Ziyosfera

Hosilani hisoblash

HOSILANI HISOBLASH QOIDALARI

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

berilgan boʻlib, x0{\displaystyle x_0}∈{\displaystyle \in}(a, b) nuqtada f′(x0){\displaystyle f'(x_0)} va g′(x0){\displaystyle g'(x_0)} hosilalarga ega boʻlsin.

Hosila taʼrifiga koʻra

limn→∞⁡{\displaystyle \lim_{n \to \infty}}f(x)−f(x0)x−x0=f′(x0){\displaystyle {f(x)-f(x_0)\over x-x_0}=f'(x_0)}, (1)


limn→∞⁡{\displaystyle \lim_{n \to \infty}} g(x)−g(x0)x−x0=g′(x0){\displaystyle {g(x)-g(x_0) \over x-x_0}=g'(x_0)} (2)

boʻladi.

1) f(x)±g(x){\displaystyle f(x)\pm g(x)} funksiya x0{\displaystyle x_0} nuqtada hosilaga ega boʻlib,

(f(x)±g(x))x′=f′(x0)±g′(x0){\displaystyle (f(x)\pm g(x))'_x=f' (x_0)\pm g'(x_0)}

(f(x)±g(x))x′=f′(x0)±g′(x0){\displaystyle (f(x)\pm g(x))'_x=f' (x_0)\pm g'(x_0)}

boʻladi.

Failed to parse (Unknown function \blacktriangleleft): {\displaystyle \blacktriangleleft F(x)=f(x)\pm g(x)} deb topamiz:

F(x)−F(x0)x−x0=f(x)−f(x0)x−x0±g(x)−g(x0)x−x0.{\displaystyle {F(x)-F(x_0) \over x-x_0}={f(x)-f(x_0) \over x-x_0}\pm {g(x)-g(x_0)\over x-x_0}.}


Bu tenglikda x→x0{\displaystyle x\rightarrow x_0} da limitga oʻtib, yuqoridagi munosabatlarni eʼtiborga olsak. Unda


limn→x0⁡F(x)−F(x0)x−x0=limn→x0⁡=f(x)+f(x0)x−x0±{\displaystyle \lim_{n \to x_0}{F(x)-F(x_0) \over x-x_0}=\lim_{n \to x_0}={f(x)+f(x_0) \over x-x_0}\pm}


±limn→x0⁡g(x)−g(x0)x−x0=f′(x0)±g′(x0){\displaystyle \pm\lim_{n \to x_0}{g(x)-g(x_0) \over x-x_0} =f'(x_0)\pm g'(x_0)}


boʻlishi kelib chiqadi. Demak,


Failed to parse (Unknown function \blacktriangleright): {\displaystyle F'(x_0)=(f(x)\pm g(x))'_x=f'(x_0)\pm g'(x_0).\blacktriangleright}


2) f(x)⋅g(x){\displaystyle f(x)\cdot g(x)} funksiya x0{\displaystyle x_0} nuqtada hosilaga ega boʻlib,


(f(x)⋅g(x))x′=f′(x0)⋅g(x0)±f(x0)⋅g′(x0){\displaystyle (f(x)\cdot g(x))'_x=f'(x_0)\cdot g(x_0)\pm f(x_0)\cdot g'(x_0)}


boʻladi.


Failed to parse (Unknown function \blacktriangleleft): {\displaystyle \blacktriangleleft} F(x)=f(x)⋅g(x){\displaystyle (x)=f(x)\cdot g(x)} deb


F′(x)−F(x0)x−x0{\displaystyle {F'(x)-F(x_0) \over x-x_0}}

nisbatini quyidagicha


F(x)−F(x0)x−x0=f(x)−f(x0)x−x0⋅g(x0)+g(x)−g(x0)x−x0⋅f(x){\displaystyle {F(x)-F(x_0)\over x-x_0}={f(x)-f(x_0) \over x-x_0}\cdot g(x_0)+{g(x)-g(x_0) \over x-x_0}\cdot f(x)}


yozib olamiz. Soʻng x⟶x0{\displaystyle x\longrightarrow x_0} da limitga oʻtib topamiz.


limn→x0⁡F(x)−F(x0)x−x0=g(x0)⋅limn→x0⁡f(x)−f(x0)x−x0+limn→x0⁡g(x)−g(x0)x−x0⋅f(x)={\displaystyle \lim_{n \to x_0}{F(x)-F(x_0) \over x-x_0}=g(x_0)\cdot \lim_{n \to x_0}{f(x)-f(x_0) \over x-x_0}+\lim_{n \to x_0}{g(x)-g(x_0) \over x-x_0}\cdot f(x)=}


=f′(x0)⋅g(x0)+f(x0)⋅g′(x0).{\displaystyle =f'(x_0)\cdot g(x_0)+f(x_0)\cdot g'(x_0).}


Demak,


Failed to parse (Unknown function \blacktriangleright): {\displaystyle F'(x_0)=(f(x)\cdot g(x))'_{x0}=f'(x_0)\cdot g(x_0)+ f(x_0)\cdot g'(x_0) \blacktriangleright}


3) f(x)g(x){\displaystyle {f(x) \over g(x)}} funksiya (g(x0)≠0){\displaystyle (g(x_0)\neq0)} x0{\displaystyle x_0} nuqtada ho silaga ega boʻlib,


(f(x)g(x))x0′=f′(x0)⋅g(x0)−f(x0)⋅g′(x0)g2(x0){\displaystyle \biggl({f(x) \over g(x)}\biggr)'_{x0}={f'(x_0)\cdot g(x_0)-f(x_0)\cdot g'(x_0) \over g^2(x_0)}}


boʻladi.


