dao ham, tich phan ung dung duoc gi

Chuyn san EXP Nhm dch thut

Khoa Ton hc, trng i hc Khoa hc, Tp.HCM

1

Tc gi: Murray Bourne, ngi s hu trang www.intmath.com.

Bin dch: V Hong Trng, thnh vin chuyn san EXP, sinh vin khoa Ton Tin hc, trng

i hc Khoa hc T nhin, i hc Quc gia Tp. H Ch Minh.

Chnh sa: ng Phc Thin Quc, ch nhim chuyn san EXP, sinh vin khoa Ton Tin hc,

trng i hc Khoa hc T nhin, i hc Quc gia Tp. H Ch Minh.

Trnh by ba: Cng ty trch nhim hu hn Cng ngh Thit k DUKES, 30 Nguyn Vn Dung,

Phng 6, Qun G Vp, Tp. H Ch Minh.



2

- Cun sch ny c dch t 2 phn: Differentiation v Integration trn trang web

www.intmath.com, tiu o hm, Tch phn ng dng c g? do ngi bin dch t t.

(i) Bn quyn vi trang web IntMath:

- Xut bn theo s cho php ca tc gi thng qua th in t vo ngy 18 thng 1 nm 2015.

Email xin php dch thut t thnh vin ca chuyn san EXP, V Hong Trng:

Ive known this site since i was in high school and im very impressed. Your site so helpful for me.

So, I want to translate some lessons of your site (like differentiation, intergral, etc...) into

Vietnamese for studying and sharing to anyone who need. The production is a book or a file type

.PDF upload on the internet and sharing for free. No operation will be made. But first, I need your

agreement (for copyright). So, can I do this?

Email chp thun dch thut t qun l trang web IntMath, Murray Bourne:

Hello Trong

Thank you for your interest (and kind words) about IntMath and for requesting permission before

going ahead.

Id like to support you on this, but Id be more comfortable if the translated document was

published on IntMath, rather than somewhere else. Where did you hope to upload it to?

I was in your country a week ago. I love Vietnam!

Bng chng:

(ii) Bn quyn vi Chuyn san EXP:

- Ti, ng Phc Thin Quc, ch nhim Chuyn san EXP, trc thuc Cu lc b Hc thut khoa

Ton Tin hc, trng i hc Khoa hc T nhin, i hc Quc Gia Thnh ph H Ch Minh

ng chnh sa cun sch ca tc gi Murray Bourne do thnh vin V Hong Trng bin dch



3

theo tiu chun ca Chuyn san EXP.

- Cun sch ny c s dng min ph n bt k ai c nhu cu c. Chng ti khng ng h mi

hnh vi kinh doanh c lin quan n cun sch (bn ting Vit) ny m cha thng qua kin ca

Chuyn san EXP.

- Cc chnh sa bao gm:

(i) Thay i mu sc theo tiu chun ca EXP.

(ii) nh s, nh dng li paragraph cho ton vn bn.

(iii) Canh chnh kch thc hnh nh, ng khung,

(iv) Sa li cc nh dng Ton hc c, MathType sang nh dng Ton hc mi, Equation.

(v) nh dng li cc biu thc tng tc hon ton vi phn mm Microsoft Mathematics (c

th sao chp - dn trc tip cng thc m khng cn nh my li).

(vi) Kim tra chnh t, li tnh ton, li nh my st.

(vii) Tnh ton li, nh dng sai s 9 ch s thp phn (quy c cho ton b bi).

- Nhm chng ti hoan nghnh mi s gp , bnh lun ca bn cho cun sch c hon thin

hn. Mi phn hi v cun sch ny (phn ting Vit), c gi c th gi email v a ch:

[email protected]

tiu ghi [Phn hi o hm, Tch phn ng dng c g?].

- Trn trng cm n!

Thnh ph H Ch Minh, ngy 08 thng 05 nm 2015.



4

o hm, Tch phn ng dng c g? .............................................................................................. 1

Bn quyn ............................................................................................................................................ 2

Mc lc ................................................................................................................................................ 4

Li ni u ........................................................................................................................................... 7

Chng 1: Tng quan v ngnh vi tch phn ....................................................................................... 8

Chng 2: Vi phn ............................................................................................................................. 11

Phn 2.1: Vi phn (tm o hm) .............................................................................................. 11

Bi 2.1.1 M u ........................................................................................................................ 11

Bi 2.1.2 Gii hn v vi phn .................................................................................................... 14

Bi 2.1.3 dc ca tip tuyn vi ng cong (tnh ton gi tr) .......................................... 18

Bi 2.1.4 Nguyn l c bn tnh o hm ............................................................................. 21

Bi 2.1.5 o hm vi tc thay i tc thi ......................................................................... 25

Bi 2.1.6 o hm a thc ......................................................................................................... 28

Bi 2.1.7 o hm tch v thng ............................................................................................. 32

Bi 2.1.8 Vi phn hm s c ly tha ........................................................................................ 35

Bi 2.1.9 Vi phn hm n ........................................................................................................... 38

Bi 2.1.10 o hm cp cao ...................................................................................................... 41

Bi 2.1.11 o hm ring .......................................................................................................... 44

Phn 2.2: ng dng ca vi phn ............................................................................................... 48

Bi 2.2.1 Gii thiu v vi phn ng dng .................................................................................. 48

Bi 2.2.2 Tip tuyn v php tuyn ........................................................................................... 50

Bi 2.2.3 Cng thc Newton ...................................................................................................... 53

Bi 2.2.4 Chuyn ng cong ...................................................................................................... 57

Bi 2.2.5 Tc lin quan ......................................................................................................... 64

Bi 2.2.6 S dng vi phn v th ...................................................................................... 67

Bi 2.2.7 p dng vi phn x l nhng vn cc tr ......................................................... 77

Bi 2.2.8 Bn knh cong ............................................................................................................. 80

Phn 2.3: o hm hm s siu vit ......................................................................................... 89

Bi 2.3.1 M u ........................................................................................................................ 89

Bi 2.3.2 o hm hm s lng gic v ng dng .................................................................. 90



5

Bi 2.3.3 o hm hm s logarithm, hm m v ng dng ..................................................... 97

Chng 3: Tch phn ....................................................................................................................... 107

Phn 3.1: Tch phn ................................................................................................................ 107

Bi 3.1.1: M u..................................................................................................................... 107

Bi 3.1.2 Vi phn ..................................................................................................................... 109

Bi 3.1.3 Nguyn hm v tch phn bt nh ........................................................................... 111

Bi 3.1.4 Din tch di ng cong ....................................................................................... 117

Bi 3.1.5 Tch phn xc nh ................................................................................................... 123

Bi 3.1.6 Quy tc hnh thang .................................................................................................... 130

Bi 3.1.7 Quy tc Simpson....................................................................................................... 133

Phn 3.2 ng dng ca tch phn ........................................................................................... 139

Bi 3.2.1 M u ...................................................................................................................... 139

Bi 3.2.2 ng dng ca tch phn bt nh .............................................................................. 140

Bi 3.2.3 Dng tch phn tnh din tch di ng cong ....................................................... 143

Bi 3.2.4 Dng tch phn tnh din tch di 2 ng cong .................................................... 148

Bi 3.2.5 Th tch khi trn xoay ............................................................................................. 153

Bi 3.2.6 Trng tm b mt...................................................................................................... 163

Bi 3.2.7 Moment qun tnh..................................................................................................... 170

Bi 3.2.8 Cng sinh ra bi lc bin thin ................................................................................ 173

Bi 3.2.9 in tch ................................................................................................................... 177

Bi 3.2.10 Gi tr trung bnh .................................................................................................... 178

Bi 3.2.11 Tiu chun chn thng u (HIC): Ch s nghim trng ..................................... 179

Bi 3.2.12 Tiu chun chn thng u (HIC): Ch s HIC, v d ......................................... 183

Bi 3.2.13 Lc ca p sut cht lng ....................................................................................... 186

Bi 3.2.14 S dng tch phn tnh di ng cong ............................................................. 189

Bi 3.2.15 di ng cong: phng trnh tham s, ta cc .......................................... 195

Phn 3.3: Cc cng thc tnh tch phn .................................................................................. 200

Bi 3.3.1 M u ...................................................................................................................... 200

Bi 3.3.2 Cng thc tnh tch phn hm ly tha tng qut .................................................... 201

Bi 3.3.3 Cng thc tnh tch phn hm logarithm c bn ...................................................... 205

Bi 3.3.4 Cng thc tnh tch phn hm m ............................................................................ 208

Bi 3.3.5 Cng thc tnh tch phn hm lng gic c bn ..................................................... 211



6

Bi 3.3.6 Mt s cng thc khc tnh tch phn hm lng gic............................................. 215

Bi 3.3.7 Cng thc tnh tch phn hm lng gic ngc ..................................................... 220

Bi 3.3.8 Tch phn tng phn ................................................................................................. 223

Bi 3.3.9 Tnh tch phn bng cch t n lng gic ............................................................. 229

Bi 3.3.10 Bng mt s tch phn thng gp ......................................................................... 233

Bi 3.3.11 Tnh tch phn bng cch dng bng ...................................................................... 235

Bi 3.3.12 Tnh tch phn bng cng thc quy ................................................................... 236

Bi 3.3.13 Tnh tch phn bng phn s ring phn ................................................................. 238

Chng 4: Bi c thm .................................................................................................................. 243

Bi 4.1 Archimedes v din tch mt phn hnh parabola ....................................................... 243

Bi 4.2 Th tch mt dy chuyn.............................................................................................. 248

Bi 4.3 Newton ni g v vi tch phn? ............................................................................... 252

Bi 4.4 Tng Riemann ............................................................................................................. 257

Bi 4.5: nh l c bn ca vi tch phn .................................................................................. 261

Bi 4.5 Cng thc Tanzalin tnh tch phn tng phn ............................................................. 265

Gii thiu trang www.intmath.com ................................................................................................. 268



7

- Cho bn, ti tn V Hong Trng. Khi hon tt cun sch ny, ti l sinh vin nm 2, khoa Ton

Tin hc, trng i hc Khoa hc T nhin, i hc Quc gia Thnh ph H Ch Minh.

