( ) vuvu

ИЗВОД ФУНКЦИЈЕ
ТАБЛИЦА ИЗВОДА ЕЛЕМЕНТАРНИХ ФУНКЦИЈА
1. y = c
2. y = x
y′ = 0
y′ = 1
3. y = x
y′ =
4. y = x n
y ′ = n ⋅ x n−1
12. y = ctg x
5. y = a x
y ′ = a x ln a
13. y = arcsin x
6. y = e x
y′ = e x
7. y = log a x
y′ =
8. y = ln x
9. y = sin x
10. y = cos x
1
2 x
1
x ln a
1
y′ =
x
11. y = tg x
14. y = arccos x
15. y = arctg x
16. y = arcctg x
y′ = −
y ′ = cos x
y ′ = − sin x
1
y′ =
cos 2 x
1
y′ = −
sin 2 x
1
y′ =
1− x2
1
1− x2
1
1+ x2
1
y′ = −
1+ x2
y′ =
ПРАВИЛА: Ако функције u (x ) и v (x ) имају изводе у тачки x , тада важи:
(c ⋅ u )′ = c ⋅ (u )′
c − konstanta
(u ± v )′ = u ′ ± v ′
(u ⋅ v )′ = u ′ ⋅ v + u ⋅ v ′
′
 u  u ′ ⋅ v − u ⋅ v′
  =
v2
v
1
ИЗВОД ЗБИРА И РАЗЛИКЕ
1. y = 5 y ′ = 0
y ′ = 4 ⋅ x 4 −1 = 4 ⋅ x 3
2. y = x 4
′
′
3. y = 3 ⋅ x 5 y ′ = (3) ⋅ x 5 + 3 ⋅ x 5 = 0 ⋅ x 5 + 3 ⋅ 5 ⋅ x 5−1 = 15 x 4
( )
( )′ = 3 ⋅ 5 ⋅ x
5
или краће y ′ = 3 ⋅ x
4. y = 4 ⋅ x 8
5−1
= 15 x 4
y ′ = 4 ⋅ 8 ⋅ x 8−1 = 32 x 7
5. y = 2 x 4 − x 3 + 4 x − 2 y ′ = 2 ⋅ 4 ⋅ x 4−1 − 3 ⋅ x 3−1 + 4 ⋅ 1 − 0 = 8 x 3 − 3 x 2 + 4
6. y =
1 3 3 2
1
x + x −x−
3
4
10
7. y =
1
2
3
+
−
x2 x3 x4
y′ =
1
3
3
⋅ 3 x 3−1 + ⋅ 2 x 2−1 − 1 − 0 = x 2 + x − 1
3
4
2
Функцију можемо написати у облику
y = x −2 + 2 x −3 − 3x −4
y ′ = −2 x − 2−1 + 2 ⋅ (− 3)x −3−1 − 3 ⋅ (− 4 )x − 4−1 = −2 x −3 − 6 x − 4 + 12 x −5 = −
2
6 12
−
+
x3 x 4 x5
8. y = 2 x − 33 x 2 + 44 x 3
Функцију можемо написати у облику
1
2
2
3
y = 2 x − 3x + 4 x
3
4
2
2
1
3
1 2 −1
2 3 −1
3 4 −1
′
y = 2 ⋅ x − 3⋅ x + 4 ⋅ x =
2
3
4
1
1
1
−
−
−
1
2
3
1
2
3
3
2
= x − 2 x + 3x 4 = 1 − 1 + 1 =
−3 +4
x
x
x
x2 x3 x4
9. y = 1 − 3
1
x
+ 2x x
Функцију можемо написати у облику
y = 1−
1
x
= 1− x
−
1
3
+ 2x
1+
1
2
= 1− x
1
