Funksiyaning hosilasi va differensiali

4⁰. Elementar funksiyalarning hosilalari. Funksiya hosilasi ta’rifidan foydalanib elementar funksiyalarning hosilalarini topamiz.
1). y  x^ (x  0) darajali funksiyaning hosilasi. Bu funksiya uchun quyidagiga egamiz:

 x  
y  (x  x)  x  x1  1
_ x  
va
 x ^   x 
x1  1 1
y  x   1 
  x.
x x
x
(1 x)^ 1
Ma’lumki, lim  (qaralsin 5–bob) unda
x0 x
 x ^ x ^
1 1 1 1
lim y  lim x1    x1 lim   x1
x0 _x x0x0
x x
bo’ladi.
Demak, (x^)  x^1. Umuman, bu formula y  x^ funksiyaning aniqlanish sohasidagi ixtiyoriy x uchun o’rinlidir. Xususan,  1 bo’lganda

 1 1
    ₂ , x  0.
 x  x
2). y  a^x (a  0, a 1) ko’rsatkichli funksiyaning hosilasi. Bu funksiya uchun quyidagiga egamiz:
y  axx ax  ax(ax 1)
va
y a^x (a^x 1)
 .
x x
a^x 1
Ma’lumki, lim  lna (qaralsin 5-bob). Unda
x0 x
lim y  lim ax  ax 1  ax lim ax 1  ax lna
x0 x x0 x x0 x
bo’ladi. Demak,
y  (a^x)  a^x ln a.
Xususan, (e^x)  e^x.
3). y  log_a x (a  0, a 1, x  0) logarifmik funksiyaning hosilasi.
Bu funksiya uchun quyidagiga egamiz:
 x
y  log_a (x  x)log_a x  log_a1 
 x 
va
 x 
yx  1x log 1 xx   1x loga 1 xx x  . a
      
Ma’lumki, lim log^{a(1 x)}  log_a e (qaralsin 5-bob). Unda
x0 x
 x 
lim yx  lim 1x loga 1 xx x   1x loga e x 0  x 0 
 
boladi.
Demak,
1
y  (log_a x)  log_a e. x
Xususan,
1
( ln x )  . x
4). Trigonometrik funksiyalarning hosilalari. Ushbu y  sin x funksiya uchun quyidagiga egamiz:
x x
y  sin(x  x)sin x  2sin cos(x  )
2 2
va
x sin
y  cos(x  x) 2 .
x 2 x
2
Keyingi tenglikda x 0 da limitga otib topamiz: sin x
lim ^y _ lim cos(x _ ^x) lim ² _ cos x .
x0 x x0 2 x0 x
2
Demak,
y  (sin x)  cos x.
Shunga o’hshash (6-bob, 1-§ ga qarang) (cos x)  sin x formula ham isbotlanadi.
Endi y  tgx funksiyaning hosilasini tgx  ^{sin x} nisbatning hosilasi cos x
firmulasidan foydalanib topamiz:
 ^{2 2}
 sin x  (sin x)cos x sin x(cos x) cos x sin x 1
y  (tgx)     ₂  ₂  ₂ .
 cos x  cos x cos x cos x
Demak,
1
(tgx)  2 .
cos x
Huddi shunga o’hshash quyidagi formulalar ham isbotlanadi:

1. 
   sin x cos x
   

(ctgx)   2 , (secx)  2 , (cosecx)   .

2. 
   sin x cos x sin x
   

5) Teskari trigonometrik funksiyalaning hosilalari. Teskari funksi-yaning hosilasini topish qoidasidan foydalanib, teskari trigonometrik funksiyalaning hosilalarini hisoblaymiz. Ushbu y  arcsin x funksiyani olaylik.
Bu funksiya x  siny funksiyaga teskari bo’lib, uni (^ , ^ ) intervalda
2 2 qarasak,
1 1 1 1
y  (arcsin x)    
(sin y) cos y 1sin² y 1 x²
kelib chiqadi.
Demak,
(arcsin x)  (1 x 1 ).
Xuddi shunga o’hshash quyidagi formulalar ham isbotlanadi:
(arccos x)   (1 x 1 ),

1. 
   (arctgx)  , (arcctgx)   .
   
2. 
   2
   

1 x 1 x
6). Giperbolik funksiyalarning hosilalari. Endi giperbo’lik funksiya- larning hosilalarini hisoblaymiz. Bunda hosila hisoblashdagi sodda qoida- lardan va ko’rsatkichli funksiya hosilasi formulasidan foydalanamiz. Sodda hisoblashlar yordamida y  shx funsiya uchun topamiz:

e 
y  (shx)  12 (e^x e^{ x})^  ¹2  ^x e1_x   12 (ex e x )  chx.

