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Mc lc
1 Hm s lng gic v phng trnh lng gic 2
2 T hp -xc sut 3
3 Dy s. Cp s cng v cp s nhn 4
4 Gii hn 5
5 o hm 6
5.1 nh ngha v ngha ca o hm . . . . . . . . . . . . . . . . . . . . 6
5.1.1 nh ngha o hm . . . . . . . . . . . . . . . . . . . . . . . . . 6
5.1.2 Quy tc tnh o hm bng nh ngha . . . . . . . . . . . . . . 6
5.1.3 Quan h gia tnh lin tc v s c o hm . . . . . . . . . . . 6
5.1.4 ngha hnh hc ca o hm . . . . . . . . . . . . . . . . . . . 7
5.1.5 ngha vt l ca o hm . . . . . . . . . . . . . . . . . . . . . 7
5.2 Quy tc tnh o hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2.1 Cc cng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2.2 Php ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2.3 o hm ca hm s hp . . . . . . . . . . . . . . . . . . . . . . 12
5.3 o hm ca hm s lng gic . . . . . . . . . . . . . . . . . . . . . . . 15
5.3.1 Cc gii hn cn nh . . . . . . . . . . . . . . . . . . . . . . . . . 15
5.3.2 Cc cng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5.4 Vi phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.4.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.5 o hm cp hai, o hm cp cao . . . . . . . . . . . . . . . . . . . . . 18
5.5.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
5.5.2 ngha c hc ca o hm cp hai . . . . . . . . . . . . . . . . 19
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Chng 1
Hm s lng gic v phng trnh
lng gic
2
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Chng 2
T hp -xc sut
3
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Chng 3
Dy s. Cp s cng v cp s nhn
4
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Chng 4
Gii hn
5
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Chng 5
o hm
5.1 nh ngha v ngha ca o hm
A. Tm tt l thuyt
5.1.1 nh ngha o hm
Cho hm s y = f(x) xc nh trn khong (a; b), x0 (a, b) , x0+x (a; b) . Nu tnti, gii hn (hu hn)
limx0
f(x0 +x) f(x0)x
c gi l o hm ca hm s f(x) ti x0, k hiu f (x0) hay y(x0).
Vy f (x0) = limx0
f(x0 +x) f(x0)x
= limxx0
f(x) f(x0)x x0 .
5.1.2 Quy tc tnh o hm bng nh ngha
Bc 1: Vi x l mt s gia ca i s ti x0, tnh y = f (x0 +x) f(x0);Bc 2: Lp t s
y
x;
Bc 3: Tnh limx0
y
x.
5.1.3 Quan h gia tnh lin tc v s c o hm
a) Hm s f(x) c o hm ti x0 th hm s f(x) lin tc ti x0;
b) Hm s f(x) lin tc ti x0 th cha hn f(x) c o hm ti x0.
6
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5.1 nh ngha v ngha ca o hm 7
5.1.4 ngha hnh hc ca o hm
Nu tn ti, f (x0) l h s gc ca tip tuyn ca th hm s y = f(x) ti
M0 (x0; f(x0)) . Khi phng trnh tip tuyn ca th hm s ti M0 l
y = f (x0)(x x0) + f(x0).
5.1.5 ngha vt l ca o hm
v(t) = s(t) l vn tc tc thi ca chuyn ng s = s(t) ti thi im t.
B. Bi tp minh ha
Dng ton 1. Dng nh ngha tnh o hm ca hm s y = f(x) ti x0Phng php
Bc 1: Cho x0 mt s gia l x; tnh y = f (x0 +x) f(x0);Bc 2: Lp t s
y
x;
Bc 3: Tnh limx0
y
xhoc tnh lim
xx0
f(x) f(x0)x x0 .
Bi 5.1. Dng nh ngha tnh o hm ca cc hm s sau ti x0:
a) y = f(x) = x2 3x+ 3 ti x0 = 1;b) y = f(x) =
x+ 1
x 1 ti x0 = 0;c) y = f(x) =
7 2x ti x0 = 3.
Gii
a) Cch 1
Cho x0 = 1 mt s gia l x; y = f (1 + x) f(1) = (x)2 x;T s
y
x= x 1;
Tnh limx0
yx = 1.