−
1
3
+ 2x
3
1
3
1
2
+ 2x ⋅ x =
3
2
4
1
3 −1 1 −
1
1
 1  − −1
y′ = 0 −  −  x 3 + 2 ⋅ x 2 = x 3 + 3x 2 = 4 + 3 x =
+3 x
4
3
2
3
 3
3 x
3x 3
3
ИЗВОД ПРОИЗВОДА ДВЕ ФУНКЦИЈЕ
(
)
1. y = 2 x 2 + 4 x − 3 ⋅ (3x − 2)
′
′
y ′ = 2 x 2 + 4 x − 3 ⋅ (3x − 2) + 2 x 2 + 4 x − 3 ⋅ (3x − 2) =
(
)
(
)
(
)
= (4 x + 4 − 0) ⋅ (3 x − 2 ) + 2 x 2 + 4 x − 3 ⋅ (3 − 0 ) =
= 12 x 2 − 8 x + 12 x − 8 + 6 x 2 + 12 x − 9 = 18 x 2 + 16 x − 17
2. y = e x x 2
′
′
y′ = e x x 2 + e x x 2 = e x x 2 + e x ⋅ 2 x = xe x ( x + 2)
( )
( )
3. y = y = (x − 1)e x
′
′
y ′ = (x − 1) e x + (x − 1) e x = (1 − 0 )e x + (x − 1)e x = e x + xe x − e x = xe x
( )
4. y = x ln x y ′ =
( x )′ ln x +
1
1
′
x (ln x ) =
ln x + x
x
2 x
5. y = sin x ⋅ cos x
′
′
y′ = (sin x ) ⋅ cos x + sin x ⋅ (cos x ) =
= cos x ⋅ cos x = sin x ⋅ (− sin x ) = cos 2 x − sin 2 x
6. y = x 2 cos x
′
′
y ′ = x 2 cos x + x 2 (cos x ) = 2 x cos x + x 2 (− sin x ) = x(2 cos x − x sin x )
( )
7. y = x 3 sin x
′
′
y ′ = x 3 sin x + x 3 (sin x ) = 3 x 2 sin x + x 3 cos x = x 2 (3 sin x + x cos x )
( )
4
ИЗВОД КОЛИЧНИКА ДВЕ ФУНКЦИЈЕ
1. y =
5 − 3x
5 + 2x
′
′
(
5 − 3 x ) (5 + 2 x ) − (5 − 3 x )(5 + 2 x )
y′ =
(5 + 2 x )2
=
− 15 − 6 x − 10 + 6 x
(5 + 2 x )2
=
=
(0 − 3)(5 + 2 x ) − (5 − 3x )(0 + 2) =
(5 + 2 x )2
− 25
(5 + 2 x )2
x2 −1
2. y = 2
x +1
′
′
(
2 x − 0) x 2 + 1 − x 2 − 1 (2 x + 0)
x2 −1 x2 +1 − x2 −1 x2 +1
=
=
y′ =
2
2
2
2
x +1
x +1
(
)(
) (
(
=
)(
)
2x 3 + 2x − 2x 3 + 2x
3. y =
(x
2
)
+1
2
)
(
) (
(
=
)
)
4x
(x
2
)
+1
2
ln x
x
1
⋅ x − ln x ⋅1
(
ln x ) ( x ) − (ln x )( x )
1 − ln x
x
′
y =
=
=
x2
x2
x2
′
′
5
4. y =
1 + ln x
1 − ln x
′
y′ =
′
(1 + ln x ) (1 − ln x ) − (1 + ln x )(1 − ln x )
(1 − ln x )2
1
 1