- Ti hin ang l thnh vin Chuyn san EXP. y l mt trong cc sn phm ca nhm chuyn

san EXP, trc thuc CLB hc thut, khoa Ton - Tin hc, i hc Khoa hc T nhin i hc

Quc gia Tp. H Ch Minh. Trong hn 3 nm qua nhm chng ti thc hin cc d n quy m

nh nhm ci thin tnh trng gio dc Vit Nam, ht li cht xm t nc ngoi tr v, v hin i

ha cc cng c Ton hc trong nc.

- Ti t nhn ti l mt a thch Ton. Khi ti hc cp 3, ti c th t gii nhng bi ton kh

trn lp m khng ai trong lp gii c cng nh chng c ai hng dn ti cch lm, nht l tch

phn. Vo thi im y, ti c th ngi hng gi lin ch gii mt bi tch phn no v ngy

hm sau em ln lp np ly im 10. Khi y, ti bit kh nhiu cch gii cc bi tch phn, t

m c, tm kim trn mng cng c, ng nhin ti ly lm t ho lm.

- Vo cui nm 12, ti t hi: Khng bit nc ngoi h hc o hm, tch phn nh th no?.

Vi bn tnh t m, ti ln Google tm kim v ti tip cn trang www.intmath.com. Cng vi

trang tra t trc tuyn tratu.soha.vn dch t vng, ti t m xem cch m trang web ny ni v

o hm, tch phn v sau ti b cun ht, khng phi v trang ny c nhng cch gii hay,

nhiu phng php mi m l nhng ng dng trong i sng hng ngy ca o hm, tch phn, v

d nh chn ch ngi d quan st nht trong rp phim, cch thit k khc cua ca con ng, xc

nh trng tm ca vt th, tnh cng sinh ra, Ngoi ra, ti cn bit c bn cht thc s ca

tch phn l g, du t u m ra hay mang ngha g. Cch hng dn ca trang web ny

song hnh l thuyt ln ng dng thc tin, to c s thu ht i vi ti v ti quyt nh dch

cc bi trong trong trang web nhm lm ngun ti liu cho ring mnh cng nh chia s cho bt

k ai c nhu cu c v tm hiu nhng ng dng ca o hm, tch phn trong cuc sng.

- Trc kia, ti ngh tch phn l ci g gh gm m ch cc b c thin ti mi ngh ra c,

nhng sau khi bit c lch s hnh thnh ca chng, ti ngh sai. S tht th tng hnh thnh

khi nim tch phn rt n gin v ti tin ngay c nhng hc sinh lp 6, lp 7 cng c th hiu

c tng ny. c bit hn, nhng iu m ti ni trn him khi c cp trong nhng tit

ton trn lp. Cn vic tnh tch phn ? Trong lc ti cn khng bit nn tnh tch phn tng phn

hay t n nh th no th ngi ta nghin cu ra phng php lp trnh trn my tnh v gii ra

p s cho bt k bi tch phn no vi chnh xc n kinh ngc. Ngi ta y chnh l

nhng ngi sng cch y gn c th k. Qua , ti thy rng trnh ton ca mnh tt

hu xa so vi Th gii.

- Ti nghe nhiu bn hi rng: o hm, tch phn c ng dng g trong cuc sng? ng tic

y l phn th v v hp dn nht li c cp qu t trong sch gio khoa. Hi vng rng qua

cun sch ny, bn s c cu tr li.

- Li cui cng, ti chn thnh cm n ng Murray Bourne, tc gi trang www.intmath.com cho

php ti dch ngun ti liu t trang web ny.

- Cn by gi, mi bn bt u hnh trnh khm ph vi tch phn.



8

- Ngnh vi tch phn nghin cu v nhng i lng bin thin phi tuyn tnh, c s dng rng

ri trong cc ngnh khoa hc v k thut, xut pht t nhng vn m chng ta c hc (nh vn

tc, gia tc, dng in trong mch) trong thc t khng h n gin, gn gng, p . Nu nhng

i lng thay i 1 cch lin tc, chng ta cn php vi tch phn tm hiu xem chuyn g xy

ra vi i lng y.

- Ngnh vi tch phn c pht trin bi mt nh khoa hc ngi Anh tn Issac Newton v mt nh

khoa hc ngi c l Gottfried Lebniz, 2 nh khoa hc ny nghin cu 1 cch c lp vi nhau v

nhng i lng bin thin vo khong cui th k 17. c 1 cuc tranh ci rng ai l ngi u

tin pht trin ngnh vi tch phn, nhng do 2 nh khoa hc ny nghin cu c lp vi nhau nn

chng ta c s ha ln khng c nh v k hiu v cch din t khi dng vi tch phn. T

Lebniz ta c k hiu

v .

Isaac Newton (1642 1726) Gottfried Wilhelm von Leibniz (1646 1716)

- S pht trin ca ng h chy chnh xc tng giy vo th k 17 mang li nhiu ngha quan

trng trong khoa hc ni chung v ton hc ni ring, v nh cao ca s pht trin l ngnh vi

tch phn.

- i vi cc nh khoa hc th y l iu rt quan trng c th d on v tr ca nhng ngi

sao, qua h tr cho ngnh hng hi. Th thch ln nht ca cc thy th khi i bin chnh l xc

nh kinh ca con tu ngoi khi, bt k quc gia no a tu c n Th Gii Mi u s

mang v rt nhiu vng bc chu bu, thc phm, qua quc gia trn s tr nn giu c.

- Newton v Lebniz xy dng trn cc php ton i s v hnh hc ca Rene Descartes, ngi pht

trin h ta Descartes m chng ta gp trong chng trnh ph thng.

- Ngnh vi tch phn ny c 2 mng chnh:

+ Vi phn (hay o hm) gip chng ta tm ra tc thay i ca 1 i lng vi 1 i lng

khc.

+ Tch phn, ngc vi vi phn. Chng ta c th c cho trc 1 gi tr bin thin no v ta

phi lm iu ngc li, tc tm mi quan h ban u (hay phng trnh ban u) gia 2 i lng.



9

Th tch thng ru l mt trong nhng vn

c gii quyt bng cch s dng phng php vi tch phn

I. Vi tch phn trong hnh ng 1:

- Mt thp nng lng cung cp in t mt tri bng cch thit lp hng ngn tm gng c kh

nng iu chnh c, gi l gng nh nht, mi tm gng c t trn nh thp, thu nng

lng nhit t mt tri v ct gi trong b cha nhng ht mui c nu chy (nm bn phi

thp) vi nhit hn 500.

- Khi cn dng in, nng lng trong b c dng to hi nc truyn chuyn ng cho

turbine sinh ra in ( bn tri thp).

- Vi tch phn (c th trong trng hp ny l o hm) dng lm tng ti a cng sut qu trnh

ny.

Solar Two phc v cho n nng lng California

II. Vi tch phn trong hnh ng 2:

- Vi tch phn dng pht trin nng sut cng v nhng thnh phn khc ca my tnh.



10

III. Mc lc:

Chng 2: Vi phn:

- Chng ny c 3 phn gm:

+ Phn 2.1 Vi phn: Gii thiu s nt v o hm v mt s v d c bn v k thut tnh vi phn.

+ Phn 2.2 ng dng ca vi phn: Ni ta s khm ph mt s ng dng c bn, bao gm c tm

tip tuyn, nhng vn v chuyn ng cong cng nh ti u ha.

+ Phn 2.3 Vi phn hm s siu vit: Ta s khm ph cch tm o hm ca mt s hm s nh

hm sine, cosine, logarithms v hm s m.

Chng 3: Tch phn:

- Ba phn trong chng ny l:

+ Phn 3.1 Tch phn: Ta s khm ph mt s nt c bn ca tch phn.

+ Phn 3.2 ng dng ca tch phn: Ni ta s thy vi ng dng c bn ca tch phn gm tnh

din tch, th tch, trng tm, moment qun tnh, np in tch v gi tr trung bnh. Mt iu th v

l Archimedes nm c vi yu t hnh thnh nn vi tch phn trc c Newton v Leibniz

tn 2000 nm!

+ Phn 3.3 Cng thc tnh tch phn: Phn ny s cho cc bn thy vi k thut tnh tch phn.

Chng 4: Bi c thm

- Nhng cu chuyn lch s v mt s cch tnh vi tch phn khc s c nu trong chng ny.



11

- Ni dung trong phn 2.1 ny:

+ Bi 2.1.1 M u.

+ Bi 2.1.2 Gii hn v vi phn.

+ Bi 2.1.3 dc ca tip tuyn vi ng cong (tnh ton s).

+ Bi 2.1.4 Nguyn l c bn tnh o hm.

+ Bi 2.1.5 o hm vi tc thay i tc thi.

+ Bi 2.1.6 o hm a thc.

