22
23
Eslatish joizki,
x
miqdor
x
dan
x+h
gacha o‘zgarganda
y=f
(
x
) miqdor
o‘zgarishining
o‘rtacha tezligi
0
(
)
( )
lim
h
f x h
f x
h
→
+
−
ayirmali nisbatga teng.
Bundan
0
(
)
( )
lim
h
f x h
f x
h
→
+
−
ifoda
y=f
(
x
) miqdor o‘zgarishining
oniy tezligini
bildiradi
.
Mashqlar
13.
Quyidagi funksiyaning hosilasi nimaga teng?
a)
f
(
x
)=
x
3
; b)
f
(
x
)=
x
–1
; c)
f
(
x
)=
1
2
x
;
d)
f
(
x
)=
c.
14
. Jadvalni daftaringizga ko‘chiring va to‘ldiring:
a)
f
(
x
)
f
ʹ(
x
)
x
1
x
2
x
3
x
–1
1
2
x
b) Fikringizcha,
y=x
n
funksiya hosilasi nimaga teng (bu
yerda
n
–
ratsional son) ?
15.
Ta’rifdan foydalanib, funksiya hosilasini toping:
a)
f
(
x
)=2
x
+ 3;
b)
f
(
x
)=3
x
2
+ 5
x
+ 1; c)
f
(
x
)=2
x
3
+ 4
x
2
+6
x
– 1.
16*.
Daftaringizga ko‘chiring va to‘ldiring:
a)
f
(
x
)=
ax
+
b
uchun
f
ʹ(
x
) = ...;
b)
f
(
x
)=
ax
2
+ bx +c
uchun
f
ʹ(
x
) = ...;
c)
f
(
x
)=
ax
3
+ bx
2
+ cx + d
uchun
f
ʹ(
x
) = ...
17*.
Quyidagi tasdiqlarni isbotlang:
a)
f
(
x
) =
cg
(
x
)
bo‘lsa,
u holda
f
ʹ(
x
) =
cg
ʹ(
x
);
b)
f
(
x
) =
g
(
x
)
+ h
(
x
)
bo‘lsa, u holda
f
ʹ(
x
) =
g
ʹ(
x
) +
h
ʹ(
x
).
22
23
18*.
Funksiya grafigiga qarab hosilalar qiymatlarini solishtiring:
a)
f
ʹ(–7) va
f
ʹ(–2);
c)
f
ʹ(–9) va
f
ʹ(0);
b)
f
ʹ(–4) va
f
ʹ(2);
d)
f
ʹ(–1) va
f
ʹ(5).
19.
1) Yuqoridagi funksiya grafigiga qarab ushbu
shartlarni qanoat lanti-
radigan
x
1
,
x
2
nuqtalarni toping (
x
1
,
x
2
–
Ox
o‘qidagi nuqtalar: –9, –8, ...,
5, 6):
a)
f
ʹ(
x
1
) > 0,
f
ʹ(
x
2
) > 0;
b)
f
ʹ(
x
1
) < 0,
f
ʹ(
x
2
) > 0;
c)
f
ʹ(
x
1
) < 0,
f
ʹ(
x
2
) < 0;
d)
f
ʹ(
x
1
) > 0,
f
ʹ(
x
2
) < 0.
2) Grafikka qarab ushbu savollarga javob bering:
a)
funksiya qaysi oraliqda o‘suvchi? qaysi oraliqda kamayuvchi?
b) funksiyaning [0; 3], [3; 6], [–9; –6] oraliqlaridagi orttirmalarini
hisoblang.
3) Funksiya qaysi nuqtada eng katta, qaysi nuqtada eng kichik qiymatni
qabul qiladi?
4) Funksiya qaysi nuqtalarda nolga aylanyapti?
5) Qaysi oraliqda funksiya musbat qiymatlarni qabul qilyapti?
6) Qaysi oraliqda funksiya manfiy qiymatlarni qabul qilyapti?
24
25
7–9
HOSILANI HISOBLASH QOIDALARI
Agar
f
(
x
)
va
g
(
x
)
funksiyalarning har biri hosilaga ega bo‘lsa,
u holda
quyidagi differensiallash qoidalari o‘rinlidir:
1.
Yig‘indining hosilasi hosilalar yig‘indisiga teng:
(
f
(
x
)
+ g
(
x
))'
= f
'(
x
)
+ g
'(
x
)
.
(1)
2.
Ayirmaning hosilasi hosilalar ayirmasiga teng:
(
f
(
x
)
– g
(
x
))'
= f
'(
x
)
– g
'(
x
).
(2)
1-misol.
Funksiyaning hosilasini toping:
1)
f
(
x
)=
x
3
+
x
2
–
x
+10; 2)
1
( )
.
f x
x
x
=
−
Hosilani topishda 1, 2-qoidalaridan va hosilalar jadvalining 1, 3-
bandlaridan foydalanamiz, ya’ni:
1)
3
2
2
( ) ( ) ( ) ( ) 10 3
2 1;
f x
x
x
x
x
x
′
′
′
′
=
+
−
+
=
+
−
2)
1
1
1
3
2
2
2
2
1
1
'( )
.
2
2
f x
x
x
x
x
−
−
−
′
′
=
−
=
+
Javob:
1)
3
2
2
( ) ( ) ( ) ( ) 10 3
2 1;
f x
x
x
x
x
x
′
′
′
′
=
+
−
+
=
+
−
2)
1
1
1
3
2
2
2
2
1
1
'( )
.
2
2
f x
x
x
x
x
−
−
−
′
′
=
−
=
+
▲
3.
O‘zgarmas ko‘paytuvchini hosila belgisidan
tashqariga chiqarish
mumkin:
(
cf
(
x
))'
=c∙f
'
(
x
),
c
– o‘zgarmas son.
(3)
2-misol.
Funksiyaning hosilasini toping:
1)
f
(
x
) = 7
x
3
–5
x
2
+4; 2)
( )
3
3
5
=
+ −
f x
x
x
x
3
( )
3
3
5
=
+ −
f x
x
x
x
.
Hosilani topishda 1, 2, 3-qoidalaridan va hosilalar jadvalining 1, 3-
bandlaridan
foydalanamiz, ya’ni:
1)
3
2
3
2
2
( ) (7
5
4)' (7 )' (5 )' (4)' 21
10 ;
f x
x
x
x
x
x
x
′
=
−
+
=
−
+
=
−
2)
( )
( )
'
'
'
3
3
5
1
3
3
5·
( )'
=
+ −
=
+
−
′
f x
x
x
x
x
x
x
=
2
2
5 3
2
,
3
−
−
x
x x
.
Javob:
1)
3
2
3
2
2
( ) (7
5
4)' (7 )' (5 )' (4)' 21
10 ;
f x
x
x
x
x
x
x
′
=
−
+
=
−
+
=
−
; 2)
2
2
5 3
2
,
3
−
−
x
x x
.
▲