Failed to parse (Unknown function \blacktriangleleft): {\displaystyle \blacktriangleleft} Modomiki, g(x0)≠0{\displaystyle g(x_0)\neq0} ekan, unda x0{\displaystyle x_0} nuqtaning biror atrofidagi x{\displaystyle x} larda g(x)≠0{\displaystyle g(x)\neq0} boʻladi.

SHuni etiborga olib topamiz:


f(x)g(x)−f(x0)g(x0)x−x0=f(x)⋅g(x0)−f(x0)⋅g(x0)+f(x0)⋅g(x0)−f(x0)⋅g(x)g(x)⋅g(x0)⋅(x−x0)={\displaystyle {{f(x) \over g(x){f(x_0) \over g(x_0)} \over x-x_0}={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) \over g(x)\cdot g(x_0)\cdot (x-x_0)}=}


Failed to parse (Unknown function \centerdot): {\displaystyle ={1 \over g(x)\cdot g(x_0)}\left [ \frac{f(x)-f(x_0)}{x-x_0}\centerdot g(x_0)-f(x_0)\cdot \frac{g(x)-g(x_0)}{x-x_0} \right ].}


Bu tenglikda x⟶x0{\displaystyle x\longrightarrow x_0} da limitga oʻtib, Ushbu


(f(x)g(x))x0′=f′(x0)⋅g(x0)−f(x0)⋅g′(x0)g2(x0){\displaystyle \left ( \frac{f(x)}{g(x)} \right )'_{x0}={f'(x_0)\cdot g(x_0)-f(x_0)\cdot g'(x_0) \over g^2(x_0)}}


tenglikka kelamiz.Failed to parse (Unknown function \blacktriangleright): {\displaystyle \blacktriangleright}

1-natija. Agar f(x){\displaystyle f(x)} funksiya x0{\displaystyle x_0} nuqtada f′(x0){\displaystyle f'(x_0)} hosilaga ega boʻlsa, c⋅f(x){\displaystyle c\cdot f(x)} funksiya (c=const){\displaystyle ( c=const)} x0{\displaystyle x_0} nuqtada hosilaga ega boʻlib,


(c⋅f(x))x0′=c⋅f′(x0){\displaystyle (c\cdot f(x))'_{x0}=c\cdot f'(x_0)}


boʻladi, yaʼni oʻzgarmas sonni hosila ishorasidan tashqariga chiqarish mumkin.

2-natija. Agar f1(x),f2(x),...,fn(x){\displaystyle f_1(x), f_2(x),...,f_n(x)} funksiyalar x0{\displaystyle x_0} nuqtada hosilalarga ega boʻlib, c1,c2,...,cn{\displaystyle c_1, c_2, ...,c_n} oʻzgarmas sonlar boʻlsa

u holda

(c1f1(x)+c2f2(x)+...+cnfn(x)x0′=c1f1′(x0)+c2f2′(x0)+...+cnfn′(x0){\displaystyle (c_1f_1(x)+c_2f_2(x)+...+c_nf_n(x)'_{x0}=c_1f_1'(x_0)+c_2f_2'(x_0)+...+c_nf_n'(x_0)}

boʻladi.

Murakkab funksiyaning hosilasi

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

berilgan boʻlib, x0∈X{\displaystyle x_0\in X} nuqtada f′(x0){\displaystyle f'(x_0)} hosilaga, y0∈{f(x)|x∈X}{\displaystyle y_0\in \{f(x)|x\in X\}} nuqtada (y0=f(x0)){\displaystyle (y_0=f(x_0))} g′(y0){\displaystyle g'(y_0)} hosilaga ega boʻlsin. U holda

g(f(x)){\displaystyle g(f (x))} murakkab funksiya xo{\displaystyle x_o} hosilaga ega boʻlib,


(g(f(x)))x0′=g′(f(x0))⋅f′(x0){\displaystyle (g(f(x)))'_{x0}=g'(f(x_0))\cdot f'(x_0)}

boʻladi.