+ Bi 2.1.7 o hm tch v thng.

+ Bi 2.1.8 Vi phn hm s c lu tha.

+ Bi 2.1.9 Vi phn hm n.

+ Bi 2.1.10 o hm cp cao.

+ Bi 2.1.11 o hm ring.

I. Vi phn l g?

- Php vi phn ch yu tm tc thay i ca i lng ny vi i lng khc. Chng ta cn php

vi phn khi tc thay i khng c gi tr c nh, iu ny c ngha l g?

II. Tc thay i c nh:

- u tin, ta s kho st mt chic xe chuyn ng vi tc 60 , th qung ng thi

gian s nh th ny:

- Chng ta cn lu rng qung ng tnh t im xut pht tng vi hng s c nh l 60

mi gi, v vy sau 5 chic xe i c 300. Ch rng dc (gradient) lun l 300

5= 60

trong ton b th. y chnh l tc thay i c nh ca qung ng theo thi gian, dc

lun dng (v th i ln khi bn i t tri sang phi).

III. Tc thay i khng c nh:

- By gi ta qung qu bng ln tri. Di tc dng ca trng lc th qu bng di chuyn chm dn,

sau bt u i ngc chiu chuyn ng ban u v rt xung. Trong sut qu trnh chuyn ng

th vn tc qu bng thay i t dng (khi qu bng i ln), chm v 0, sau v m (qu bng



12

ri xung). Trong qu trnh i ln, qu bng c gia tc m v khi n ri xung th c gia tc dng.

- Ta c th mi lin h gia cao () v thi gian ().

- Lc ny dc ca th thay i trong sut qu trnh chuyn ng. Ban u dc kh ln, c

gi tr dng (biu th vn tc ln khi ta nm bng), sau khi qu bng chm dn, dc ngy

cng t v bng 0 (khi qu bng im cao nht v vn tc lc bng 0). Sau qu bng bt u

rt xung v dc chuyn sang m (ng vi gia tc m) sau ngy cng dc hn khi vn tc

tng ln.

- dc ca mt ng cong ti 1 im cho ta bit tc thay i ca i lng ti im .

IV. Khi nim quan trng: tnh xp x ca ng cong:

- By gi ta hy phng to mt phn th gn v tr = 1 (ni ti nh du hnh ch nht pha

trn), quan st mt on ngn gia v tr = 0.9 v = 1.1, n s trng ging nh th ny:

- Lu rng khi ta phng to gn ng cong, n bt u ging nh ng thng. Chng ta c

th tm gi tr xp x dc ca ng cong ti v tr = 1 (chnh l dc ca tip tuyn ca



13

ng cong c v mu ) bng cch quan st nhng im m ng cong i qua gn = 1

(tip tuyn l 1 ng thng tip xc vi ng cong ti duy nht 1 im).

- Quan st th, ta thy rng ng cong y i qua (0.9; 36.2) v (1.1; 42). Vy dc ca tip

tuyn ti v tr = 1 khong: 2 12 1

=42 36.2

1.1 0.9= 29

- n v l ging nh vn tc, vy chng ta tm c tc thay i bng cch nhn vo

dc.

- R rng, nu chng ta phng to gn hn, ng cong s thng hn v ta s c gi tr xp x ng

hn cho dc ca ng cong.

- tng ca vic phng to vo th v tm gi tr xp x ng nht ca dc ng cong

(cho ta bit c tc thay i) dn n s pht trin ca vi phn.

V. S pht trin ca php tnh vi phn:

- Cho n thi i ca Newton v Lebniz th vn cha c 1 cch chc chn d on hay miu t

v hng s bin i ca vn tc. C 1 s cn thit thc t hiu lm nh th no ta c th phn

tch v d on cc i lng c hng s bin thin. l l do h pht trin php tnh vi phn.

VI. Ti sao phi nghin cu php tnh vi phn?

- C rt nhiu ng dng ca php vi phn trong khoa hc v k thut.

- Vi phn cn c dng trong vic phn tch v ti chnh cng nh kinh t.

- Mt ng dng quan trng ca vi phn l ti u ha phm vi, tc tm iu kin gi tr ln nht

(hay nh nht) xy ra. iu ny rt quan trng trong kinh doanh (tit kim chi tiu, gia tng li ch)

v k thut ( di ln nht, gi tin nh nht).

VII. V d v ti u ha:

- Mt hp c y hnh vung c m mt trn. Nu s dng vt liu 64 2 th th tch ln nht

c th ca hp l bao nhiu?

- Chng ta s gii quyt vn ny trong phn sau: ng dng ca vi phn.

VIII. Tnh gn ng m chng ta s dng:

- Nhng tnh gn ng di y u c gi tr rt quan trng:

+ Tr s gn ng tm dc.

+ i s gn ng tm dc.

+ Tp hp nhng quy lut ca vi phn.

- Bn c th b qua phn ng dng nu bn ch cn quan tm n cch tnh vi phn, nhng y s l

mt thiu st ln v bn s khng bit c ti sao li c cch .



14

- Tip theo bi M u, hiu r hn v ngnh ny, trc tin chng ta phi hiu v gii hn.

I. Gii hn:

- Trong vic nghin cu v ngnh vi tch phn, chng ta s cm thy th v v iu g s xy ra vi

mt hm s khi cc gi tr khc nhau thay vo hm th hm n gn mt gi tr c th. Chng

ta bt gp iu ny trong bi Vi phn (o hm) khi phng to ng cong tm gi tr xp x

ca dc ng cong.

II. Gii hn khi tin n mt con s c th:

- Thnh thong vic tm gi tr gii hn ca mt biu thc ch n gin l th s.

V d 1: Tm gii hn khi tin n 10 ca biu thc = 3 + 10.

Tr li v d 1:

- S dng k hiu gii hn, ta vit nh sau:

lim10

(3 + 7)

- V d ny khng kh khn g c, ta ch th s 10 vo biu thc v vit:

lim10

(3 + 7) = 37

- iu ny hp l v hm () = 3 + 7 l hm lin tc

- Tuy nhin c mt vi trng hp ta khng th p dng cch ny.

V d 2: Trong biu thc sau th hin nhin khng th bng 3 (do mu s phi khc 0), hy tm

gii hn biu thc khi tin n 3:

() =2 2 3

3

Tr li v d 2:

- Chng ta c th thy hm s tin n gn mt gi tr c th khi tin n 3 t bn tri:

2.5 2.6 2.7 2.8 2.9

() 3.5 3.6 3.7 3.8 3.9

- Tip tc tin gn n gi tr = 3:

2.9 2.92 2.94 2.96 2.97 2.98 2.99

() 3.9 3.92 3.94 3.96 3.97 3.98 3.99

- Tng t, tin n 3 t bn phi cho ta gi tr gii hn tng t:

3.5 3.1 3.01 3.00001

() 4.5 4.1 4.01 4.00001

- Ta nhn thy rng cc gi tr hm tin gn n 4:

- Ta vit:

lim3

2 2 3

3= 4

- Ch : Ta c th tm gi tr gii hn ny bng cch phn tch thnh nhn t:

lim3

2 2 3

3= lim3

( + 1)( 3)

3= lim3

+ 1 = 4



15

- Cch lm ny ng v ta c 3.

- y l v d c bn nhm gii thiu vic nghin cu gii hn. N c v kh ng ngn v nhng g

ta lm ch khc g bi ton cp 2, nhng li rt quan trng v n th hin rng hm khng tn ti gi

tr thc no khi = 3, nhng khi ta cho ngy cng dn ti 3 th gi tr hm cng i v mt gi tr

thc (nh trong v d trn l 4).

III. Gii hn khi tin n 0:

- Chng ta phi nh rng chng ta khng th chia cho s 0.

- Nhng c mt vi iu rt th v v quan trng, l gii hn khi tin n 0 v ni m gi tr

gii hn xut hin khi ta c mu s bng 0.

V d 3: Tm gii hn khi tin n 0 ca sin()

.

Tr li v d 3:

- Ta khng th thay s 0 vo biu thc v sin(0)

0 khng xc nh.

- Khng c phng php i s no tm gii hn ny, nhng ta c th tm bng cch cho tin

gn n 0 t bn tri v phi v c kt lun rng:

lim0

sin()

= 1

- Mt cch kim chng kt qu ny l da vo th v ta thy rng gi tr hm s khi gn

n 0 l 1.

- C ch trng ni = 0 trong th nhng n qu nh chng ta thy c.

IV. Gii hn khi tin n v cc:

V d 4: Cho biu thc 5

, chuyn g s xy ra vi biu thc khi tin ra v cc?

Tr li v d 4:

- R rng khi gi tr cng ln th gi tr biu thc ngy cng nh cho n khi n st gi tr 0, ta

ni rng gii hn ca 5

khi tin ra v cc l 0.

V. Gii hn khi gi tr bin thin mu:

- Mt cch tng qut:

lim

(1

) = 0

- Tng t:

lim

(1

2) = 0



16

- Ta dng nhng gi tr gii hn ny khi cn c lng gii hn ca cc hm s v c bit hu ch

khi ta v th ng cong.

V d 5: Tm gii hn:

lim

(5 3

6 + 1)

Tr li v d 5:

- Bi ny khng my r rng gi tr gii hn l bao nhiu. Ta c th thay gi tr cng ln dn vo

biu thc cho n khi ta pht hin iu g kh quan (hy th vi 100, ri 1,000, ri 1,000,000 v

c th).

- Hoc ta c th sp xp biu thc v dng cng thc:

lim

(1

) = 0

tm gi tr gii hn.

