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,
x
0
{\displaystyle x_{0}}
∈
{\displaystyle \in }
(a, b) nuqtada
f
′
(
x
0
)
{\displaystyle f'(x_{0})}
va
g
′
(
x
0
)
{\displaystyle g'(x_{0})}
hosilalarga ega boʻlsin.
Hosila taʼrifiga koʻra
lim
n
→
∞
{\displaystyle \lim _{n\to \infty }}
f
(
x
)
−
f
(
x
0
)
x
−
x
0
=
f
′
(
x
0
)
{\displaystyle {f(x)-f(x_{0}) \over x-x_{0}}=f'(x_{0})}
, (1)
lim
n
→
∞
{\displaystyle \lim _{n\to \infty }}
g
(
x
)
−
g
(
x
0
)
x
−
x
0
=
g
′
(
x
0
)
{\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
x
0
{\displaystyle x_{0}}
nuqtada hosilaga ega boʻlib,
(
f
(
x
)
±
g
(
x
)
)
x
′
=
f
′
(
x
0
)
±
g
′
(
x
0
)
{\displaystyle (f(x)\pm g(x))'_{x}=f'(x_{0})\pm g'(x_{0})}
(
f
(
x
)
±
g
(
x
)
)
x
′
=
f
′
(
x
0
)
±
g
′
(
x
0
)
{\displaystyle (f(x)\pm g(x))'_{x}=f'(x_{0})\pm g'(x_{0})}
boʻladi.
◂
F
(
x
)
=
f
(
x
)
±
g
(
x
)
{\displaystyle \blacktriangleleft F(x)=f(x)\pm g(x)}
deb topamiz:
F
(
x
)
−
F
(
x
0
)
x
−
x
0
=
f
(
x
)
−
f
(
x
0
)
x
−
x
0
±
g
(
x
)
−
g
(
x
0
)
x
−
x
0
.
{\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
→
x
0
{\displaystyle x\rightarrow x_{0}}
da limitga oʻtib, yuqoridagi munosabatlarni eʼtiborga olsak. Unda
lim
n
→
x
0
F
(
x
)
−
F
(
x
0
)
x
−
x
0
=
lim
n
→
x
0
=
f
(
x
)
+
f
(
x
0
)
x
−
x
0
±
{\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 }
±
lim
n
→
x
0
g
(
x
)
−
g
(
x
0
)
x
−
x
0
=
f
′
(
x
0
)
±
g
′
(
x
0
)
{\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,
F
′
(
x
0
)
=
(
f
(
x
)
±
g
(
x
)
)
x
′
=
f
′
(
x
0
)
±
g
′
(
x
0
)
.
▸
{\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
x
0
{\displaystyle x_{0}}
nuqtada hosilaga ega boʻlib,
(
f
(
x
)
⋅
g
(
x
)
)
x
′
=
f
′
(
x
0
)
⋅
g
(
x
0
)
±
f
(
x
0
)
⋅
g
′
(
x
0
)
{\displaystyle (f(x)\cdot g(x))'_{x}=f'(x_{0})\cdot g(x_{0})\pm f(x_{0})\cdot g'(x_{0})}
boʻladi.
◂
{\displaystyle \blacktriangleleft }
F
(
x
)
=
f
(
x
)
⋅
g
(
x
)
{\displaystyle (x)=f(x)\cdot g(x)}
deb
F
′
(
x
)
−
F
(
x
0
)
x
−
x
0
{\displaystyle {F'(x)-F(x_{0}) \over x-x_{0}}}
nisbatini quyidagicha
F
(
x
)
−
F
(
x
0
)
x
−
x
0
=
f
(
x
)
−
f
(
x
0
)
x
−
x
0
⋅
g
(
x
0
)
+
g
(
x
)
−
g
(
x
0
)
x
−
x
0
⋅
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
⟶
x
0
{\displaystyle x\longrightarrow x_{0}}
da limitga oʻtib topamiz.
lim
n
→
x
0
F
(
x
)
−
F
(
x
0
)
x
−
x
0
=
g
(
x
0
)
⋅
lim
n
→
x
0
f
(
x
)
−
f
(
x
0
)
x
−
x
0
+
lim
n
→
x
0
g
(
x
)
−
g
(
x
0
)
x
−
x
0
⋅
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
′
(
x
0
)
⋅
g
(
x
0
)
+
f
(
x
0
)
⋅
g
′
(
x
0
)
.