Cch 2
limx1
f(x) f(1)x 1 = 1
b) limx0
f(x) f(0)x
= 2;
c) limx3
f(x) f(3)x 3 = 1.
Dng ton 2. Vit phng trnh tip tuyn ca th (C)
Loi 1: Vit phng trnh tip tuyn ca th (C) ti mt im M0 (C)Phng php
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5.1 nh ngha v ngha ca o hm 8
Phng trnh tip tuyn ca th hm s ti M0 l
y = f (x0)(x x0) + f(x0).
Bi 5.2. Cho hm s y = x3 3x+ 1 c th (C). Vit phng trnh tip tuyn ca th (C) ti im c honh x0 = 3.
Gii
y = f (x) = 3x2 3.y (3) = f(3) = 19, f (3) = 24.
Vy phng trnh tip tuyn l y = 24x 53.
Bi 5.3. Cho hm s y =x2 + x
x 2 , (C).a) Hy tnh bng nh ngha o hm ca hm s cho ti x = 1;
b) Vit phng trnh tip tuyn ca (C) ti im A(1;-2).
Gii
a) Vi x l mt s gia ca i s ti x = 1, ta c
y =(1 + x)
2+ (1 + x)
1 + x 2 1 + 1
1 2 =5x+2x
x 1 ;
y
x=
5 +x
x 1;
limx0
y
x= 5.
Vy y(1) = 5.b) Phng tnh tip tuyn y = 5x+ 3.Loi 2: Vit phng trnh tip tuyn ca th (C) bit h s gc k
Phng php
Gi x0 l honh tip im. Khi f (x0) = k x0, tnh y (x0) = f(x0).Vy phng trnh tip tuyn cn tm l y = k (x x0) + y (x0) .
Bi 5.4. Cho hm s y = x3 3x+ 1 c th (C). Vit phng trnh tip tuyn ca th (C) bit tip tuyn c h s gc bng 9.
Gii
Gi x0 l honh tip im. Khi f (x0) = 9 3x20 3 = 9 x0 = 2.Vi x0 = 2 y (2) = 3. Phng trnh tip tuyn l y = 9x 15.Vi x0 = 2 y (2) = 1. Phng trnh tip tuyn l y = 9x+ 17.
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5.1 nh ngha v ngha ca o hm 9
Bi 5.5. Cho hm s y = x2 2x+ 3 c th (C). Vit phng trnh tip tuyn ca th (C):
a) Bit tip tuyn song song vi ng thng 4x 2y + 5 = 0;b) Bit tip tuyn vung gc vi ng thng x+ 4y + 5 = 0
Gii
a) Gi x0 l honh tip im. Khi f (x0) = 2 2x0 2 = 2 x0 = 2.x0 = 2 y (2) = 3. Phng trnh tip tuyn l y = 2x 1.b) Gi x0 l honh tip im. Khi f (x0) = 4 2x0 2 = 4 x0 = 3.x0 = 3 y (3) = 6. Phng trnh tip tuyn l y = 4x 6.Loi 3: Quan h gia hm s lin tc v o hm
Phng php
+ Hm s y = f(x) c o hm trn khong K l iu kin hm s lin tc trn
khong K hay ni cch khc l hm s lin tc iu kin cn hm s c o hm.
Ch : Hm s f(x) lin tc ti x0 th cha hn f(x) c o hm ti x0.
Bi 5.6. Chng minh rng hm s f(x) =
{(x 1)2, x 0(x+ 1)
2, x < 0
khng c o hm ti x = 0 nhng hm s lin tc ti .
Gii
a) Ta c f(0) = 1.
limx0
f(x) f(0)x 0 = limx0 (x+ 2) = 2;
limx0+
f(x) f(0)x 0 = limx0+ (x 2) = 2.
iu chng t hm s y = f(x) khng c o hm ti x = 0.
b) limx0
f(x) = limx0+
f(x) = 1 v f(0) = 1. Vy hm s lin tc ti x = 0.
Bi 5.7. Chng minh rng hm s f(x) =
{cosx, x 0sinx, x < 0
khng c o hm ti x = 0.