- Ta chia c t v mu cho to ra biu thc m ta c th nh gi c gi tr gii hn.

lim

(5 3

6 + 1) = lim

(

5 3

6 +1

) =0 3

6 + 0=

1

2

- Ch rng ta khng thay k hiu vo biu thc 5

3

6+1

v n khng c ngha trong ton hc.

- ng vit 53

6+1, iu ny khng ng u nh!

V d 6: Tm gii hn:

lim

(1 2

82 + 5)

Tr li v d 6:

- Cch th s: Thay cc gi tr ln dn vo biu thc nh 100, ri 10,000, ri 1,000,000, v ta

nhn thy biu thc tin v 1

8.

- Cch i s: Chia t v mu cho 2 ri ly gii hn:

lim

(1 2

82 + 5) = lim

(

12 1

8 +52

) = 1

8

VI. Tnh lin tc v vi phn:

- Trong phn ny ta s ly vi phn ca a thc, sau ta s gii quyt nhiu hm kh hn, c khi ta

khng th ly vi phn c. Ta cn phi hiu iu kin no mt hm c th ly vi phn.

- Mt hm s nh () = 3 62 + 30 l hm lin tc vi mi gi tr ca nn c th ly vi

phn vi mi gi tr ca .



17

- Tuy nhin, hm s nh () =2

2 khng xc nh ti = 0 v = 1.

- Hm khng lin tc ti 2 im , v vy ta khng th ly vi phn vi nhng gi tr nh vy.

VII. Hm s nhiu phng trnh v vi phn:

- Hm s nhiu phng trnh ly c vi phn vi mi nu hm s y lin tc vi mi

V d 7:

() = {2 + 3 < 12 + 2 1

- Hm s ny khng lin tc ti = 1, nhng vn tn ti gi tr ti = 1 (c th (1) = 1). Hm

s ny c vi phn vi mi tr gi tr = 1 v hm khng lin tc ti im trn.



18

- Trong bi vit ny, ti s cho bn thy mt trong nhng vn c t lu, l tm dc tip

tuyn ca ng cong. Vn ny c trc khi vi phn ra i.

- Khi chng ta m hnh ha nhiu vn vt l bng cch s dng ng cong, ta phi hiu v

dc ca ng cong nhiu im khc nhau v ngha ca dc trong nhng ng dng thc t.

- Hy nh rng: Ta ang c gng tm tc thay i ca 1 i lng ny so vi i lng khc.

- Nhng ng dng bao gm:

+ Nhit thay i trong thi gian nht nh

+ Vt tc ca 1 vt th ri t do trong khong thi gian nht nh.

+ Dng in qua mch trong thi gian nht nh.

+ S bin thin ca th trng chng khon trong khong thi gian nht nh.

+ S gia tng dn s trong khong thi gian nht nh.

+ Nhit gia tng theo t trng trong bnh gas.

- Sau , ta s khm ph ra tc thay i ca nhng iu trn bng cch ly vi phn hm s v

thay th gi tr thch hp vo. By gi, ta bt u tm tc thay i mt cch gn ng (c ngha

l ta thay s vo cho n khi ta tm c gi tr xp x ph hp).

- Ta quan st trng hp tng qut v vit phng trnh ph hp bao gm n (c lp) v gi tr

(khng c lp).

- dc ca ng cong = () ti im chnh l dc tip tuyn ti . Ta cn tm dc

ny gii quyt nhiu ng dng v n cho ta bit tc thay i mt cch nhanh chng.

- Ta vit = () trn ng cong v l hm theo , tc l, nu thay i th cng thay i.

* K hiu :

- k hiu ny, ta vit:

+ Thay i theo l .

+ Thay i theo l .

- Theo nh ngha ny, dc c cho bi:

=

=2 12 1

- Ta dng cng thc ny tm nghim bng s dc ng cong.

V d: Tm dc ca ng cong = 2 ti im (2; 4) s dng phng php tnh bng s



19

Tr li v d:

- Ta bt u vi im (1; 1) v gn vi im (2; 4).

- dc ca tnh bi:

=2 12 1

=4 1

2 1= 3

- By gi ta di chuyn im quanh ng cong, tin n gn , dng im (1.5; 2.25) gn vi

(2; 4).

- D tnh dc ng cong l = 3.5.

- Ta thy y l gi tr xp x ph hp vi dc tip tuyn ti , nhng ta c th tm c gi tr

xp x tt hn.

- By gi ta di chuyn li gn hn na, gi s (1.9; 3.61).

- By gi ta c:



20

- Vy ta tnh c = 3.9.

- Ta thy ta gn tm c gi tr dc cn tm.

- By gi nu tip tc di chuyn n (1.99; 3.9601), dc l 3.99.

- Nu l (1.999; 3.996001) th dc l 3.999.

- R rng, nu 2 th dc 4, nhng ta lu rng ta khng c ly = 2 v nh vy

phn s ca c 0 mu, iu ny l v l.

- Ta tm c tc thay i ca theo l 4 ti im = 2.



21

- Trong bi ny, chng ta s tnh vi phn ca mt hm bng nguyn l c bn. Tc l chng ta s

bt u t mt m hn tp v sau dng i s tm cng thc tng qut cho dc ng cong

ng vi mi gi tr ca .

- Nguyn l c bn c th hiu l cng thc v nhiu bi vit s dng k hiu (ng vi s

thay i ca ) v (ng vi s thay i ca ). iu ny v tnh lm cho i s thm phc tp,

nn chng ta dng thay th cho , ta vn gi l cng thc .

- Ta tm kim mt cch thc i s tm dc ca = () ti theo cch thay s m ta

xem trong bi dc ca tip tuyn vi ng cong (tnh ton gi tr).

- Ta c th tnh xp x gi tr ny bng cch ly 1 im no gn (; ()), gi s nh ( +

; ( + )).

- Gi tr

l gi tr xp x ca dc tip tuyn ta yu cu.

- Ta c th vit dc ny l:

=

- Nu ta di chuyn ngy cng gn ti , ng s gn trng vi tip tuyn ti v dc ca

gn bng vi dc ta cn tm.



22

- Nu ta trng vi (tc = 0) th ta s c chnh xc dc tip tuyn.

- By gi

c th vit thnh:

=( + ) ()

- Tng ng dc l:

=2 12 1

=

=( + ) ()

- Nhng ta ang tm dc ti nn ta cho dn tin n 0 dn n tin n v

tin ti

dc ta ang tm.

I. dc ng cong theo o hm:

- Ta c th vit dc tip tuyn ti l:

= lim0

( + ) ()

- y chnh l nguyn l c bn tnh o hm (hay cng thc ) l tc thay i tc thi ca

theo .

- iu ny tng ng vi iu sau (ni trc ta dng thay cho ):

= lim0

- Bn c th vit cng thc thnh:

= lim0

( + ) ()

II. Lu v o hm:

QUAN TRNG: o hm (vi phn) c th vit theo nhiu cch, iu ny c th dn n mt s

phin phc cho nhng bn mi nghin cu vi phn:

- iu theo sau y tng ng cch vit o hm bc 1 ca = ():

hoc () hoc

V d 1: Tm

khi = 22 + 3.

Tr li v d 1:

- Ta c:

() = 22 + 3

- Nn:

( + ) = 2( + )2 + 3( + ) = 22 + 4 + 22 + 3 + 3

- By gi ta cn tm:

= lim0

( + ) ()

= lim0

(22 + 4 + 22 + 3 + 3) (22 + 3)

= lim0(4 + 2 + 3)

= 4 + 3

- Chng ta tm ra biu thc cho ta dc tip tuyn bt k ni no ca ng cong

- Nu = 2 th dc l 4(2) + 3 = 5 (ng mu trong hnh di).

- Nu = 1 th dc l 4(1) + 3 = 7 (xanh l cy).



23

- Nu = 4 th dc l 4 (4) + 3 = 19 (en).

- Ta c th thy p n l ng khi ta v ng cong ra th (hnh parabola) v nhn xt dc

tip tuyn.

- y chnh l iu lm cho vi tch phn rt hu dng, ta c th tm dc bt c u trn ng

cong (ng vi tc thay i ca hm s bt c u).

V d 2:

a) Tm ca = 2 + 4.

b) Tm dc tip tuyn ti = 1 v = 6.

c) V ng cong v c 2 tip tuyn.

Tr li v d 2:

a) Ch : tc o hm bc 1, c th vit thnh

.

- t:

() = 2 + 4

- Ta c:

( + ) = ( + )2 + 4( + ) = 2 + 2 + 2 + 4 + 4

- V vy:

= lim0

( + ) ()

= lim0

(( + )2 + 4( + )) (2 + 4)

= lim0(2 + + 4)

= 2 + 4

b) Khi = 1; = 2(1) + 4 = 6.

- Khi = 6; = 2(6) + 4 = 8.

c) th:



24



25

- o hm cho ta bit tc thay i ca mt i lng so vi i lng khc vi v tr hay im

ring bit (nn ta gi l tc thay i tc thi). Khi nim ny c nhiu ng dng trong in t

hc, ng lc hc, kinh t hc, trn cht lng, kiu mu dn s, l thuyt sp hng v cn nhiu

na.

- Bt c khi no mt s lng lun thay i gi tr, ta u c th dng vi tch phn (vi phn v tch

phn) m t trng thi ca n.

- Trong bi ny, ta s bn lun v nhng s vic xy ra trong nhng khong thi gian rt nh, nn ta

s dng thay v nh ta thy bi Nguyn l c bn tnh o hm.