Shunga o’hshash quyidagi formulalar ham isbotlanadi:
1 1
(chx)  shx, (thx)  2 , (cthx)   2 (x  0).
ch x sh x
4⁰. Hosilalar jadvali. Biz ushbu bandda elementar funksiyalar hosilalari uchun topilgan formulalarni jamlab, ularni jadval sifatida keltiramiz:
1). (x^)  x^1 (x  0);
2). (a^x)  a^x ln a (a  0, a 1);
3). (log_a x)  ¹ log_a e (x  0, a  0, a 1); x
Xususan,
1
(ln x)  (x  0); x
4). (sin x)  cos x ;
5). (cos x)  sin x ;
1 
6). (tgx)  2 (x   k, k  0, 1,  2, ) ;
cos x ₂
1
7). (ctgx)   ₂ (x  k, k  0, 1,  2, ) ; sin x
8). (arcsin x)  (1 x 1 ) ;
9). (arccos x)   (1 x 1 ) ;
1
10). (arctgx)  ₂ ;
1 x
1
11). (arcctgx)   ₂ ;
1 x
12). (shx)  chx ;
13). (chx)  shx ;
1
14). (thx)  ₂ ; ch x
1
15). (cthx)   ₂ . sh x
6.5–misol. Agar a  0, a 1, x  0 bo’lsa,

log_a x ^  ¹ xln a
bolishi isbotlansin.
◄ Aytaylik, x  0 bo’lsin. Unda x  x bo’lib,
log_a x ^  ¹ xln a
bo’ladi.
Aytaylik, x  0 bo’lsin. Unda x  x bo’lib,

log_a x  log_a(x)
bo’ladi. Murakkab funksiyaning hosilasini topish qoidasiga ko’ra
1 1
(log_a(x))  (1)  (x)ln a xln a
bo’ladi.►
6.6–misol. Agar u (x) va v (x) funksiyalar hosilaga ega bo’lib, u (x)  0 bo’lsa,
y  (u (x))^{v (x)}
funksiyaning hosilasi topilsin.
◄ Ravshanki,
ln y  v (x)lnu (x).
Murakkab funksiyaning hosilasi va ko’paytmaning hosilasi uchun tegishli formulalardan foydalanib topamiz:
1 1
y  v(x)ln u (x)  v (x) u(x), y u (x)
 1  y  y v(x)ln u (x)  v (x) u (x) u(x)
y  (u (x))^{v (x)}v(x)lnu (x)  v (x) u (x))^v(x)1u(x). ► (6.12)
6.7–misol. Ushbu
f (x)  x^x , (x)  x^xx (x  0); funksiyani hosilalari topilsin.
◄ (6.12) formulaga ko’ra
f (x)  (x^x )  x^x (ln x 1)
bo’ladi. Ravshanki,
 (x)  x^xx  x ^{f (x)}
yana (6.12) formulaga ko’ra
(x)  (x ^{f (x)} )  (x)ln x f (x)  f (x)x ^{f (x)1} bo’lib,
(x)  x^xx ln x x^x (ln x 1)  x^x  x^xx1  x^xxx1(xln x(ln x 1) 1) bo’ladi. ►
^x² sin ¹ , agar x  0 bo'lsa,
6.8–misol. Ushbu f (x)  _ x
 0, agar x  0 bo'lsa
funksiyaning hosilasi topilsin.
◄ Aytaylik, x  0 bo’lsin. Unda

1x ² cos 1x ( x1₂ )  2xsin 1x  cos 1x f (x)  2xsin  x
bo’ladi.
Aytaylik, x  0 bo’lsin. Bu holda hosila ta’rifidan foydalanib topamiz:
² sin 1 x
f (0)  lim ^x  lim xsin ¹  0
x0 x x0 x
Demak,
 1 1
f _(x) _ _2xsin x  cos x , agar x  0 bo'lsa,
 0, agar x  0 bo'lsa
bo’ladi. ►
4–§. Funksiyaning differensiali
1⁰. Funksiyaning differensiallanuvchi bo’lishi tushunchasi.
f (x) funksiya a,b intervalda aniqlangan, x₀ a,b, x₀  xa,b bo’lsin. U holda f (x) funksiya ham x₀ nuqtada y  f (x₀  x)  f (x₀ ) orttirmaga ega bo’ladi.
3–ta’rif. Agar f (x) funksiyaning x₀ a,b nuqtadagi orttirmasi y ni
y  Ax  x (6.13)
ko’rinishda ifodalash mumkin bo’lsa, f (x) funksiya x₀ nuqtada differensi-allanuvchi deb ataladi, bunda A miqdor x ga bog’liq bo’lmagan o’zgarmas,  esa x ga bog’liq va x 0 da (x) 0.
Agar x 0 da
 x  xx  o x
ekanini e’tiborga olsak, u holda yuqoridagi (6.13) ifoda ushbu
y  Ax  ox
ko’rinishni oladi. Funksiya orttirmasi uchun (6.13) formulada Ax ifoda orttirmaning bosh qismi deb yuritiladi.
3–teorema. f (x) funksiyaninng xa,b nuqtada differensiallanuvchi bo’lishi uchun uning shu nuqtada chekli hosilaga ega bo’lishi zarur va yetarli.
◄ Zarurligi. f (x) funksiya xa,b nuqtada differensiallanuvchi bo’lsin, unda
y  Ax  ox , ^y _ A_ ^ox
x x
bo’lib, lim ^y  lim^_ A  ^{ox}_  A bo’ladi. Demak, f (x)  A .
x0 x x0 x 
Yetarliligi. f (x) funksiya xa,b nuqtada chekli f (x) hosilaga ega bo’lsin:
f (x)  lim y  lim f (x  x)  f (x) .
x0 x x0 x
Agar
y
 f (x)  x
deb olsak, undan
y  f (x)x  x
ekanini topamiz. Bu tenglikdagi  miqdor x ga bog’liq va x 0 da 0. Demak, f (x) funksiya xa,b nuqtada differensiallanuvchi bo’lib,

A  f (x) bo’ladi. ►
Isbot etilgan teorema f (x) funksiyaninng xa,b nuqtada chekli f (x) hosilaga ega bo’lishi bilan uning shu nuqtada differensiallanuvchi
bo’lishi ekvivalent ekanini ko’rsatadi.