 0 + (1 − ln x ) − (1 + ln x )1 − 
x
 x =
=
(1 − ln x )2
2
1 1
1 1
− ln x + + ln x
2
x x
x
= x x
=
=
(1 − ln x )2
(1 − ln x )2 x(1 − ln x )
ex
5. y =
x +1
′
′
e x ( x + 1) − e x ( x + 1)
e x ( x + 1) − e x (1 + 0 ) xe x + e x − e x
xe x
=
=
=
y′ =
(x + 1)2
(x + 1)2
(x + 1)2
(x + 1)2
( )
( )
sin x
1 + cos x
′
′
(
sin x ) (1 + cos x ) − (sin x )(1 + cos x )
cos x(1 + cos x ) − sin x(0 − sin x )
=
=
y′ =
2
2
(1 + cos x )
(1 + cos x )
6. y =
cos x + cos 2 x + sin 2 x
cos x + 1
1
=
=
=
2
2
(1 + cos x )
(1 + cos x ) 1 + cos x
7. y =
1 − cos x
sin x
′
′
(
(0 − (− sin x ))sin x − (1 − cos x ) cos x =
1 − cos x ) sin x − (1 − cos x )(sin x )
=
y′ =
sin 2 x
sin 2 x
sin x ⋅ sin x − cos x + cos 2 x sin 2 x − cos x + cos 2 x 1 − cos x 1 − cos x
=
=
=
=
=
sin 2 x
sin 2 x
sin 2 x
1 − cos 2 x
1 − cos x
1
=
=
(1 − cos x )(1 + cos x ) 1 + cos x
6
ИЗВОД СЛОЖЕНЕ ФУНКЦИЈЕ
Сложена функција облика y = f (φ (x )) може се написати преко посредне
функције u = φ (x ) , тј. може се написати у облику y = f (u ) .
Ова функција има извод који се одређује на следећи начин:
y ′x = y u′ ⋅ u ′x
односно
y ′( x ) = f ′(u ) ⋅ u ′( x )
y = f (u ) по променљивој u
( y ′ ) , па се затим он помножи са изводом посредне функције u = φ ( x ) по
Значи прво се одреди извод основне функције
u
променљивој x (u ′x ) .
З А Д А Ц И:
1. y = sin x ⇒
y = u , u = sin x y ′ =
( u )′ ⋅ u ′ = 2 1u ⋅ (sin x )′ = 2
x
1
sin x
⋅ cos x =
cos x
2 sin x
2. y = 1 − x 2
′
1
− 2x
x
1
(
y′ =
⋅ 1− x2 =
=−
0 − 2x) =
2 1− x2
2 1− x2
2 1− x2
1− x2
(
)
3. y = x 2 − 2 x
y′ =
1
2 x 2 − 2x
(x
2
′
− 2x =
)
1
2 x 2 − 2x
(2 x − 2) =
2(x − 1)
2 x 2 − 2x
=
x −1
x 2 − 2x
7
4. y =
(4 + 3x )
3
⇒ y = (4 + 3 x )
2
2
3
1
2
2
(4 + 3x ) 3 −1 ⋅ (4 + 3x )′ = 2 (4 + 3x )− 3 ⋅ (0 + 3) =
3
3
2
2
3
=
⋅
=
1
3
4 + 3x
3(4 + 3x ) 3
y′ =
x2 − x+2
5. y = e
y′ = e x
2
− x+2
(x
2
2
2
′
− x + 2 = e x − x + 2 (2 x − 1 + 0) = (2 x − 1)e x − x + 2
)
6. y = e 3 x +1
′
y ′ = e 3 x +1 (3 x + 1) = e 3 x +1 (3 + 0 ) = 3e 3 x +1
7. y = e x
2
2
( )′ = e
y′ = e x x 2
x2
⋅ 2 x = 2 xe x
2
8. y = a 5 x
′
y′ = a 5 x ⋅ ln a ⋅ (5 x ) = a 5 x ⋅ ln a ⋅ 5
(
)
12. y = ln 1 + x 2
1
1
2x
2 ′
(
)
y′ =
⋅
1
+
x
=
⋅
0
+
2
x
=
1+ x2
1+ x2
1+ x2
(
)
8
(
)
9. y = ln x 2 + 2 x + 3
′
1
1
(2 x + 2 + 0) = 2 2 x + 2
y′ = 2
x2 + 2x + 3 = 2
x + 2x + 3
x + 2x + 3
x + 2x + 3
(
(
10. y = ln 2 x 2 + 5
y′ =
)
)
′
1
1
4x
2
(
)
2
x
+
5
=
2
⋅
2
x
+
0
=
2x2 + 5
2x2 + 5
2x2 + 5
(
)
11. y = sin 7 x
′
y′ = 7 sin 6 x ⋅ (sin x ) = 7 sin 6 x ⋅ cos x
12. y = cos 3 x
′
y′ = 3 cos 2 x ⋅ (cos x ) = 3 cos 2 x ⋅ (− sin x ) = −3 cos 2 x ⋅ sin x
13. y = cos x 3
′
y′ = − sin x 3 ⋅ x 3 = − sin x 3 ⋅ 3x 2 = −3x 2 sin x 3
( )
14. y = ctg x 3
1
3x 2
3 ′
y′ = − 2 3 ⋅ x = − 2 3
sin x
sin x
( )
15. y = tg x 4
1
1
4x3
4 ′
3
y′ =
⋅ x =
⋅ 4x =
cos 2 x 4
cos 2 x 4
cos 2 x 4
( )
9
16. y = sin 2 x
′
y′ = cos 2 x ⋅ (2 x ) = cos 2 x ⋅ 2 = 2 cos 2 x
17. y = sin 2 3 x
′
′
y ′ = 2 sin 3 x ⋅ (sin 3 x ) = 2 sin 3 x ⋅ cos 3 x ⋅ (3 x ) =
= 2 sin 3 x ⋅ cos 3 x ⋅ 3 = 6 sin 3 x ⋅ cos 3 x
(
)
′
1
1
1 