Ch : Bi vit ny l mt phn ca bi vit Tng quan v ngnh vi tch phn. Ta s nghin cu

vi quy lut d hn nhiu trong cch tnh vi phn trong bi vit tip theo o hm a thc.

I. Vn tc:

- Nh ta bit, vn tc chnh l thng s gia qung ng v thi gian vt i ht qung ng

, nhng iu ny ch ng khi vn tc l hng s c nh (hay vt chuyn ng u). Ta cn mt

cng thc khc khi vn tc thay i theo thi gian.

- Nu ta c biu thc cho (qung ng) theo (thi gian) th vn tc bt k thi im nh

no c tnh bi:

= lim0

- lm i s tr nn n gin hn, ta dng thay cho v vit:

= lim0

( + ) ()

V d: Mt vt th ri t ci gi c qung ng theo c cho bi = 4902, tnh

theo giy (), hi vn tc vt th khi = 10?

Tr li v d:

- y l th ca (qung ng) theo thi gian (tnh theo giy). Ta thy rng vn tc (tng

ng vi dc tip tuyn ca ng cong) khng c nh. Ban u, dc l 0 (ng cong

nm ngang), theo thi gian, vt th tng tc, dc ng cong tr nn dc hn (thng ng hn).

- By gi vn tc c tnh bi:

= lim0

( + ) ()



26

- Nn ta c:

= lim0

( + ) ()

= lim0

(490( + )2) (4902)

= lim0(490(2 + )) = 980

- Nn = 980 l biu thc cho ta bit vn tc vt th bt k thi im no ( 0).

- Khi = 10; = 980 10 = 9,800 .

- Nn vn tc khi = 10 l 98 .

- Ta vit vn tc l =

hay ta c th vit = .

- o hm cho ta bit:

+ Tc thay i ca mt i lng so vi i lng khc.

+ dc tip tuyn ca ng cong bt k im no.

+ Vn tc khi ta bit biu thc qung ng l =

.

+ Gia tc khi ta bit biu thc vn tc l =

.

II. Cu hi c gi:

- Mt ngi c thng hay hi:

- ngha ca

l g?

- y l cu tr li ca ti:

- Mt cch n gin,

ngha l s thay i ca so vi s thay i ca gi tr chnh xc ca

- Khi nim trn c dng khi i lng ph thuc vo mt hng s thay i. d hiu hn, ta

hy ly nhit mi trng lm v d. Gi s bn ang Melbourne, c (ni c nhit chnh

lch khc nghit), v ta mun bit by gi nhit gia tng nhanh n mc no.

- ma ng, v m, nhit thng thng l 2, ma h (6 thng sau) v m, nhit c

th ln n 26. Tc thay i trung bnh l: 26 2

6=24

6= 4 thng

- y l gi tr trung bnh xa, khng phi

.

- Nhng by gi hy ngh v mt ngy trong h. Lc 6: 00 sng nhit c th l 13, v 1: 00

chiu ln n 27, gi tr thay i trung bnh l: 27 13

7= 2

- Ta vn khng c

.

- By gi hy gi s lc 9: 00 sng l 20 v lc 10: 00 sng l 22.4 nn gi tr thay i trung

bnh l:

22.4 20

60= 0.04 pht (tng ng 2.4 )

- Ta c th tin n nhng khong thi gian nh hn (nh ;; v hn na) d on s thay

i nhit lc 10: 00 sng. S d on ny c biu din bi khi nim

.

- Ni mt cch lch s, nhng g ti m t trong dc ca tip tuyn vi ng cong (tnh ton

gi tr) chnh l nhng iu ngi xa lm trc khi Newton v Leibniz cho ta php tnh vi



27

phn.

- Trong bi Nguyn l c bn tnh o hm, ta thy cch tip cn i s m Newton v Leibniz

pht trin. By gi ta c th tm gi tr d on ca

bng cc quy trnh ton hc da trn hm

s m khng cn phi thay s trong mi v tr.

- bi vit tip theo ta s thy nhiu quy lut d hn cho vi phn. Ta s t dng nguyn l c bn

nhng s rt l tt nm r vi phn xut pht t u v n gip ch g cho ta.



28

- Ta c th tm o hm mt a thc m khng cn dng cng thc ta gp trong bi Nguyn l

c bn tnh o hm.

- Isaac Newton v Gottfried Leibniz thu c nhng quy lut di y vo u th k 18, h

theo ci c bn tin n vi phn, t lm cho cuc sng chng ta tr nn thun tin hn.

Hng s

= 0

y l iu c bn, tc l nu mt

i lng c gi tr hng s c

nh, hin nhin tc thay i

ca n bng 0.

Ly tha

bc ca

() = 1 Chng minh theo cng thc .

Tch c

hng s

() =

() =

y l mt hm s theo , c

ngha khi ta tm o hm ca mt

hng s nhn vi hm s th

cng ging nh tm o hm

hm s trc, sau

nhn cho hng s.

o hm tng ( + )

=

+

y v l hm s theo ,

o hm ca tng th bng vi

o hm ca ci u tin cng vi

o hm ca ci th hai. Nhng iu

ny s khng cn ng vi

o hm tch 2 s m ta

s gp bi sau.

I. V d:

V d 1: Tnh o hm:

= 76

Tr li v d 1:

- S dng quy tc sau:

() =

- Ta a 7 ra pha trc:

(76) = 7

(6)

- V:

() = 1

- Cho ta:

7

(6) = 7 65 = 425

Ch : Ta c th lm bc sau:



29

= 425

- Ta c th vit:

= 425 hoc = 425 u c ngha nh nhau.

V d 2: Tnh o hm:

= 35 1

Tr li v d 2:

= 35 1

- By gi:

(35) = 3 54 = 154

- V

= 0 nn ta vit:

(1) = 0

- Vy:

=

(35) +

(1) = 154

V d 3: Tnh o hm:

= 134 63 1

Tr li v d 3:

- By gi ta tnh theo th t:

(134) = 523, (

() = 1)

(63) = 182, (

() = 1)

() = 1, ( = 1 1;

() = 1)

(1) = 0, (

= 0)

- Vy:

= 523 182 1

V d 4: Tnh o hm:

= 1

48 +

1

24 32

Tr li v d 4:

= 1

48 +

1

24 32

- Ly vi phn tng phn mt, ta c:

(1

48) =

8

47 = 27

(1

24) =

4

23 = 23



30

(32) = 0

- Vy:

=

(1

48 +

1

24 32) = 27 + 23

V d 5: Xc nh o hm:

= 4 92 5, ti im (3; 15)

Tr li v d 5:

= 4 92 5

- Vy:

= 43 18 5

- Ti im c = 3 th gi tr o hm l:

|=3

= 4 33 18 3 5 = 49

- iu ny c ngha dc ca ng cong = 4 92 5 ti = 3 l 49.

V d 6: Tm o hm hm s:

= 1 4 2

Tr li v d 6:

- trng hp ny ta c phn s v s m ly tha ca (nn y khng phi a thc).

- Quy lut vi phn vn c p dng:

= 1 4 2

- Ta c th vit li thnh:

= 1 4 21

- Vi phn cho ta:

=1

4(1 4 )1 2 (1)11

=1

4

34 + 22

=1

434 +

2

2

II. Bi tp:

- Vit phng trnh tip tuyn ca ng cong = 3 3 ti = 2.

Tr li:

- Ta c:

= 3 3

= 3 32

- V gi tr o hm vi = 2 l:

|=2

= 3 3 22 = 9



31

- V = 3 3 nn khi = 2 th = 2.

- Nn ta cn tm phng trnh ng thng i qua (2;2) c dc 9.

- S dng cng thc tng qut ca phng trnh ng thng:

1 = ( 1)

- Ta c:

+ 2 = 9( 2)

- Vy phng trnh cn tm l:

= 9 + 16

- Hay vit di dng tng qut:

9 + 16 = 0



32

I. Cng thc tch:

- Nu v l 2 hm s theo th o hm tch c xc nh bi:

()

=

+

- Pht biu thnh li:

Mun o hm tch hai hm s, ta ly hm s th nht nhn vi o hm hm s th hai cng vi

hm s th hai nhn o hm hm s th nht

- Cng thc ny t u m ra? Nh nhiu cng thc vi phn ta gp u c chng minh da

vo Nguyn l c bn tnh o hm.

V d 1:

- Nu ta c tch 2 hm s:

= (22 + 6)(23 + 52)

- Ta c th tnh o hm trc tip m khng cn phi ph ngoc nhn phn phi

Tr li v d 1:

- Ta c 2 hm s = 22 + 6 v = 23 + 52.

- Ta dng cng thc tch:

()

=

+

- u tin ta tnh:

= 62 + 10

- V:

= 4 + 6

- Sau ta vit:

()

=

+

= (22 + 6)(62 + 10) + (23 + 52)(4 + 6)

= 204 + 883 + 902

V d 2: Tnh o hm:

= (3 6)(2 43)

Tr li v d 2:

- Nhn thy dng , c th:

{ = 3 6

= 2 43

()

=

+

= (3 6)(12) + (2 43)(32 6)

= 245 + 963 + 62 12

Ch : Ta c th vit cng thc tch ny theo nhiu cch:

()

=

+



33

- Hay:

(()())

= ()

()

+ ()

()

- Hoc:

() = +

II. Cng thc thng (phn s):

- Nu v l 2 hm s theo th o hm thng

c xc nh bi:

(

) =

2

- Pht biu thnh li:

o hm thng s bng mu s nhn o hm t s tr t s nhn o hm mu s tt c chia

cho mu s bnh phng.