Gii
limx0+
g(x) = limx0+
cosx = 1;
limx0
g(x) = limx0
sinx = 0;
f(0) = cos0 = 1
nn hm s y = f(x) gin on ti x = 0.Vy hm s khng c o hm.
C. Bi tp t luyn
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5.1 nh ngha v ngha ca o hm 10
Bi 5.8. Dng nh ngha tnh o hm ca hm s:
a) y = f(x) = 2x2 3x+ 1 ti x0 = 1;b) y = f(x) =
4x 32 3x ti x0 = 1;
c) y = f(x) =5 2x ti x0 = 2;
d) f(x) =
1 cosxx
, x 6= 00, x = 0
, x0 = 0;
e) f(x) =
x
2 sin1
x, x 6= 0
0, x = 0, x0 = 0;
f) f(x) = sin 2x ti x0 =pi
4.
Hng dn
a) 1;
b) 125;
c) -1;
d)1
2;
e) 0;
f) 0.
Bi 5.9. Dng nh ngha tnh o hm ca cc hm s sau:
a) y = f(x) = x2 4x+ 1;b) f(x) =
1
2x 3 .
Hng dn
a) f (x) = 2x 4;b) f (x) =
2(2x+ 3)
2 .
Bi 5.10. Vit phng trnh tip tuyn ca th ca cc hm s sau:
a) y =x2 + 4x+ 5
x+ 2ti im c honh x = 0;
b) y = x3 3x2 + 2 ti im A(1;2);c) y =
2x+ 1 bit h s gc ca tip tuyn l
1
3.
Hng dn
a) y =3
4x+
5
2;
b) y = 9x+ 7;;
c) y =1
3x+
5
3.
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5.1 nh ngha v ngha ca o hm 11
Bi 5.11. Vit phng trnh tip tuyn ca th ca cc hm s sau:
a) y =2x+ 1
x 1 ti im c honh x = 2;b) y = x3 + x+ 3 ti im A(1; 1);c) y =
3x 2 ti im c honh x = 2.
Hng dn
a) y = 3x+ 11;b) y = 4x+ 5;
c) y =3
4x+
1
2.
Bi 5.12. Cho parabol (P ) c phng trnh
y = x2.
Tm h s gc ca tip tuyn ca parabol (P ):
a) Ti im A(-2;4);
b) Ti giao im ca (P ) vi ng thng y = 3x 2.
Hng dn
a) -4;
b) 2 v 1.
Bi 5.13. Cho hm s y = x3 c th (C).
a) Ti nhng im no ca (C) th tip tuyn ca (C) c h s gc bng 1;
b) Liu c tip tuyn no ca (C) m tip tuyn c h s gc m?
Hng dn
a)
(3
3;
3
9
),
(3
3;3
9
);
b) Khng c.
Bi 5.14. a) Cho hm s y =x+ 1
x 1 c th (C). Vit phng trnh tip tuyn ca th (C) bit tip tuyn song song vi ng thng y = 2x+ 1;b) Cho hm s y =
x+ 2
x 2 c th (C). Vit phng trnh tip tuyn ca th (C)bit tip tuyn vung gc vi ng thng y = x+ 2;
c) Cho hm s y = 13x3+3x2 5x+1 c th (C). Tm tip tuyn c h s gc ln
nht ca th (C).
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5.2 Quy tc tnh o hm 12
Hng dn
a) C hai phng trnh tip tuyn y = 2x 1, y = 2x+ 7;b) C hai phng trnh tip tuyn y = x 1, y = x+ 7;c) Gi x0 l honh tip im, y (x0) 4. T suy ra h s gc ln nht k = 4ng vi x0 = 3, y(3) = 4. Vy phng trnh tip tuyn y = 4x 8.
Bi 5.15. Chng minh rng hm s y = |x 2| khng c o hm ti x = 2 nhnglin tc ti im .
5.2 Quy tc tnh o hm
A. Tm tt l thuyt
5.2.1 Cc cng thc
a) (c) = 0;
b) (xn) = n.xn1, n 1, n N, x R;c) (
x)
=1
2x, x > 0.
