[VNMATH.COM]-Tóm tắt lí thuyết và bài tập chương 5 -đạo hàm, giải tích 11

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.

Thc s Trn Vn Khnh

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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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