V d 3: Tnh o hm:

=23

4

Tr li v d 3:

- Nhn thy hm s c dng

, vi = 23 v = 4 .

- Dng cng thc thng, ta c:

= 62

- V:

= 1

- Li c:

(

) =

2

=(4 )(62) (23)(1)

(4 )2

=242 43

(4 )2

V d 4: Tm

ca:

=42

3 + 3

Tr li v d 4:

- Ta t = 42 v = 3 + 3.

- Dng cng thc thng, ta c:

= 8

= 32

- Li c:



34

(

) =

2

=(3 + 3)(8) (42)(32)

(3 + 3)2

=44 + 24

(3 + 3)2

Ch : Ta c th vit cng thc thng ny theo nhiu cch:

(()

()) =

()()

()()

(())2

(

) =

2

(

)

=

2



35

I. Hm hp:

- Nu l hm s theo , cn l hm s theo th ta ni: l hm hp theo .

V d 1: Hy m t phng trnh:

= (5 + 7)12

Tr li v d 1:

- Nu ta gi = 5 + 7 (biu thc trong ngoc) th phng trnh trn vit li thnh:

= 12

- Ta vit l hm s theo , v tng t l hm s theo .

- y l khi nim quan trng trong vi phn. Nhng phng trnh ta gp n by gi s l phng

trnh trong phng trnh v ta cn phi nhn din chng c th tnh vi phn mt cch chnh xc.

II. Quy tc xch:

- tm o hm hm hp, ta cn s dng quy tc xch:

=

- iu ny c ngha ta cn phi:

(i) Nhn din (lun lun chn biu thc nm trong cng, thng nm trong ngoc hay di du

cn).

(ii) Sau ta cn ghi li biu thc theo .

(iii) o hm (theo ) sau ta biu din li mi th theo .

(iv) Bc tip theo ta tm

.

(v) Nhn

vi

.

V d 2: Tm

ca:

= (2 + 3)5

Tr li v d 2:

- Trong trng hp ny ta t = 2 + 3, t ta c = 5.

- Ta nhn thy rng:

+ l hm s theo .

+ l hm s theo .

- Theo quy tc xch, u tin ta cn tm

v

.

= 54 = 5(2 + 3)4

= 2

- Vy:

=

= 5(2 + 3)4(2)

= 10(2 + 3)4

V d 3: Tm

ca:



36

= 42

Tr li v d 3:

- Trong trng hp ny, ta t = 42 , t ta c = = 1 2 .

- Mt ln na:

+ l hm s theo .

+ l hm s theo .

- S dng quy tc xch, ta cn tm:

=1

21 2 =

1

2=

1

242

- V:

= 8 1

- V vy:

=

=

8 1

242

III. o hm hm s c ly tha:

- M rng ra vi quy tc xch chnh l cng thc ly tha cho vi phn. Ta ang tm o hm ca

(ly tha ca hm s):

() = 1

V d 4: phng trnh sau:

= (23 1)4

- Ta c hm s c ly tha.

Tr li v d 4:

- Nu ta t = 23 1 th = 4.

- V vy:

+ l hm s n ly tha.

+ l hm s n ( = ()).

- tnh o hm biu thc trn, ta c th dng cng thc mi:

() = 1

- Vi = 23 1 v = 4.

- V vy:

() = 1

= (4(23 1)3)(62)

= 242(23 1)3

- ng nhin ta c th dng quy tc xch:

=

IV. Th thch:

- Tm o hm hm s:



37

=2(3 + 1)

4 + 2

Tr li th thch:

- v d ny, ta c phn s vi t s l php nhn.

- Mt ln na, ta c =

, vi = 2(3 + 1) v = 4 + 2.

- Cng thc phn s yu cu dng

nhng li l tch.

- t = vi = 2 v = 3 + 1.

=

+

= 2 3 + (3 + 1)(2)

= 92 + 2

- Ngoi ra ta c:

= 43

- V vy:

= ()

=

2

=(4 + 2)(92 + 2) (2)(3 + 1)(43)

(4 + 2)2

=36 25 + 182 + 4

(4 + 2)2



38

- Chc hn cc bn tng gp nhng phng trnh m khng th biu din theo ch bng cch

chuyn v, chng hn:

4 + 222 + 62 = 7

- tnh

theo nhng cch thng thng trc y th rt phc tp bin i theo , thm ch

l khng th.

- Vy ta phi c mt cch no tnh vi phn nhm xc nh tc thay i ca khi thay

i. lm c iu ny th chng ta cn bit n vi phn hm n.

- Hy xem qua mt s v d sau:

V d 1: Tm biu thc

nu:

4 + 5 72 51 = 0

Tr li v d 1:

4 + 5 72 51 = 0

- v d ny ta d dng phn tch theo , t tnh vi phn mt cch d dng. Th nhng ta hy

s dng mt cch khc tm vi phn xem.

Phn A: Tm o hm vi ca 4.

- vi phn biu thc ny, ta coi nh l hm theo v s dng o hm hm s c ly tha.

- C bn: Tin hnh cc bc tnh o hm:

() =

(2) = 2

(3) = 32

- Tng t:

(4) = 43

Phn B: Tm o hm theo ca:

5 72 5

- y l cch tnh vi phn thun ty:

(5 72

5

) = 54 14 +

5

2

Phn C:

- v phi ca phng trnh, o hm ca 0 l 0.

- By gi kt hp phn ; ; .

43

+ 54 14 +

5

2= 0

- Chuyn v ta c kt qu:

=54 + 14

52

43



39

V d 2: Tm dc tip tuyn ti im (2;1) ca ng cong:

2 + 5 2 3 = 0

Tr li v d 2:

- i t tri sang phi, ta c:

- o hm ca 2:

(2) = 2

- o hm ca 5 l 0.

- o hm ca 2 l 2.

- o hm ca 3:

(3) = 32

- Kt hp li, vi phn hm n trn cho ta:

2

2 32

= 0

- Thu gn:

(2 32)

= 2

- Vy:

=

2

2 32

- Vy khi = 2 v = 1:

=

2 2

2 3 (1)2= 4

- Vy dc tip tuyn ti (2;1) l 4.

- Hy quan st chng ta lm g, ta v th ng cong:

2 + 5 2 3 = 0

- Sau ta v tip tuyn ti (2;1), qu thc dc tip tuyn l 4.

V d 3: (Bao gm cng thc tch)

- Tnh

ca:

4 + 222 + 62 = 7



40

(y chnh l v d nu u bi)

Tr li v d 3:

- n gin ha vn , ta chia nh cu hi thnh nhiu phn.

Phn A: Tm o hm theo ca 4.

(4) = 43

Phn B: Tm o hm theo ca 222.

- tnh o hm ca 222 theo ta cn nhn thy y l mt tch.

- Nu ta t = 22 v = 2 th ta c:

(222) =

+

= (22) (2

) + (2)(4) = 42

+ 42

Phn C:

- By gi:

(62) = 12

- V:

(7) = 0

- By gi tm

ca ton b biu thc:

4 + 222 + 62 = 7

- Tin hnh t tri sang phi, s dng cu tr li t nhng phn trn:

(43

) + (42

+ 42) + (12) = 0

- Rt gn, chuyn v, ta c kt qu:

=

2 + 3

3 + 2



41

- Ta c th tnh o hm ca o hm, c ngha l:

+ o hm cp 2 bng cch o hm ca o hm u tin.

+ o hm cp 3 bng cch o hm ca o hm cp 2.

V d 1: Cho biu thc:

5 + 33 2 + 7

- Hi o hm cp cao hn ca biu thc ny l g?

Tr li v d 1:

- o hm cp 1:

= = 54 + 92 2

- By gi tm o hm cp 2, ta ch vic vi phn phng trnh o hm cp 1:

2

2= = 203 + 18

- Tip tc tm o hm cp 3, cp 4:

3

3= = 602 + 18

4

4= (4) = 120

- o hm cp 5 l:

(5) = 120

- o hm cp 6; 7; 8; u c kt qu o hm l 0 do o hm ca mt hng s bng 0.

I. ng dng: Gia tc:

- Nh ta bit gia tc chnh l tc thay i ca vn tc.

=

- Nhng ng thi vn tc cng chnh l tc thay i ca dch chuyn:

=

- V vy o hm cp hai ca dch chuyn s cho ta gia tc:

=2

2

V d 2: Cho phng trnh chuyn ng (tnh theo ) theo thi gian (tnh theo ) ca mt vt th

l:

= 43 + 72 2

- Tnh gia tc vt th ti = 10.

Tr li v d 2:

= 43 + 72 2

=

= 122 + 14 2

=2

2= 24 + 14



42

- Ti thi im = 10 th vt c gia tc l:

= 24 10 + 14 = 254 2

II. o hm cp cao ca hm n:

V d 3:

a) Tm o hm cp 2 ca hm n:

+ 2 = 4

b) Tm gi tr o hm cp 2 ca hm n phn a) vi = 2 v > 0.

Tr li v d 3 cu a):

- o hm cp 1:

+ Ta c l tch nn ta dng cng thc tch lm:

() = +

+ Ta nghin cu v vi phn hm n bi trc:

(2) = 2

- Ta c th vit li l:

(2) = 2

+ Rp li v ta c o hm bc 1 ca phng trnh:

+ + 2 = 0

+ ( y ti s dng y thay cho

thun tin hn trong vic c v vit)

+ Ti s dng cng thc tch (cho tch xy) v quy tc xch cho 2.