5.2.2 Php ton
a) (u v w) = u v w;b) (u.v) = uv + uv;
c) (ku) = k.u;
d)(uv
)
=uv uv
v2;
e)(1
v
)
= v
v2.
5.2.3 o hm ca hm s hp
(yx)= (yu)
(ux)
.
B. Bi tp minh ha
Dng ton 1. Tnh o hm ca hm s y = f(x)
Phng php
Vn dng cc cng thc v cc php ton tnh o hm.
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5.2 Quy tc tnh o hm 13
Bi 5.16. Tnh o hm cc hm s sau:
a) y = f(x) = x4 3x3 + 5x2 4x+ 1;b) y = f(x) =
1
x2 1x3
;
c) y = f(x) =(x3 + 2
)(x+ 1);
d) y = f(x) =x 43x + 5
;
e) y = f(x) =x2 + x+ 1
2x 3 .
Gii
a) y = f (x) = 4x3 9x2 + 10x 4;b) y = f (x) =
3
x4 2x3;
c) y = f (x) = 4x3 + 3x2 + 2;
d) y = f (x) =17
(3x + 5)2;
e) y = f (x) =2x2 6x 5(2x 3)2
.
Bi 5.17. Tnh o hm cc hm s sau:
a) y =(4x3 2x2 5x) (x2 7x);
b) y =(2
x+ 3x
)(x 1);
c) y =x2 + 2x+ 3
x3 2 ;d) y = (x 2)x2 + 1.
Gii
a)
y =(4x3 2x2 5x) (x2 7x)+ (4x3 2x2 5x) (x2 7x)
=20x4 120x3 + 27x2 + 70x;
b) y =( 2x2
+ 3
)(x 1) + 1
xx+
3
2x;
c)
y =
(x2 + 2x + 3) (x3 2) (x2 + 2x+ 3) (x3 2)(x3 2)2
=x4 4x3 9x2 + 4x 4
(x3 2)2;
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5.2 Quy tc tnh o hm 14
d) y =2x2 2x+ 1
x2 + 1.
C. Bi tp t luyn
Bi 5.18. Tnh o hm cc hm s sau:
a) y =(x2 1)6;
b) y = x(x+ 2)4;
c) y =(x 1)2(x+ 1)3
.
Hng dn
a) y = 12x(x2 1)5;
b) y = (5x+ 2) (x+ 2)3;
c) y =(5 x) (x 1)
(x+ 1)4 .
Bi 5.19. Tnh o hm cc hm s sau:
a) y =x2 + 6x 7;
b) y =x+ 2 +
4 x;
c) y = x6 x.
Hng dn
a) y =x+ 3
x2 + 6x 7 ;
b) y =1
2
(1x+ 2
14 x
);
c)3 (4 x)26 x .
Bi 5.20. Tnh o hm cc hm s sau:
a) y =x2 + 4 +
(x2 1)2;
b) y = (x+ 1)x2 + x+ 1;
c) y =
x2 + x+ 3
2x+ 1.
Hng dn
a) y = 4x(x2 1)+ x
x2 + 4;
b) y =4x2 + 5x+ 3
2x2 + x+ 1
;
c) y =11
2(2x+ 1)2x2 + x+ 3
.
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5.3 o hm ca hm s lng gic 15
Bi 5.21. Tnh o hm cc hm s sau:
a) y = (9 2x) (2x3 9x2 + 1);b) y =
(x2 + 1
) (x3 + 1
)2(x4 + 1
)3;
c) y =(a+
b
x+
c
x2
)4.
Hng dn
a) y = 16x3 + 108x2 162x 2;b) y = 2x
(x3 + 1
)4(x4 + 1
)3+ 6x2
(x4 + 1
)3 (x2 + 1
) (x3 + 1
)+12x3
(x4 + 1
)2 (x2 + 1
) (x3 + 1
)2;c) y = 4
(a +
b
x+
c
x2
)3(b
x2+
2c
x3
).
Bi 5.22. a) Cho f(x) = x5 + x3 2x 3. Chng minh rng f (1) + f (1) = 4f(0);b) Cho f(x) = 2x3 + x2, g(x) = 3x2 + x+2. Gii bt phng trnh f (x) > g(x);c) Cho f(x) =
2
x, g(x) =
x2
2 x
3
3. Gii bt phng trnh f (x) g(x).
Hng dn
b) (; 0) (1; +);c) [1; 0).