{\displaystyle =f'(x_{0})\cdot g(x_{0})+f(x_{0})\cdot g'(x_{0}).}
Demak,
F
′
(
x
0
)
=
(
f
(
x
)
⋅
g
(
x
)
)
x
0
′
=
f
′
(
x
0
)
⋅
g
(
x
0
)
+
f
(
x
0
)
⋅
g
′
(
x
0
)
▸
{\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
(
x
0
)
≠
0
)
{\displaystyle (g(x_{0})\neq 0)}
x
0
{\displaystyle x_{0}}
nuqtada ho silaga ega boʻlib,
(
f
(
x
)
g
(
x
)
)
x
0
′
=
f
′
(
x
0
)
⋅
g
(
x
0
)
−
f
(
x
0
)
⋅
g
′
(
x
0
)
g
2
(
x
0
)
{\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.
◂
{\displaystyle \blacktriangleleft }
Modomiki,
g
(
x
0
)
≠
0
{\displaystyle g(x_{0})\neq 0}
ekan, unda
x
0
{\displaystyle x_{0}}
nuqtaning biror atrofidagi
x
{\displaystyle x}
larda
g
(
x
)
≠
0
{\displaystyle g(x)\neq 0}
boʻladi.
SHuni etiborga olib topamiz:
f
(
x
)
g
(
x
)
−
f
(
x
0
)
g
(
x
0
)
x
−
x
0
=
f
(
x
)
⋅
g
(
x
0
)
−
f
(
x
0
)
⋅
g
(
x
0
)
+
f
(
x
0
)
⋅
g
(
x
0
)
−
f
(
x
0
)
⋅
g
(
x
)
g
(
x
)
⋅
g
(
x
0
)
⋅
(
x
−
x
0
)
=
{\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})}=}
=
1
g
(
x
)
⋅
g
(
x
0
)
[
f
(
x
)
−
f
(
x
0
)
x
−
x
0
⋅
g
(
x
0
)
−
f
(
x
0
)
⋅
g
(
x
)
−
g
(
x
0
)
x
−
x
0
]
.
{\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
⟶
x
0
{\displaystyle x\longrightarrow x_{0}}
da limitga oʻtib, Ushbu
(
f
(
x
)
g
(
x
)
)
x
0
′
=
f
′
(
x
0
)
⋅
g
(
x
0
)
−
f
(
x
0
)
⋅
g
′
(
x
0
)
g
2
(
x
0
)
{\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.
▸
{\displaystyle \blacktriangleright }
1-natija. Agar
f
(
x
)
{\displaystyle f(x)}
funksiya
x
0
{\displaystyle x_{0}}
nuqtada
f
′
(
x
0
)
{\displaystyle f'(x_{0})}
hosilaga ega boʻlsa,
c
⋅
f
(
x
)
{\displaystyle c\cdot f(x)}
funksiya
(
c
=
c
o
n
s
t
)
{\displaystyle (c=const)}
x
0
{\displaystyle x_{0}}
nuqtada hosilaga ega boʻlib,
(
c
⋅
f
(
x
)
)
x
0
′
=
c
⋅
f
′
(
x
0
)
{\displaystyle (c\cdot f(x))'_{x0}=c\cdot f'(x_{0})}
boʻladi, yaʼni oʻzgarmas sonni hosila ishorasidan tashqariga chiqarish mumkin.