- o hm cp 2:

( + ) + () + (2 + (2)) = 0

+ n gin ha, ta c:

( + 2) + 2 + 2()2 = 0

+ Ta c th gii theo .

= 2 + 2()2

+ 2

Tr li v d 3 cu b):

- Ta cn tm vi = 2.

- Thay vo phng trnh, ta c:

2 + 2 = 4

- Gii phng trnh bc hai ny, kt hp iu kin > 0, ta c:

= 1 + 5

- Ta cng cn tm gi tr

khi = 2.

- Ta tm phng trnh o hm u tin l:

+ + 2

= 0

- Gii theo

, ta c:



43

=

=

+ 2

- Thay = 2; = 1 + 5 ta c kt qu (xp x):

=

0.276

- Tip tc thay vo phng trnh o hm cp hai tm phn a) tm ra cu tr li:

= 2 + 2()2

+ 2 0.0894



44

- Ta nghin cu v cc hm ch duy nht 1 bin s. Tuy nhin, nhiu phng trnh trong ton

hc xut hin 2 hay nhiu bin. Trong bi ny ta s nghin cu cch tnh o hm ca hm s c

nhiu hn 1 n.

- Bi vit ny c lin quan nhng khng ging vi mt bi vit ta gp trc l vi phn hm

n.

V d 1: Hm s c 2 bin:

- y l hm s c 2 bin s l v :

(; ) = + 6 sin() + 52

- v th hm s ny ta cn n ta khng gian .

I. Vi phn ring theo :

- o hm ring theo c ngha Xem tt c cc k t khc nh hng s v ch vi phn phn c

.

- v d trn (cng nh nhng phng trnh c cha 2 bin) th o hm ring lin quan n c

ngha l (cng nh trong thc t) ta c th xoay th v nhn t trc . Ta ang nhn vo mt

phng .

- Ta thy rng ng cong hm sin di chuyn theo trc , iu ny xut pht t 6 sin() trong

phng trnh.

(; ) = + 6 sin() + 52

- Phn ta c th xem nh l hng s (trong trng hp ny c th xem l 0).

- By gi ta o hm ring ca:

(; ) = + 6 sin() + 52

- tnh theo :

= 6 cos()

- o hm ca 6 sin() l 6 cos(), cn o hm ca l 0 do c xem nh l hng s.

- Cn lu rng ta dng k hiu biu hin cho vi phn ring trong khi dng k hiu cho

php vi phn thng thng.

II. Vi phn ring theo :

- Thut ng: Vi phn ring theo ngha l: Gi s tt c cc k t l hng s ngoi tr vi



45

phn.

- Nh ta lm phn trn, ta xoay th v nhn theo trc , v vy ta thy mt phng .

- Ta thy mt hnh parabola, iu ny xy ra do s hng 2 v trong:

(; ) = + 6 sin() + 52

cn 6 sin() by gi c xem nh l hng s:

- By gi o hm ring ca (; ) = + 6 sin() + 52

theo :

= 6 cos()

theo .

= 1 + 10

- o hm ca phn t c l 1 + 10. o hm ca 6 sin() l 0 v phn t ny c xem nh l

hng s khi ta vi phn theo .

III. o hm ring bc :

- Ta c th tm 4 o hm ring bc 2 khc nhau. Hy xem qua nhng v d sau:

V d 2: Vi phng trnh:

= 6 cos()

= 1 + 10

- Hy xc nh:

) 2

)

2

)

2

2 )

2

2

Tr li v d 2 cu a):

- Ta vit li:

2

=

(

)

- Biu thc ny c ngha: u tin ta tm o hm ring theo ca hm (phn trong ngoc), sau

tm o hm ring theo ca kt qu va thu c.



46

- v d (; ) = + 6 sin() + 52 trn, ta tm c:

= 6 cos()

- tm 2

, ta cn tm o hm ring theo ca

:

2

=

(

) =

(6 cos()) = 0

- V cos() l hng s (khi ta tnh vi phn ring theo ) nn o hm ca n bng 0.

Tr li v d 2 cu b):

- Ta vit li:

2

=

(

)

- Biu thc ny c ngha: Tm o hm ring theo ca o hm ring theo

- v d (; ) = + 6 sin() + 52 trn, ta tm c:

= 1 + 10

- tm 2

, ta cn tm o hm ring theo ca

:

2

=

(

) =

(1 + 10) = 0

- Do l hng s (khi ta tnh vi phn ring theo ) nn o hm ca n bng 0.

Tr li v d 2 cu c):

- Ta vit li:

2

2=

(

)

- Biu thc ny c ngha: Tm o hm ring theo ca o hm ring theo .

- v d (; ) = + 6 sin() + 52 trn, ta tm c:

= 6 cos()

- tm 2

2, ta cn tm o hm ring theo ca

:

2

2=

(

) =

(6 cos()) = 6 sin()

Tr li v d 2 cu d):

- Ta vit li:

2

2=

(

)

- Biu thc ny c ngha: Tm o hm ring theo ca o hm ring theo .

- v d (; ) = + 6 sin() + 52 trn, ta tm c:

= 1 + 10

- tm 2

2 ta cn tm o hm ring theo ca

:



47

2

2=

(

) =

(1 + 10) = 10



48

- Trong thi i ca Issac Newton, mt vn ng quan tm l c qu t phng tin di chuyn

bng ng bin.

- Nn m tu hay xy ra v con tu khng theo ng mun ca thuyn trng. Khi cha c s

hiu bit nhiu v s tng quan gia Tri t, ngi sao v cc hnh tinh chuyn ng tng tc

ln nhau.

Trc khi ngnh vi tch phn pht trin,

cc ngi sao nh hng n s sng cn

ca ngnh hng hi.

- Vi tch phn (vi phn v tch phn) c pht trin thc y s hiu bit ny.

- Vi phn v tch phn c th gip chng ta gii quyt nhiu vn trong th gii thc.

- Ta dng o hm xc nh gi tr ln nht, nh nht ca cc hm ring bit (v d nh gi tin,

di, s lng vt liu dng cho xy dng, li ch, tn tht,...).

- Ta d bt gp php tnh o hm trong cc vn lin quan n c kh v tin hc, c bit khi ta

lm m hnh c im ca mt vt th ang chuyn ng.

- Trong phn 2.2 ny:

+ Bi 2.2.1 Tip tuyn v php tuyn: Nhng th rt quan trng tron vt l (nh lc ca chic xe

hi ang r).

+ Bi 2.2.2 Cng thc Newton: p dng cho nhng phng trnh xu m bn khng th gii i

thun bng i s.

+ Bi 2.2.3 Chuyn ng cong: Bn s bit c cch tm vn tc v gia tc ca mt vt th ang

chuyn ng theo ng cong.

+ Bi 2.2.4 Tc lin quan: Ni 2 bin lun thay i theo thi gian v gia chng c mt mi

quan h no .

+ Bi 2.2.5 S dng vi phn v th: Ta s bit cch m hnh ha trng thi ca bin.

+ Bi 2.2.6 p dng vi phn x l nhng vn cc tr: Mt trong nhng ng dng rt ln ca

vi phn.

+ Bi 2.2.7 Bn knh cong: Bn s nghin cu trng thi ca ng cong l mt phn ca ng



49

trn trong min ln cn no .

- Chng ta bt u nghin cu ng dng ca vi phn vi chng tip tuyn v php tuyn.



50

- Tip tuyn v php tuyn c vai tr quan trng trong ngnh vt l (nh lc ca chic xe khi r

khc cua).

- Thnh thong ta cn tm tip tuyn v php tuyn ca mt ng cong khi ta phn tch 1 lc tc

dng ln 1 vt th ang chuyn ng.

I. Tip tuyn:

- Tip tuyn ca mt ng cong l ng thng tip xc vi ng cong ti 1 im nm trn

ng cong , tip tuyn c dc bng vi dc ca ng cong ti im .

II. Php tuyn:

- Php tuyn ca ng cong l ng thng vung gc vi tip tuyn ca ng cong.

Ch 1: Nh nghin cu trong bi dc ca tip tuyn vi ng cong (tnh ton gi tr), ta

c th tm dc ca tip tuyn ti bt k im no (; ) thng qua

.

Ch 2: tm phng trnh tip tuyn, ta cn nh iu kin 2 ng thng c dc ln lt

l 1; 2 vung gc nhau.

1 2 = 1

III. ng dng:

1. Tip tuyn:

(i) Gi s ta i du lch trn 1 chic xe hi quanh khc cua, bt cht ta ng vo mt th g trn

trt trn ng (c th l du, bng, nc hay ct mm) v xe ca ta bt u trt, th chic xe s

di chuyn theo hng tip tuyn vi khc cua .

Mt chic xe trt sau khi r khc cua

to ra vt tip tuyn vi vch k ng i mu vng

(ii) Tng t, nu ta cm tri banh v nm chng quanh 1 vt th ang xoay trn, tri banh ngay lp

tc bay ra theo phng tip tuyn vt th xoay trn .



51

2. Php tuyn:

(i) Khi bn li xe nhanh theo ng trn, lc khin bn cm thy nh xe mnh sp ri khi ng

trn chnh l php tuyn ca ng cong con ng. Mt iu kh th v l lc gip bn di

chuyn vng quanh khc cua hng thng v tm ng trn, php tuyn vi ng trn.

