Bµi tËp
B¶ng c«ng thøc
(x )
n '
= nx
( x) = 2
'
n −1
1
( ku )
( u.v )
(x > 0)
x
( u + v − w)
'
( n ∈ N, n > 1)
'
= u '.v + v'.u
u '.v − v '.u
u
=
v2
v
(( u ) ) = nu .u '
u'
( u) = 2 u
'
c
( k lµ h»ng sè )
'
n '
( c lµ h»ng sè )
= u '+ v'− w'
= k .u '
'
( c) ' = 0
'
( cx ) =
n −1
⇒
'
− v'
1
=
2
v
v
KiÓm tra bµi cò
C©u hái : Lêi gi¶i sau sai ë ®©u ?
(x
2
(x
2
)
'
− 3x + 2 x − 5 = 2 x − 3 −
)
'
− 3x + 2 x − 5 = 2 x − 3 +
2
x
1
x
Bµi 1 : TÝnh ®¹o hµm cña c¸c
hµm sè sau :
a)
x x3
y= +
− 0,5 x 4
2 3
(
b) y = 1 - 2x + 3x
(
)(
)
2 20
c) y = x − 4 5 − 2 x
2
3
)
Gi¶i :
'
'
'
x x
x x
4
a ) y' = +
− 0,5 x = + − 0,5 x 4
2 3
2 3
1 3 2
1
3
2
3
= + x − 0,5.4 x = + x − 2 x
2 3
2
3
3
(
)
'
Bµi 1 : TÝnh ®¹o hµm cña c¸c
hµm sè sau :
a)
x x3
y= +
− 0,5 x 4
2 3
(
b) y = 1 - 2x + 3x
(
)(
)
2 20
c) y = x − 4 5 − 2 x
2
3
)
((
b) y' = 1 − 2 x + 3 x
(
= 20 1 - 2x + 3x
= 20(1 - 2x )
19
)
)
'
20
2
) (1 − 2 x + 3x )
2 19
( − 2 + 6x)
= - 40(1 - 2x ) (3 x − 1)
19
2 '
Bµi 1 : TÝnh ®¹o hµm cña c¸c
hµm sè sau :
a)
x x3
y= +
− 0,5 x 4
2 3
(
b) y = 1 - 2x + 3x
(
)(
)
2 20
c) y = x − 4 5 − 2 x
2
3
)
(
)(
'
) (
) (x
) − 6 x ( x − 4)
c) y = x − 4 5 − 2 x + 5 − 2 x
2
(
= 2x 5 − 2x
3
3
2
2
= 10x - 4x − 6 x + 24 x
4
4
= - 10x + 24 x + 10 x
4
2
2
3 '
2
−4
)
Ta cã
(
y = −2 x 5 + 8 x 3 + 5 x 2 − 20
)
'
y ' = − 2 x + 8 x + 5 x − 20 = −10 x + 24 x + 10 x
5
3
2
4
2
x − x +1
Cho hµm sè : y = 2
x + x +1
2
a) T×m y’
a) T×m ®¹o hµm cña hµm sè :
u= x
a) T×m ®¹o hµm cña hµm sè : y =
x−
x +1
x+
x +1
Ta cã :
(
2 x − 1) ( x 2 + x + 1) − (2 x + 1)( x 2 − x + 1)
y' =
( x 2 + x + 1) 2
2x 2 − 2
= 2
( x + x + 1) 2
Ta cã :
y = 1−
x2
2x
+ x +1
− 2( x + x + 1) − (2 x + 1)2 x
2x − 2
Nªn : y ' =
= 2
2
2
( x + x + 1)
( x + x + 1) 2
2
2
x − x +1
Bµi 2 : Cho hµm sè : y = 2
x + x +1
2
a) T×m y’
a) T×m ®¹o hµm cña hµm sè :
u= x
a) T×m ®¹o hµm cña hµm sè : y =
x−
x +1
x+
x +1
Ta cã :
(u) '= (
)
'
x =
1
2 x
x − x +1
y= 2
x + x +1
2
Bµi 2 : Cho hµm sè :
a) T×m y’
a) T×m ®¹o hµm cña hµm sè :
u= x
a) T×m ®¹o hµm cña hµm sè : y =
x−
x +1
x+
x +1
Ta cã :
(
)
1
1
1 −
x + x + 1 − 1 +
( x − x + 1)
2 x
2 x
y' =
2
x + x +1
(
)
x 1
1
x 1
1
x + x +1−
− −
− x − x +1+
− +
2 2 2 x
2 2 2 x
=
2
x + x +1
1
x −1
=
.
x ( x + x + 1) 2
(
)
x − x +1
y= 2
x + x +1
2
Bµi 2 : Cho hµm sè :
a) T×m y’
a) T×m ®¹o hµm cña hµm sè :
u= x
a) T×m ®¹o hµm cña hµm sè : y =
x−
x +1
x+
x +1
§Æt u= x hµm sè ®· cho trë thµnh :
u − u +1
2u
y= 2
= 1− 2
u + u +1
u + u +1
2
y' =
(
)
− 2u ' u 2 + u + 1 − ( 2u.u '+u ')( − 2u )
= 2u '.
(u
u −1
2
)
+ u +1
2
=
2u 2 .u '−2u '
(u
2
)
+ u +1
2
(u
2
)
+ u +1
Thay l¹i biÕn x ta cã :
2
y' =
1
.
x −1
x ( x + x + 1)
2
2
Bµi tËp vÒ nhµ :
Gi¶i bÊt ph¬ng tr×nh y’>0 biÕt :
3x + 1
a) y =
1- x
x
b) y = 2
x +4
c) y = x − 2 x + 3
4
2
1
d ) y = 4x - 1 +
x -1