(
1+ x) =
⋅ 0 +
=
(1 + x )⋅ ln 2  2 x 
x )⋅ ln 2
18. y = log 2 1 + x
y′ =
=
(1 +
1
1
⋅
1 + x ⋅ ln 2 2 x
(
)
(
19. y = ln 1 + e 2 x
)
1
1
1
2e 2 x
′
2x ′
2x
2x


y′ =
⋅ 1+ e
=
⋅ e ⋅2 =
⋅  0 + e ⋅ (2 x )  =
 1 + e2 x
1 + e2 x
1 + e2 x 
1 + e2 x
(
)
(
)
20. y = x 2 e ax
′
′
′
y′ = x 2 e ax + x 2 e ax = 2 xe ax + x 2 e ax ⋅ (ax ) =
( )
( )
(
= 2 xe ax + x 2 e ax ⋅ a = e ax 2 x + ax 2
)
10
21. y =
3− x
3+ x
′
′
′
(
1
3− x
1
3 − x ) (3 + x ) − (3 − x )(3 + x )

⋅
⋅
=
y′ =
 =
2
(
3 + x)
3− x 3+ x
3− x
2
2
3+ x
3+ x
(0 − 1)(3 + x ) − (3 − x )(0 + 1) =
1
⋅
=
2
(
3 + x)
3− x
2
3+ x
−3− x −3+ x
−6
1
1
=
⋅
=
⋅
=
2
2
(
3 + x)
3− x
3 − x (3 + x )
2
2
3+ x
3+ x
−3
−3
=
=
=
3
3− x
4
−
+
(
3
)(
3
)
x
x
⋅ (3 + x )
3+ x
−3
−3
=
=
(3 − x )(3 + x )(3 + x )2 (3 + x ) 9 − x 2
′
′
 1 − sin x 
1
1
1 − sin x 

 =
⋅
⋅
⋅
y′ =
 =
1 − sin x  1 + sin x 
1 − sin x
1 − sin x  1 + sin x 
2
1 + sin x
1 + sin x
1 + sin x
′
′
(
1 − sin x) (1 + sin x) − (1 − sin x)(1 + sin x)
1
1
=
=
⋅
⋅
2
(
1 − sin x
1 + sin x)
1 − sin x
2
1 + sin x
1 + sin x
(0 − cosx)(1 + sin x) − (1 − sin x)(0 + cosx) =
1
⋅
=
1 − sin x
(1 + sin x)2
2⋅
1 + sin x
− cosx
− 2 cosx
− cosx − cosx sin x − cosx + sin x cosx
1
1
=
=
⋅
=
⋅
=
2
2
1 − sin x (1 + sin x) (1 − sin x)(1 + sin x)
1 − sin x
(
)
x
+
1
sin
2⋅
2⋅
1 + sin x
1 + sin x
− cosx − cosx
1
=
=
=−
2
2
cosx
1 − sin x cos x
1 − sin x
22. y = ln
1 + sin x
1
11
ИЗВОД ИНВЕРЗНИХ ТРИГОНОМЕТРИЈСКИХ ФУНКЦИЈА
x
2
1. y = arcsin
′
1
1
x

⋅  =
⋅ =
y′ =
2
x2 2
 x 2
1
−
1−  
4
2
1
1
2. y = arcsin
4 − x2
4
⋅
1
=
2
1
4 −x2
2
⋅
1
=
2
1
4 −x 2
1
x
′
′
′
(
1
1
1) ⋅ x − 1 ⋅ ( x )

⋅  =
⋅
=
y′ =
2
2
x
x
1


1
1− 2
1−  
x
 x
1
−1
1
=
⋅ 2 =−
x2 −1 x
x x2 −1
1
0 ⋅ x − 1⋅1
=
2
2
x
x −1
x2
1
⋅
x
(
)
3. y = arcsin x 2 − 1
y′ =
1
(
2
)
1− x −1
=
2x
2x 2 − x 4
2
=
′
⋅ x2 −1 =
(
)
(
(
4
2
)
1 − x − 2x + 1
2x
x2 2 − x2
1
)
=
2x
x 2 − x2
=
⋅ (2 x − 0) =
2x
4
2
=
1 − x + 2x − 1
2
2 − x2
12
4. y = arctg
1
x
′
1
1

y′ =
=
⋅


2
 1   x  1+ 1
1+  
x2
x
1
5. y = arctg
1
1
 1 
 1 
⋅− 2  = 2
⋅− 2  = − 2
x +1
 x  x +1  x 
x2
1+ x
1− x
′
′
′
(
1 + x ) (1 − x ) − (1 + x )(1 − x )
1
1+ x 
y′ =
⋅
⋅
=
 =
2 
2
2
−
x
1
(1 − x )
(1 + x )

1+ x  
1+
1+ 

2
(1 − x )
1− x 
(0 + 1)(1 − x ) − (1 + x )(0 − 1) =
1
=
⋅
(1 − x )2 + (1 + x )2
(1 − x )2
1
(1 − x )2
=
1
⋅ (1 − x + 1 + x ) =
1 − 2x + x + 1 + 2x + x 2
2
1
1
=
⋅
=
=
2
2 1+ x2
1+ x2
2 + 2x 2
2
(
)
13