2-natija. Agar
f
1
(
x
)
,
f
2
(
x
)
,
.
.
.
,
f
n
(
x
)
{\displaystyle f_{1}(x),f_{2}(x),...,f_{n}(x)}
funksiyalar
x
0
{\displaystyle x_{0}}
nuqtada hosilalarga ega boʻlib,
c
1
,
c
2
,
.
.
.
,
c
n
{\displaystyle 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
)
x
0
′
=
c
1
f
1
′
(
x
0
)
+
c
2
f
2
′
(
x
0
)
+
.
.
.
+
c
n
f
n
′
(
x
0
)
{\displaystyle (c_{1}f_{1}(x)+c_{2}f_{2}(x)+...+c_{n}f_{n}(x)'_{x0}=c_{1}f_{1}'(x_{0})+c_{2}f_{2}'(x_{0})+...+c_{n}f_{n}'(x_{0})}
boʻladi.
Murakkab funksiyaning hosilasi
2
0
{\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,
x
0
∈
X
{\displaystyle x_{0}\in X}
nuqtada
f
′
(
x
0
)
{\displaystyle f'(x_{0})}
hosilaga,
y
0
∈
{
f
(
x
)
|
x
∈
X
}
{\displaystyle y_{0}\in \{f(x)|x\in X\}}
nuqtada
(
y
0
=
f
(
x
0
)
)
{\displaystyle (y_{0}=f(x_{0}))}
g
′
(
y
0
)
{\displaystyle g'(y_{0})}
hosilaga ega boʻlsin. U holda
g
(
f
(
x
)
)
{\displaystyle g(f(x))}
murakkab funksiya
x
o
{\displaystyle x_{o}}
hosilaga ega boʻlib,
(
g
(
f
(
x
)
)
)
x
0
′
=
g
′
(
f
(
x
0
)
)
⋅
f
′
(
x
0
)
{\displaystyle (g(f(x)))'_{x0}=g'(f(x_{0}))\cdot f'(x_{0})}
boʻladi.
◂
g
(
y
)
{\displaystyle \blacktriangleleft g(y)}
funksiyaning
y
0
{\displaystyle y_{0}}
nuqtada
g
′
(
y
0
)
{\displaystyle g'(y_{0})}
hosilaga ega boʻlganligidan
g
(
y
)
−
g
(
y
0
)
=
g
′
(
y
0
)
⋅
(
y
−
y
0
)
+
α
⋅
(
y
−
y
0
)
{\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
)
,
y
0
=
f
(
x
0
)
{\displaystyle y=f(x),y_{0}=f(x_{0})}
va
y
⟶
y
0
{\displaystyle y\longrightarrow y_{0}}
da
α
⟶
0
{\displaystyle \alpha \longrightarrow 0}
.
Keyingi tenglikning xar ikki tomonini
x
−
x
0
{\displaystyle x-x_{0}}
ga bo'lib topamiz;
g
(
f
(
x
)
)
−
g
(
x
0
)
)
x
−
x
0
=
g
′
(
f
(
x
0
)
)
⋅
f
(
x
)
−
f
(
x
0
)
x
−
x
0
+
a
f
(
x
)
−
f
(
x
0
)
x
−
x
0
{\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
−
x
0
{\displaystyle x-x_{0}}
da limitga o'tib,
(
g
(
f
(
x
)
)
)
x
0
′
=
g
′
(
f
(
x
0
)
)
⋅
f
′
(
x
0
)
{\displaystyle (g(f(x)))'_{x0}=g'(f(x_{0}))\cdot f'(x_{0})}
tenglikga kelamiz
▸
{\displaystyle \blacktriangleright }
HOSILALAR JADVALI
Quyidagi sodda funksiyalarning hosilalarini ifodalovchi formulalarni keltiramiz
MANBALAR
{Худойберганов_Г_ва_бошқалар_Математик_анализдан_маърузалар_1 toʻplami} dan
olindi
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