Hosila olish

orsatkichli funksiya:

y  a^x ,
bunda
a  R,a  0,a  1. Bu funksiyaning

orttirmasi
y  a^xx  a^x  a^x (a^x 1)
ga teng bo„lib,
^y  a^x 
x
a^x  1


x
bo„ladi.

Bundan
x  0 da
a^x 1 xln a
ni hisobga olib, topamiz:

lim
^y  lim a^x 
a^x  1

 a^x lim

x ln a

 a^x ln a.

Demak,
x0 _x
x0 _x
x0 _x

Xususan,
(e^x )^  e^x .
(a^x )^  a^x ln a.

4. Logorifmik funksiya:

y  log _a x , bunda
a  R,
a  0,
a  1.
y  log _a x

funksiya U holda
x  a^y
funksiyaga teskari funksiya. Bunda
x^( y)  a^y ln a .

Demak,
y^(x) 
1


x^( y)
 ¹ 
a ^y ln a
1 _.
x ln a

Xususan,

(ln x)^  ¹ .

x
(log _a
x)^ 
1 _.
x ln a

5. Trigonometrik funksiyalar.

y  sin x
funksiyaning orttirmasi

y  sin( x  x)  sin x  2sin ^x cos^ x  ^{x }


bo„lib,




₂  ₂ 

2sin ^x cos^ x  ^{x }

^y 2 ^ 2 ^
_   _.
x x

Bu tenglikdan

x  0 da

sin ^x
2
_~ x
2

ni hisobga olib, topamiz:

2 ^x cos^ x  ^{x }

^y 2 ^ 2 ^
_ x _

lim
 lim ^{ }  lim cos_ x 
_  cos(x  0)  cos x.

x0 _x
Demak,
x0 _x
x0 _{ 2 }

y  cosx
(sin x)^  cos x.
funksiyaning hosilasini murakkab funksiyaning hosilasi

formulasidan foydalanib topamiz:




^  ^  ^
 ^   ^{ }

 ^ 

^{(cosx) } ^sin_ ^{ x} _
 cos_  x _  _  x ^
 cos_  x _  (1)  sin x.

_  ² _
 ²   ^{2 }
 ² 

Demak,
y  tgx

(cosx)^  sin x.

funksiyaning hosilasini bo„linmaning hosilasi formulasidan

foydalanib topamiz:
_{ } sin x _^ (sin x)^cos x  (cosx)^sin x cos² x  sin ² x 1
(tgx)  _{ }    .

Demak,
_ cos x _
cos² x
(tgx)^ 
1 _.
cos² x
cos² x
cos² x

y  ctgx funksiyaning hosilasini topishda murakkab funksiyaning hosilasi
formulasidan foydalanamiz:


 _  _ ^ 1 1

^{(ctgx) } ^tg_ ^{ x} _ ^
2
 (1)   .
 ^  ^{sin 2 x}

Demak,
_  _
cos² _
 2

• 
  x _
  



(ctgx)^  
1 _.
sin ² x

6. Teskari trigonometrik funksiyalar.

y  arcsin x
funksiya
x  sin y

funksiyaga teskari. Bunda U holda
x^( y)  cos y   .

Demak,
y^(x) 
1 _ 1 _.
x^( y)

(arcsin x)^  ¹ .

y  arccosx


funksiyaning hosilasini
arcsin x  arccosx  ^
2


formuladan

foydalanib topamiz:
 _{ } 


_ 1


(arccosx)
 _  arcsin x _
 ² 
 (arcsin x)
  .
1  x²

Demak,


(arccosx)^   ¹ .

y  arctgx funksiyaning hosilasini teskari funksiyaning hosilasi formulasidan foydalanib topamiz:

Demak,
(arctgx)^ 
1


(tgy)^

 cos² y 

1

1  tg ² y

 ¹ .
1  x²

(arctgx)^ 
1 _.
1 x²

arctgx va arcctgx funksiyalar Bundan
arctgx  arcctgx  ^
2

bog„lanishga ega.

Demak,


(arcctgx)^
_{ } 

 ₂



• 
  
  
  arcctgx ^
  



 (arctgx)^

  ¹ .
1  x²

Misol
(arcctgx)^  
1 _.
1 x²

Differensiallash qoidalari va y  e^x
funksiyaning hosilasidan foydalanib,

giperbolik funksiyalarning hosilalarini topamiz:

_  ^{e e}


x _  x ^{
(shx) }  
 _ex _{ e} x

 chx,

ya‟ni

(shx)^  chx;

_ 2 _ 2

_  ^{e e}


x _  x ^{
(chx) }  
 _ex _{ e} x

 shx,

ya‟ni

(chx)^  shx;

_ 2 _ 2


_{ } shx ^ (shx)^chx  shx(chx)^ ch² x  sh ² x 1
(thx)  _{ }    ;

_ chx _
ch² x
ch² x
ch² x


_{ } chx ^
sh ² x  ch² x 1
_ 1

(cthx)  _{ } 
  , ya‟ni
(cthx)   .

