Algebra va analiz asoslari

30
31
10–12
MURAKKAB FUNKSIYANING HOSILASI
Murakkab funksiya.
y 
= (
x
2 
+ 3
x
)
4
funksiyani qaraylik. Agar biz 
g
(
x
) = 
x
2 
+ 3
x
,
f
(
x
) = 
x
4
belgilashlarni kiritsak, 
y 
= (
x
2 
+ 3
x
)
4
funksiya 
y 
= 
f
(
g
(
x
)) 
ko‘rinishini oladi. Biz 
y 
= 
f
(
g
(
x
)) funksiyani 
murakkab funksiya 
deymiz.
1-misol.
 
Agar 
 f
(
x
) = 
x
2
va 
2
( )
3
x
y g x
x
−
=
=
+
bo‘lsa, quyidagilarni toping:
1) 
f
(
g
(2));
2) 
f
(
g
(–4));
3) 
g
(
f
(1));
4) 
f
((–4)); 5) 
f
(
f
(1))
6) 
g
(
g
(–1)).

Berilgan funksiyalardan foydalanib, hisoblashlarni bajaramiz:
1) 
f g x
f x
x
( ( ))
=
−
−






2
3
, bundan 
2
2 2
( (2))
(0) 0
0
2 3
f g
f
f
−


=
=
=
=


+


;
2) 
2
4 2
( ( 4))
(6) 6
36
4 3
f g
f
f
− −


−
=
=
=
=


− +


;
3) 
2
1 2
1
( (1))
(1 )
(1)
1 3
4
g f
g
g
−
=
=
=
= −
+
;
4) 
2
16 2 14
( ( 4))
((–4) )
(16)
16 3 19
g f
g
g
−
−
=
=
=
=
+
;
5) 
2
2
( (1))
(1 )
(1) 1 1
f f
f
f
=
=
= =
;
6) 
3 2
1 2
3
3,5
7
2
( ( 1))
3
1 3
2
1,5
3
3
2
g g
g
g
− −
− −
−




−
=
=
−
=
=
= −




− +



 − +
.
Javob: 
1) 0; 2) 36; 3) 
1
4
−
; 4) 
14
19
; 5) 1; 6) 
7
3
−
. 
▲
Murakkab funksiyaning hosilasi 
uchun
 
ushbu formula o‘rinli:
(
f 
(
g
(
x
)))=
f 
(
g
(
x
))·
g
'(
x
) (1)

---

30
31
2-misol.
 
Funksiyaning hosilasini toping
 
(
k, b – 
o‘zgar mas sonlar):
1) 
f
(
x
) = (
kx 
+ 
b
)
n
;
2)
 f
(
x
) 
= 
sin(
kx + b
);
3) 
f
(
x
) 
= 
cos(
kx + b
);

4)
 f
(
x
) 
= 
tg(
kx + b
)
.

 
1) 
f
(
t
) 
= 
t
n
 
va
 t
(
x
) 
= kx + b 
funksiyalarga (1) formulani qo‘llaymiz:
((
kx
+
b
)
n
)′ = (
t
n
)′ 
∙ 
(
kx+b
)′ 
= n
t
n–
1
∙k = n∙k∙
(
kx + b
)
n
–1
 
.
2)
 f
(
t
)
=
sin
t 
va
 t
(
x
) 
=kx+b 
funksiyalarga (1) formulani qo‘llaymiz:
(sin(
kx+b
))′ = (sin
t
)′ 
∙
(
kx+b
)′ 
= k ∙ 
cos
t = k ∙ 
cos(
kx + b
)
.
3)
 f
(
t
) 
= 
cos
t
va
t
(
x
) 
= kx + b 
funksiyalarga (1) formulani qo‘llaymiz: 
(cos(
kx + b
))′ 
= 
(cos
t
)′
∙ 
(
kx+b
)′ 
= –k ∙ 
sin
t = – k ∙ 
sin(
kx + b
)
.
4)
 f
(
t
) 
= 
tg
t 
va
 t
(
x
) 
= kx+b
funksiyalarga (1) formulani qo‘llaymiz:
(tg(
kx + b
))
′ 
= 
(tg
t
)
′
∙ 
(
kx + b
)
′ 
(
)
2
2
1
cos
cos
k
k
t
kx b
=
⋅ =
+
.
Javob: 
1) ((
kx 
+ 
b
)
n
)′ 
= 
n∙k∙
(
kx + b
)
n
–1
;

2) (sin(
kx + b
))
′ 
= k∙
cos(
kx+b
); 
3)
 
(cos(
kx+b
)
′
=–k∙
sin(
kx+b
); 4)
 
(tg(
kx+b
)
′
= 
(
)
(
)
'
2
(
cos
k
tg kx b
kx b
+
=
+
.
▲
3-misol. 
f
(
x
) 
= 
sin8
x∙e
(3
x
+2)
 
funksiya hosilasini toping. 