5.3 o hm ca hm s lng gic
A. Tm tt l thuyt
5.3.1 Cc gii hn cn nh
a) limx0
sin x
x= 1;
b) limx0
sinx
x= 1, 6= 0.
5.3.2 Cc cng thc
a) (sinx) = cosx;
b) (cosx) = sinx;c) (tanx) =
1
cos2x;
d) (cotx) = 1sin2x
.
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5.3 o hm ca hm s lng gic 16
B. Bi tp minh ha
Dng ton 1. Tnh o hm cc hm s lng gic
Phng php
Dng gii hn ca hm s lng gic v cc cng thc tnh o hm ca cc hm s
lng gic.
Bi 5.23. Tnh cc o hm ca cc hm s sau:
a) y = sin 3x+ cosx
5+ tan
x;
b) y = sin(x2 5x+ 1)+ tan a
x;
c) y =x cot 2x;
d) y = 3sin2xcosx+ cos2x.
Gii
a) y = 3cos3x 15sin
x
5+
1
2xcos2
x;
b) y = (2x 5) cos (x2 5x + 1) ax2cos2
a
x
;
c) y =1
2xcot 2x 2
x
sin22x;
d) y = sinx(6cos2x 3sin2x 2cosx).
Bi 5.24. Tnh cc o hm ca cc hm s sau:
a) y = sin4x+ cos4x;
b) y = sin(2 sin x);
c) y = sin2(cos3x);
d) y =x
1 cosx .
Gii
a) y = 4sin3xcosx+ 4cos3x( sinx) = sin 4xb) y = 2cosxcos(2 sin x);
c) y = 3 sin 3x sin(2cos3x);d) y =
1 cosx x sinx(1 cosx)2
.
C. Bi tp t luyn
Bi 5.25. Tnh cc o hm ca cc hm s sau:
a) y = 3sin2x+ sin3x;
b) y = cosx cos3x;
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5.4 Vi phn 17
c) y = xcosx sinx;d) y =
1 + sinx
1 sinx ;
e) y =sinx cosxsin x+ cosx
Hng dn
a) y = 3 sin xcosx (2 sinx) ;b) y = sinx
(3cos2x 1);
c) y = x sinx;d) y =
2cosx
(1 sinx)2;
e) y =2
(sin x+ cosx)2.
5.4 Vi phn
A. Tm tt l thuyt
5.4.1 nh ngha
Cho hm s y = f(x) xc nh trn khong (a; b) v c o hm ti x (a; b). Gi sx l mt s gia ca x sao cho x+x (a; b).Tch f (x)x hay yx c gi l vi phn ca hm s f(x) ti x, ng vi s gia x,
k hiu l df(x) hay dy.
Ch
V dx = x nn dy = df(x) = f (x)dx.
B. Bi tp minh ha
Dng ton 1. Tm vi phn ca hm s
Phng php
+ Tnh o hm ca hm s y = f(x);
+ dy = df(x) = f (x)dx.
Bi 5.26. Tm vi phn ca cc hm s:
a) y = sinx x.cosx;b) y =
1
x3;
c) y =x3 + 1
x3 1.
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5.5 o hm cp hai, o hm cp cao 18
Gii
a) Ta c y = x sinx, do dy = x. sinxdx;
b) Ta c y = 3x4, do dy = 3
x4dx;
c) Ta c y = 6x2
(x3 1)2, do y = 6x
2
(x3 1)2dx.
Dng ton 2. Tnh gn ng
Phng php
ng dng ca vi phn vo tnh gn ng
f (x0 +x) f (x0) + f (x0)x.Bi 5.27. Tnh s gn ng sau (ly 5 ch s thp phn trong kt qu):
a)0, 99998;
b) sin(0, 00002)Gii
a) Xt hm s y =x, vi x0 = 1,x = 0, 00002. Vy
0, 99998 0, 99999;
b) Xt hm s y = sinx, vi x0 = 0,x = 0, 00002. Vy sin(0, 00002) 0, 00002.C. Bi tp t luyn
Bi 5.28. Tm vi phn ca cc hm s:
a) y = 2xsinx+(2 x2) .cosx;
b) y = sin(cos2x
).cos
(sin2x
);
c) y =sinx xcosxx sinx+ cosx
.