_ shx _
sh ² x
sh ² x
sh ² x

Differensiallash qoidalari va hosilalar jadvali

Keltirib chiqarilgan differensiallash qoidalarini va asosiy elementar funksiyalarning hosilalari formulalarini jadval ko„rinishida yozamiz.
Amalda ko‟pincha murakkab funksiyalarning hosilalarini topishga to„g„ri

keladi. Shu sababli quyida keltiriladigan formulalarda "x" argument argumentga almashtiriladi.
"u" oraliq

Differensiallash qoidalari:

1. 
   (u  v)^  u^  v^, u  u(x),v  v(x)  differensiallanuvchi funksiyalar;
   

2. (u  v)^  u^v  uv^, xususan (Cu)^  Cu^, C o„zgarmas son;




5. y^  y^u^, agar y  f (u) va u  (x) .
x u x

Asosiy elementar funksiyalarning hosilalar jadvali:

Keltirilgan diferensiallash qoidalari va asosiy elementar funksiyalarning hosilalar jadvali bir o„zgaruvchi funksiyasi differensial hisobining asosini tashkil qiladi, ya‟ni ularni bilgan holda qiyinchilik darajasi qanday bo„lishidan qat‟iy nazar har qanday elementar funksiyaning hosilasini topish mumkin. Bunda yana elementar funksiya hosil bo„ladi. Shunday qilib, differensiallash jarayonida
elementar funksiyalar sinfidan tashqariga chiqilmaydi.

Misol.
f (x)  5^x  arcsin x  x ln x
funksiyaning hosilasini topamiz:

f ^(x)  (5^x  arcsin x  xln x)^  (5^x )^  (arcsin x)^  (x ln x)^  5^x ln 5 

 ¹  x^ln x  x(ln x)^  5^x ln 5  ¹
 ln x  x  ¹ 
x

¹  5^x ln 5  ¹
x

• 
  ln x  1.
  

Hosilani topishda differensiallashning 1,2 qoidalari va 3,4,9 formulalaridan foydalanildi.

4.1.5. Logarifmik differensiallash

Ayrim hollarda funksiyaning hosilasini topish uchun avval berilgan funksiyani logarifmlash, so„ngra differensiallash maqsadga muvofiq bo„ladi. Bu jarayonga logarifmik differensiallash deyiladi.

Misol. y 


(x  4)³
funksiyaning hosilasini topamiz. Bu

hosilani differensiallash qoidalari va formulalari orqali topish mumkin. Ammo bu jaroyon katta qiyinchilikka ega. Logarifmik differensiallashni qo„llaymiz.
Funksiyani logarifmlaymiz:
ln y  ln( x³  1)  ⁴ ln( x  2)  x ln 2  3ln( x  4).
5

Bu tenglikni x bo„yicha differensiallaymiz:

y^ni topamiz:
¹  y^ 
y
1


x³  1
 3x²  ⁴ 
5
1


x  2
 ln 2  3 
1 _.
x  4




y^  y  
3x² _
⁴  ln 2  ^{3 }

ya‟ni


^ x³  1





,
5(x  2)

^ x³  1
^ 3x² 4
x  4 ^


3 ^

^{y } (x  4)³
  

5(x  2)

• 
  ln 2 
  

x  4 ^.

Shunday funksiyalar borki, ularning hosilalari faqat logarifmik differensiallash orqali topiladi. Bunday funksiyalarning qatoriga dararajali-

kor‘satkichli funksiya deb ataluvchi
y  u^v
funksiya kiradi, bu yerda
u  u(x) ,

v  v(x) 
x ning differensiallanuvchi berilgan funksiyalari.

Bu funksilaning hosilasini topamiz:

ln y  v  ln u,
¹ y^  v^  ln u  v  ¹  u^, y^  y^ v^ln u  ^{vu } , ya‟ni

 
y u _ u _

y^  u^{v } v^ln u  ^{vu }
yoki

u
 
 



u^v ^  u^v  ln u  v^  v  u^v1u^.
(9)

(9) formulani eslab qolish qoidasini ifodalaymiz: dararajali-ko‟rsatkichli

funksiyaning hosilasi
u  const
shartidagi ko„rsatkichli funksiya hosilasi bilan

v  const
Misol
shartidagi darajali funksiya hosilasining yig„indisiga teng.

_{y  x}cos3 x
yoki
funksiyaning hosilasini (9) formula bilan topamiz:
y^  x^cos3x  ln x  (sin 3x)  3  cos3x  x^cos3x1 1
y^  x^cos3x1  cos3x  3x^cos3x  ln x  sin 3x.

Mavzuni mustahkamlash uchun savollar:

1. 
   To‟g‟ri chiziqli notekis harakatning o‟rtacha tezligiga ta‟rif.
   
2. 
   Oniy tezlik nima?
   
3. 
   Funksiya hosilasini ta‟rifini bering. Hosilani belgilanishlari.
   
4. 
   Hosila qanday geometrik va mexanik manoga ega?
   
5. 
   Qanday funksiyaning hosilasi nol bo‟ladi?
   
6. 
   Hosila olish qoidalari.
   
7. 
   Asosiy elementar funksiyalarning hosila jadvalini yozing.