Failed to parse (Unknown function \blacktriangleleft): {\displaystyle \blacktriangleleft g(y)} funksiyaning y0{\displaystyle y_0} nuqtada g′(y0){\displaystyle g'(y_0)} hosilaga ega boʻlganligidan


g(y)−g(y0)=g′(y0)⋅(y−y0)+α⋅(y−y0){\displaystyle g(y)-g(y_0)=g'(y_0)\cdot (y-y_0)+\alpha\cdot(y-y_0)}

boʻlishi kelib chiqadi. Bunda


y=f(x),y0=f(x0){\displaystyle y=f(x), y_0=f(x_0)} va y⟶y0{\displaystyle y\longrightarrow y_0} da α⟶0{\displaystyle \alpha\longrightarrow 0} .

Keyingi tenglikning xar ikki tomonini x−x0{\displaystyle x-x_0} ga bo'lib topamiz;

g(f(x))−g(x0))x−x0=g′(f(x0))⋅f(x)−f(x0)x−x0+af(x)−f(x0)x−x0{\displaystyle {g(f(x))-g(x_0)) \over x-x_0}=g'(f(x_0))\cdot{f(x)-f(x_0) \over x-x_0}+a{f(x)-f(x_0) \over x-x_0}}

Bundan x−x0{\displaystyle x-x_0} da limitga o'tib,

(g(f(x)))x0′=g′(f(x0))⋅f′(x0){\displaystyle (g(f(x)))'_{x0}=g'(f(x_0))\cdot f'(x_0)}

tenglikga kelamizFailed to parse (Unknown function \blacktriangleright): {\displaystyle \blacktriangleright}

HOSILALAR JADVALI

Quyidagi sodda funksiyalarning hosilalarini ifodalovchi formulalarni keltiramiz

  1. (C)′,C=const.{\displaystyle (C)', C=const.}
  2. (xa)′=a⋅xa−1,{\displaystyle (x_a)'=a\cdot x^{a-1},} a∈R,x>0{\displaystyle a\in R, x>0} (xn)′=nxn−1{\displaystyle (x^n)' =nx^{n-1}} n∈N,x∈R{\displaystyle n\in N, x\in R}
  3. (ax)=axlna,a>0,a≠1,x∈R{\displaystyle (a^x)=a^x lna, a>0, a\neq 1, x\in R} (ex)′=ex,x∈R{\displaystyle (e^x)'=e^x , x\in R}
  4. (logax)′=1xlna,{\displaystyle (log_a x)' ={1 \over x ln a},} a>0,a≠1,x≠0{\displaystyle a>0, a\neq1, x\neq0} (loga|x|)′=1xlna{\displaystyle (log_a \left\vert x \right\vert)' ={1\over xln a}} a>0,a≠1,x≠0.{\displaystyle a>0, a\neq 1, x\neq 0.} (lnx)′=1x{\displaystyle (ln x)' ={1\over x}} x>0.{\displaystyle x>0.} (ln|x|)′=1x{\displaystyle (ln\left\vert x \right\vert)' ={1 \over x}} x≠0{\displaystyle x\neq 0}
  5. (sinx)′=cosx{\displaystyle (sin x)' =cos x} x∈R{\displaystyle x\in R}
  6. (cosx)′=−sinx,{\displaystyle (cos x)' =-sin x,} x∈R{\displaystyle x\in R}.
  7. (tgx)′=1cos2x,{\displaystyle (tgx)' ={1 \over cos^2x},} x≠π2{\displaystyle x\neq {\pi \over 2}} n∈R{\displaystyle n\in R}
  8. (ctgx)′−1sin2x{\displaystyle (ctg x)' -{1 \over sin^2x}} x≠nπ{\displaystyle x\neq n\pi} n∈R{\displaystyle n\in R}
  9. (arcsin⁡x)′=1⁡\sqrt{1-x^2}⁡{\displaystyle (\arcsin x)' ={\operatorname{1}\!\over\operatorname{\sqrt{1-x^2}}\!}}, |x|<1{\displaystyle \left\vert x \right\vert<1}
  10. (arccos⁡x)′=−11−x2{\displaystyle (\arccos x)' =-{1 \over \sqrt{1-x^2}}}, |x|<1{\displaystyle \left\vert x \right\vert<1}
  11. (arctan⁡x)′=11+x2,{\displaystyle (\arctan x)' ={1 \over \sqrt{1+x^2}},} x∈R{\displaystyle x\in R}
  12. (arccot⁡)′=−11+x2{\displaystyle (\arccot)' =-{1\over \sqrt{1+x_2}}} x∈R{\displaystyle x\in R}
  13. (shx)′=chx,{\displaystyle (shx)' =chx,} x∈R{\displaystyle x\in R}
  14. (chx)′=shx,{\displaystyle (chx)' =shx,} x∈R{\displaystyle x\in R}
  15. (thx)′=1ch2x,{\displaystyle (thx)' ={1\over ch^2 x},} x∈R{\displaystyle x \in R}
  16. (cthx)′=−1sh2x,{\displaystyle (cthx)' =-{1\over sh^2 x},} x≠0{\displaystyle x\neq 0}

MANBALAR


{Худойберганов_Г_ва_бошқалар_Математик_анализдан_маърузалар_1 toʻplami} dan olindi 

"https://new.ziyosfera.uz/ziyo/Hosilani_hisoblash" dan olindi