Hng dn
a) dy = x2. sinxdx;
b) dy = sin 2xcos (cos2x) dx;c) dy =
x2
(cosx+ x sin x)2 dx.
5.5 o hm cp hai, o hm cp cao
A. Tm tt l thuyt
5.5.1 nh ngha
Gi s hm s f(x) c o hm f (x). Nu f (x) cng c o hm th ta gi o hm
ca n l o hm cp hai ca f(x) v k hiu f (x):
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5.5 o hm cp hai, o hm cp cao 19
(f (x)
)
= f (x).
Tng t (f (x)
)
= f (x);(f (n1)(x)
)
= f (n)(x), n N
y k hiu f (0)(x) = f(x); f (n)(x) l o hm cp n ca hm s f(x).
5.5.2 ngha c hc ca o hm cp hai
o hm cp hai f (x) l gia tc tc thi ca chuyn ng s = f(t) ti thi im t.
B. Bi tp minh ha
Dng ton 1. Tnh o hm cp hai ca hm s
Phng php
y = f (x) = (f (x))
Bi 5.29. Tnh o hm cp hai ca cc hm s sau:
a) y = xx2 + 1;
b) y = tan x;
c) y = x. sin 2x;
d) y = sinx. sin 2x. sin 3x.
Gii
a) y =1 + 2x21 + x2
y = x(3 + 2x2
)(1 + x2)
1 + x2
;
b) y =1
cos2x y = 2 sin x
cos3x, x 6= pi
2+ kpi, k Z;
c) y = 4 (cos2x x sin 2x);d) y =
1
4sin 2x+
1
4sin 4x 1
4sin 6x y = sin 2x 4 sin 4x+ 9 sin 6x.
C. Bi tp t luyn
Bi 5.30. Tnh o hm cp hai ca cc hm s sau:
a) y =x
x2 1;
b) y =x+ 1
x 2.
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5.5 o hm cp hai, o hm cp cao 20
Hng dn
a)y =1
2
(1
x+ 1+
1
x 1) y = 1
2
( 1(x+ 1)2
1(x 1)2
);
y =
(1
(x+ 1)3+
1
(x 1)3)
b) y = 1 +3
x+ 2 y = 3
(x 2)2;
y =6
(x 2)3
Bi 5.31. Tnh o hm cp n ca cc hm s sau:
a) y =1
1 x ;
b) y =1
1 + x;
Hng dn
a) y =1
(1 x)2; y =
2
(1 x)3
y(n) =n!
(1 x)n+1. Chng minh bng phng php quy np;
b)y(n) =(1)nn!
(1 x)n+1. Chng minh bng phng php quy np.
Bi 5.32. Cho n l s nguyn dng. Chng minh rng:
a)(sinx)(n) = sin(x+ n
pi
2
);
b)(cosx)(n) = cos(x+ n
pi
2
).
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5.5 o hm cp hai, o hm cp cao 21
Bi tp cui chng
Bi 5.33. Dng nh ngha tnh o hm ca hm s y = tan x ti x0 DfHng dn
limxx0
tanx tan x0x x0 = limxx0
sinx sinx0x x0
1
cosx.cosx0=
1
cos2x0.
Bi 5.34. Dng nh ngha tnh o hm ca cc hm s sau: a) f(x) = sinx;
b) f(x) = cosx.
Hng dn
a) f (x) = cosx; b) f (x) = sinx.
Bi 5.35.
Hng dn
Bi 5.36.
Hng dn
Bi 5.37.
Hng dn
Bi 5.38.
Hng dn
Bi 5.39.
Hng dn
Bi 5.40.
Hng dn
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5.5 o hm cp hai, o hm cp cao 22
Bi 5.41.
Hng dn
Bi 5.42.
Hng dn
Bi 5.43.
Hng dn
Bi 5.44.
Hng dn
Bi 5.45.
Hng dn
Bi 5.46.
Hng dn
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