$ boʻladi.

SHuni etiborga olib topamiz:

$\frac{\frac{f(x)}{g(x)}-\frac{f(x_0)}{g(x_0)}}{x-x_0}=\frac{f(x)\cdot g(x_0)-f(x_0)\cdot g(x_0)+f(x_0)\cdot g(x_0)-f(x_0)\cdot g(x)}{g(x)\cdot g(x_0)\cdot (x-x_0)}=$

$=\frac{1}{g(x)\cdot g(x_0)}[\frac{f(x)-f(x_0)}{x-x_0}\cdot g(x_0)-f(x_0)\cdot \frac{g(x)-g(x_0)}{x-x_0}].$

Bu tenglikda $x⟶x_0$ da limitga oʻtib, Ushbu

$\left(\frac{f(x)}{g(x)}\right){\text{'}}_{x0}=\frac{f^{\prime }(x_0)\cdot g(x_0)-f(x_0)\cdot g^{\prime }(x_0)}{g^2(x_0)}$

tenglikka kelamiz.$▸$

1-natija. Agar $f(x)$ funksiya $x_0$ nuqtada $f^{\prime }(x_0)$ hosilaga ega boʻlsa, $c\cdot f(x)$ funksiya $(c=const)$ $x_0$ nuqtada hosilaga ega boʻlib,

$(c\cdot f(x)){\text{'}}_{x0}=c\cdot f^{\prime }(x_0)$

boʻladi, yaʼni oʻzgarmas sonni hosila ishorasidan tashqariga chiqarish mumkin.

2-natija. Agar $f_1(x),f_2(x),...,f_n(x)$ funksiyalar $x_0$ nuqtada hosilalarga ega boʻlib, $c_1,c_2,...,c_n$ oʻzgarmas sonlar boʻlsa

u holda

$(c_1{f}_1(x)+c_2{f}_2(x)+...+c_n{f}_n(x){\text{'}}_{x0}=c_1{f_1}^{\prime }(x_0)+c_2{f_2}^{\prime }(x_0)+...+c_n{f_n}^{\prime }(x_0)$

boʻladi.

Murakkab funksiyaning hosilasi

$2^0$. Murakkab funksiyaning hosilasi. Faraz qilaylik, $y=f(x)$ funksiya $X\subset R$ toʻplamda, $g(y)$ funksiya $\{f(x)|x\in R\}$ toʻplamda

berilgan boʻlib, $x_0\in X$ nuqtada $f^{\prime }(x_0)$ hosilaga, $y_0\in \{f(x)|x\in X\}$ nuqtada $(y_0=f(x_0))$ $g^{\prime }(y_0)$ hosilaga ega boʻlsin. U holda

$g(f(x))$ murakkab funksiya $x_o$ hosilaga ega boʻlib,

$(g(f(x))){\text{'}}_{x0}=g^{\prime }(f(x_0))\cdot f^{\prime }(x_0)$

boʻladi.

$◂g(y)$ funksiyaning $y_0$ nuqtada $g^{\prime }(y_0)$ hosilaga ega boʻlganligidan

$g(y)-g(y_0)=g^{\prime }(y_0)\cdot (y-y_0)+\alpha \cdot (y-y_0)$

boʻlishi kelib chiqadi. Bunda

$y=f(x),y_0=f(x_0)$ va $y⟶y_0$ da $\alpha ⟶0$ .

Keyingi tenglikning xar ikki tomonini $x-x_0$ ga bo'lib topamiz;

$\frac{g(f(x))-g(x_0))}{x-x_0}=g^{\prime }(f(x_0))\cdot \frac{f(x)-f(x_0)}{x-x_0}+a\frac{f(x)-f(x_0)}{x-x_0}$

Bundan $x-x_0$ da limitga o'tib,

$(g(f(x))){\text{'}}_{x0}=g^{\prime }(f(x_0))\cdot f^{\prime }(x_0)$

tenglikga kelamiz$▸$

Quyidagi sodda funksiyalarning hosilalarini ifodalovchi formulalarni keltiramiz

1. $(C{)}^{\prime },C=const.$
2. $(x_a{)}^{\prime }=a\cdot x^{a-1},$ $a\in R,x>0$ $(x^n{)}^{\prime }=n{x}^{n-1}$ $n\in N,x\in R$
3. $(a^x)=a^{x}lna,a>0,a\ne 1,x\in R$ $(e^x{)}^{\prime }=e^x,x\in R$
4. $(lo{g}_{a}x{)}^{\prime }=\frac{1}{xlna},$ $a>0,a\ne 1,x\ne 0$ $(lo{g}_a\left|x\right|{)}^{\prime }=\frac{1}{xlna}$ $a>0,a\ne 1,x\ne 0.$ $(lnx{)}^{\prime }=\frac{1}{x}$ $x>0.$ $(ln\left|x\right|{)}^{\prime }=\frac{1}{x}$ $x\ne 0$
5. $(sinx{)}^{\prime }=cosx$ $x\in R$
6. $(cosx{)}^{\prime }=-sinx,$ $x\in R$.
7. $(tgx{)}^{\prime }=\frac{1}{co{s}^{2}x},$ $x\ne \frac{\pi }{2}$ $n\in R$
8. $(ctgx{)}^{\prime }-\frac{1}{si{n}^{2}x}$ $x\ne n\pi$ $n\in R$
9. $(\arcsin x{)}^{\prime }=\frac{\mathrm{1}}{\sqrt{\mathrm{1}\mathrm{-}{\mathrm{x}}^{\mathrm{2}}}}$, $\left|x\right|<1$
10. $(\arccos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}$, $\left|x\right|<1$
11. $(\arctan x{)}^{\prime }=\frac{1}{\sqrt{1+x^2}},$ $x\in R$
12. $(\mathrm{arccot}{)}^{\prime }=-\frac{1}{\sqrt{1+x_2}}$ $x\in R$
13. $(shx{)}^{\prime }=chx,$ $x\in R$
14. $(chx{)}^{\prime }=shx,$ $x\in R$
15. $(thx{)}^{\prime }=\frac{1}{c{h}^{2}x},$ $x\in R$
16. $(cthx{)}^{\prime }=-\frac{1}{s{h}^{2}x},$ $x\ne 0$

MANBALAR

• Differensial hisob

{Худойберганов_Г_ва_бошқалар_Математик_анализдан_маърузалар_1 toʻplami} dan olindi