Ҳосила. Ҳосилани функцияни текширишга татбиқи мавзуларини ўқитиш методикаси режа

Ta'iif. Agar y= f (x) funksiyaning x=x_o nuqtadagi orttirmasi _ y ning argu ment orttirma si _
^{x ga nisbatining}  ^{X nolga intilga nda chek li limiti mavjud bo’lsa, bu limit f (x)}
_dy

funksiyaning x _o nuqtada gi hosilasi deb ataladi va y' yoki y'(x₀) yoki f '(x_o) yoki
dx
_df
yoki

_{ko’rinishlarda belgilanadi.}¹
dx
<sub> y


f ( x


  x ) 


f ( x )</sub>

^{Demak ta'r ifga k o’ra f '( x}o⁾⁼ _lim
 x  0
^ _lim
 x ^{ x  0}
0 0
_{ x}

Misollar.
_{ y
1 . y= f(x)= s= const bo’lsin.  y= f(x+  x) -f(x)= c-c= 0 y'= lim}  0

<sub> y
2. y= f ( X)= x bo’lsin


( x   x )  x</sub>



_{ y}
 _{l; y'= lim = 1}
^{ x  0}  x

 x  x
_{ x  0} ^{ x}

_{3. y=x}² _{funksiyaning x=3 nuqtadagi hosilasini toping;}
y_o=9; y_o+ _ y=(3+ _ x)²=9+6 _ x+( _ x)²

y’=


lim
_{ x }
 y


₀ <sup> x


</sup> lim
_{ x }
( 6   x )  x


₀ <sup> x


</sup> lim
_{ x }

( 6   x )  6

0

4. y=f(x)= x ,(x>0)


_{ y
x   x  x 1 1}

y’=
lim
<sub> x 


0</sub> ^{ x
} _{lim
 x  0} ^{ x
} lim
_{ x  0}


x   x 

x 2 x

2.Hosilaning geometrik ma'nosi. Biror (a,b) oraliqda aniqla nga n y=f (x) funksiyaning	y					M1
^g^r^a^f^i^g^i^e^g^r^i^c^h^i^z^i^qd^a^n^i^bo^r^a^t^bo^’^l^s^i^n^.^L^d^a^Mo^(^xo^,^yo^)^v^a						T
^M^j^(xo^+ ^X1^,^yo^+ ^y^)^n^uq^t^a^l^a^r^o^li^b^,^u^l^a^r^n^i^b^i^r^l^a^s^h^t^i^r^uv^c^h^i			^M0	L	^M2
^M0^M1 - ^ke^s^u^v^c^h^i^n^i^k^o^’r^ay^li^k^.^A^g^a^r^M1 ^n^u^q^t^a^n^i ^M0 ^n^u^q^t^a^g^a^c^h^e^k^s^i^z^y^a^q^i^n^l^a^s^h^t^i^r^s^a^k M0^M1 ^k^e^s^u^v^c^h^i^n^i^n^g		α	 y β		 ^x
^li^m^i^t^ho^l^a^t^i^bo^’^l^g^a^n^M0^T^t^o^’^g^’r^i^c^h^i^z^i^qq^a^L^e^g^r^i		0				x
^c^h^i^z^i^qn^i^n^g^M0 ^n^u^q^t^a^s^i^g^a^o^’^t^ka^z^il^g^a^n^u^r^i^n^m^a^d^e^y^il^a^d^i^.

Yuqorida gi chizma da n ko’rina diki O x o’qining mu sbat yo’na lishi bila n u rinma α
_{ y}

bu rchak , M ₀ M k esu vchi esa β burchak ta shk il qila di. tgα =
 x
da M_{1 } M₀ intilib β _ α .
ekanligi ma'lum. _ x  0

¹ James Stewart Calculus 7E 104-140betlar

Bundan tgα=

lim
_{ x }


tg 
0


^ lim
_{ x 
}
 ^{f’(x)  f’(x)=tgα .}

y
₀ ^{ x}

Shu nda y qilib, y= f(x) fu nk siya ning x= x_o nu qta da gi hosila sining qiymati funksiya grafigidagi shu M_o(x_o,y_o) nuqtaga o’tkazilgan urinmaning O x o’qining mu sbat yo’na lishi bilan hosil qilgan burchak tangensiga teng bo’lar ekan. Boshqa cha a ytganda urinmaning burchak koeffisiyentiga teng bo’lar ekan: k=tgα = f '( x _o)
^{Agar M}_o^(x_o^{, y}_o^{) ya 'ni M}_o^(x_o^{; f(x}_o^{) nuqta ga o’tkazilgan urinma tengla masini y=kx+b}
ko’rinishda olsak, urinma shu M₀(x₀, f(x _o)) nuqtadan o’tga ni u chu n
f(x₀ )=k x₀+b  b= f(x₀ ) -kx₀ .
Bu holda y= k x+b  y=kx+f(x_o) -kx₀  y=f(x_o)+k(x-x_o)  y= f(x_o)+ f(x_o)(x-x_o) urinma tenglama si.

₁
Misol. y=
x
tuzing.


giperbolaning x= x_o=l ya 'ni (1;1) nuqta siga o’tkazilga n urinma tengla ma sini


₁

y(x_o )= f(l)= = l; f '(x)= -

2
x
; f'(l) = -l y= l -l( x -l) = > y= 2 - x.

3. Hosilaning mexanik ma’nosi.
M o d d i y M n u q t a y = f( t ) q o n u n b o ’ y i c h a t o ’ g ’ r i c h i z i q b o ’ y l a b ha rakatla nsin. Moddiy nu qta boshla ng’ich hola tda 0 nuqta da bo’lib, t _o momentda esa Mo hola tni olib, y _o= f(t _o) ma sofa ni bossin.t=t₀+ _ t momentda esa M₁ holatni olib, bosib o’tgan yo’li y=f(t₁)= f(t_o+ _ t) bo’la di. _ t vaqtda bosib o’tgan yo’li esa _ y= f( t _o+ _ t) -f(t_o) bo’ladi.
_{ y
nisbat moddiy nu qtaning  t va qtdagi o’rta cha tezligi deyiladi.}
 t
_{ y  y}

v= _{lim
 t  0} ^{ t
— moddiy nuqtaning t momentdagi tezligini beradi.} _lim
 t  0
^{f( t}_º⁾  ^{y f(t}₁^{)]
0 M}₀  ^{t M}₁₁ ^t
_=u'.
 t

Demak, hosilaning mexanik ma 'nosi harakatlana yotga n moddiy nuqtaning ma ’lum
momentdagi tezligini ifodalar ekan.