Hosilani topishning 4-qoidasi hamda (1) formulani qo‘llab hosilani 
topamiz: 
f
′
(
x
) = (sin8
xe
(3
x
+2)
)
′ 
= (sin8
x
)
′
e
3
x
+2
+ sin8
x 
∙ (
e
3
x
+2
)
′
= cos8
xe
3
x
+2 
∙ 
(8
x
)
′
+
+ sin8
xe
3
x
+2 
∙ 
(3
x+
2)
′ 
= 
e
3
x
+2
∙ 
(8cos8
x 
+ 3sin8
x
).
Javob: 
e
3
x
+2 
∙ 
(8cos8
x 
+ 3sin8
x
).
 
▲
4-misol.
 
h
(
x
) = (
x
3 
+ 1)
5
funksiyaning 
x
0 
= 1 nuqtadagi hosilasini toping.

(1) formuladan foydalanib hosilani hisoblaymiz:
h
′
(
x
) = 5(
x
3
+1)
4
(
x
3
+1)
′ 
= 5(
x
3
+1)
4
3
x
2 
= 15
x
2
(
x
3
+1)
4
.
Demak, 
h
′
(1) = 15(1
3
+1)
4
·1
2 
= 15·16 = 240.
Javob
: 240. 
▲
5-misol. 
f
(
x
) = 2
cos
x
funksiyaning hosilasini toping.

(1) formuladan foydalanib hosilani hisoblaymiz: 
f
′
(
x
) =2
cos
x
ln2 (cos
x
)
′ 
= –sin
x
2
cos
x
ln2.
Javob
: – sin
x
2
cos
x
ln2.
▲

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32
33
6-misol.
f
(
x
) = tg
5
x
funksiyaning hosilasini toping.

(1) formuladan foydalanib hosilani hisoblaymiz:
f
′
(
x
) =5tg
4
x
(tg
x
)
′ = 5tg
4
x
1
2
cos
x
.
Javob
: 
4
2
5tg .
cos
x
x
▲
7-misol.
 
h
(
x
)=3
cos
x
·log
7
(
x
3
+2
x
) funksiyaning hosilasini toping.

f 
(
x
)=3
cos
x
va 
g
(
x
)=log
7
(
x
3
+2
x
) belgilashlarni kiritib, (1) formulani –
murakkab funksiya hosilasini topish formulasini qo‘llaymiz: 
f 
ʹ(
x
)=(3
cos
x
)ʹ=3
cos
x
ln3·(cos
x
)ʹ
=–3
cos
x
ln3·sin
x
,
g
ʹ(
x
)=(log
7
(
x
3
+2
x
))ʹ=
3
1
(
2 )ln 7
x
x
+
·(
x
3
+2
x
)ʹ=
2
3
3
2
(
2 )ln 7
x
x
x
+
+
hamda 
h
(
x
) funksiyani 2 ta funksiyaning ko‘paytmasi deb qaraymiz:
h
ʹ(
x
) = (3
cos
x 
∙
log
7
(
x
3
+2
x
))ʹ = (3
cos
x
)ʹ·log
7
(
x
3
+2
x
) +
+ 3
cos
x 
⋅ 
(log
7
(
x
3
+2
x
))ʹ 
= –3
cos
x
·ln3·sin
x
·log
7
(
x
3
+2
x
)
+
cos
2
3
3 (3
2)
(
2 )ln 7
x
x
x
x
+
+
.
Javob
: –3
cos
x
·ln3·sin
x
·log
7
(
x
3
+2
x
)
+
cos
2
3
3 (3
2)
(
2 )ln 7
x
x
x
x
+
+
. 
▲
?
Savol va topshiriqlar
1. Murakkab funksiya deb nimaga aytiladi? Misol keltiring.
2. Murakkab funksiyaning aniqlanish sohasi qanday topiladi?
3. Murakkab funksiya hosilasini topish formulasini yoza olasizmi?
4. Murakkab funksiya hosilasini topishni 1–2 ta misolda ko‘rsating.

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32
33
Mashqlar
35. 
Agar 
1
)
(
2
−
=
x
x
f
bo‘lsa, ko‘rsatilgan funksiyalarni toping:
1) 
)
1
(
x
f
; 2) 
);
2
(
x
f
)
; 3) 
);
1
(
2
−
x
f
)
; 4) 
).
1
(
)
1
(
−
−
+
x
f
x
f
)
.
36.
Agar 
1
1
)
(
−
+
=
x
x
x
f
bo‘lsa, ko‘rsatilgan funksiyalarni toping:
1)
)
1
(
x
f
; 2) 
)
1
(
2
x
f
; 3)
);
1
(
−
x
f
)
; 4)
)
1
(
+
x
f
.
37.
Agar
1
)
(
,
)
(
2
−
=
=
x
x
g
x
x
f
bo‘lsa, quyidagilarni toping:
1)
));
(
(
x
g
f
2)
));
(
(
x
f
f
3) 
));
(
(
x
g
g
4) 
)).
(
(
x
f
g
38
. Agar 
1
)
(
,
)
(
2
3
+
=
=
x
x
g
x
x
f
bo‘lsa, quyidagilarni toping:
1)
;
1
)
(
)
(
2
−
x
g
x
f
2) 
2
3
)
(
3
)
(
−
+
+
x
x
g
x
f
; 
3)
));
(
(
x
g
f
4) 
))
(
(
x
f
g
)
.
Tenglikdan foydalanib,
f
(
x
)
 
ni toping (