(ii) Cm bnh xe c t php tuyn vi ng cong bnh xe nhng im c ch cho cm xe

lin kt vi tm bnh xe.

Cm xe p c t

php tuyn vi vnh bnh xe

IV. V d:

V d 1: Tm dc ca

a) Tip tuyn

b) Php tuyn

ca ng cong = 3 22 + 5 ti im (2; 5).

Tr li v d 1:

= 32 4

- dc tip tuyn l:

1 =

|=2

= 3 22 4 2 = 4

- Vy dc ca php tuyn tnh theo cng thc 1 2 = 1.

2 = 1

4

V d 2: Vit phng trnh php tuyn V d 1 bn trn.

Tr li v d 2:

- Ta dng 1 = ( 1).

- Vi:

1 = 2

1 = 5

= 1

4

- Vy:

5 = 1

4( 2)

- Cho ta:

= 1

4 +

11

2



52

- Hay:

+ 4 22 = 0

V d 3: V th v php tuyn V d 1 trn.

Tr li v d 3:

y l th ca tip tuyn v php tuyn ca ng cong ti = 2



53

- Vi nhng phng trnh phc tp m bn khng th gii thun ty i s th bi vit ny rt hu

ch cho bn.

- Cc my tnh s dng cng thc vng lp gii phng trnh. Qu trnh ny bao gm phn on

ra cch gii ng v p dng cng thc a ra cc phn on chnh xc hn cho n khi ta tm ra

c gi tr (c th xp x) ng nht ca phng trnh.

- Nu ta mun tm () = 0 (dng bi ton ph bin) th ta on mt vi gi tr 1 gn ng

nht, t ta s tm ra gi tr xp x ph hp bng cch s dng cng thc Newton:

2 = 1 (1)

(1)

(Cng thc ny da vo phng trnh ng thng theo dc)

V d 1: Gii phng trnh:

22 2 = 0

Tr li v d 1:

- Ta c th phng trnh trn:

- t:

() = 22 2

() = 4 1

- Th 1 = 1.5.

- Ta c:

2 = 1 (1)

(1)= 1.5

(1.5)

(1.5)= 1.3

- Vy 1.3 l gi tr xp x ng hn.

- Tip tc quy trnh ny:

3 = 2 (2)

(2)= 1.3

(1.3)

(1.3)=269

210 1.280,952,381

- Ta c th lm tip quy trnh ny nhiu ln tm ra gi tr chnh xc nht

- Kim tra: S dng mt vi phn mm ton hc (nh Mathcad), ta c th phi nhp vo gi tr d

on ban u (nh = 2) v kt qu l:

root(22 2, ) = 1.280 776 406 404 4

- Ngoi ra ta c th s dng phm SHIFT + SOLVE trn my tnh Casio -570ES nhp vo gi tr

d on ban u th my tnh cng a ra kt qu ny.



54

I. Hm s c nhiu nghim:

- Nhiu hm s c nhiu nghim, nn bn cn phi hiu r vn sau cho my tnh mt gi tr

d on ban u tt nht.

V d 2: Gii phng trnh:

1 2 + 2 = 0

- (Cc phn mm khoa hc khng th tm cch gii chnh xc cho chng ta. Ta cn bit s dng

cng c hp l gii, khng hn phi dng th hay cng thc Newton. iu ny s cho ta c

nh gi ban u v nghim phng trnh).

Tr li v d 2:

- t:

= 1 2 + 2

- th hm s ():

- Ta d c nhn nh ban u rng phng trnh c hai nghim, mt nghim gn = 1 v nghim

cn li gn = 3. Tuy nhin, nu ta quan st k hn khu vc gn = 3 (bng cch phng to) th ta

pht hin ra cn mt nghim na.

- Bng cch thay s th ta c mt nghim chnh xc l = 3.

- By gi vi mt nghim gn = 3.4.

- Ta s dng cng thc Newton tm ra gi tr xp x ca nghim. Ta cn tnh vi phn

= 1 2 + 2. Bi v ta c ng vai tr ly tha trong phng trnh trn, ta cn s dng

Logarithm khi tnh vi phn (cc bn c th tham kho cch tnh vi phn hm logarithm trn Internet,

ti s trnh by r hn trong chng Vi phn hm s siu vit).

- Tnh vi phn 2.

- t = 2.



55

- Ly logarithm t nhin hai v:

ln() = ln(2)

1

= ln(2)

= ln(2) = 2 ln(2)

- Vy:

= () = 2 + 2 ln(2)

- p dng cng thc Newton, ta c:

()

()=

1 2 + 2

2 + 2 ln(2)

- Ta c con s d on ban u l 1 = 3.4.

2 = 1 (1)

(1) 3.407 615 918

- Tip tc quy trnh ny:

3 = 2 (2)

(2) 3.407 450 596

- Thm vi bc na, ta c:

4 = 3.407 450 522

5 = 3.407 450 505

- Ta c th kt lun nghim phng trnh ng vi 7 ch s l l = 3.407 450 5

II. S dng th:

- S dng cc phn mm ton hc, ta c th phng to nghim v ta c th thy (ni hm s ct trc

) th gn vi gi tr 3.407 45

- By gi ta xt trng hp m, gi s 1 = 1 l gi tr d on ban u, p dng cng thc

Newton, ta c:

2 = 1.213 076 633

3 = 1.198 322 474

4 = 1.198 250 199

- Ta c th tip tc quy trnh ny cho n khi ta t gi tr chnh xc nht.

- i chiu p s vi th, ta thy rng kt qu l = 1.198 250 197 chnh xc n 9 ch s

l.



56

III. Kt lun:

- Vy kt qu ca phng trnh 1 2 + 2 = 0 l:

= 1.198 25

= 3

= 3.407 45

- Chnh xc n 5 ch s l.



57

- Bi vit ny s cho cc bn thy cch tm vn tc v gia tc ca mt vt th chuyn ng cong.

- bi o hm vi tc thay i tc thi, ta tm ra cch xc nh vn tc theo phng trnh

chuyn ng bng cch.

=

- v gia tc theo phng trnh vn tc (hay phng trnh chuyn ng), s dng:

=

=2

2

- Cng thc trn ch thch hp vi chuyn ng thng (nh vn tc v gia tc trn ng thng),

iu ny cha ph hp vi nhiu vn trong cuc sng. V vy ta nghin cu n khi nim v

chuyn ng cong khi mt vt th di chuyn theo ng cong nh trc.

- Thng thng ta biu din thnh phn chuyn ng l v l hm s theo thi gian, gi l dng

tham s.

V d 1: Cho phng trnh chuyn ng theo tham s , hy v th:

() = cos()

() = sin()

- vi = 0 n 2 trong 0.5 qung ng u.

- u tin, ta cn thit lp bng gi tr bng cch thay mt s gi tr vo .

Tr li v d 1:

- Ta xc nh 13 im theo bng gi tr, bt u ti (0; 1) nh theo hnh di y (di chuyn theo

chiu kim ng h).

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

() 0 0.48 0.84 1.00 0.91 0.60 0.14 0.35 0.76 0.98 0.96 0.71 0.28

() 1 0.88 0.54 0.07 0.42 0.80 0.99 0.94 0.65 0.21 0.28 0.71 0.96

Ta thy rng ta to ra mt hnh trn

tm ti (0; 0) v bn knh 1 n v.

- Cn lu rng n khng xut hin trong th ny m ch c n v .

I. Cc thnh phn ngang v dc ca vn tc:

- Thnh phn ngang ca vn tc c xc nh bi:



58

=

- v thnh phn dc ca vn tc:

=

- Ta c th tm ln ca vn tc tng hp mt khi ta bit cc thnh phn ngang v dc ca

vn tc bng cch s dng:

= 2 + 2

- Phng v m vt th di chuyn c xc nh bi:

tan() =

V d 2: Cho phng trnh = 53 v = 42 vi thi gian , tm ln v phng v ca vn tc

khi = 10.

Tr li v d 2:

- Khi = 10, ta c ta im l (5,000; 400).

- y l th ca chuyn ng.

Lu :

- Trc ta l v (khng c km theo ).

- Cc im chuyn ng nhanh dn theo thi gian.

- Ta c:

= 53

- V vy:

= 152

- Vi = 10, vn tc theo chiu trc l:

= = 1,500

- Tng t, = 42 nn vn tc theo chiu trc khi = 10 l:

= = 80

- Vy ln ca vn tc s l:



59

= 2 + 2 = 1,520.1

- By gi ta xc nh phng v ca vn tc (tnh theo gc hp vi trc dng)

tan() == 0.053

- Vy = 0.053 rad = 3.05

V d 3: Cho:

=20

2 + 1

- V

= 0.1(2 + )

- theo thi gian . Xc nh ln v phng v ca vn tc khi = 2. V th ng cong.

Tr li v d 3:

- Khi = 2 ta c ta im l (8; 0.6).

=20

2 + 1

- Vy,

=

20

(2 + 1)2

- Vi = 2:

= = 0.8

- Tng t, vi = 0.1(2 + ) v = 2, ta c:

= = 0.1(2 + 1) = 0.5

- Vy:

= 2 + 2 = 0.943

- By gi ta xc nh phng v:

tan() == 0.625

- Vy:

= arctan(0.625) = 0.558 rad



60

II. Gia tc vt th khi chuyn ng cong:

- Biu thc ca gia tc c cch xc nh tng t nh cch xc nh vn tc.

- Thnh phn ngang ca gia tc:

=

- Thnh phn dc ca gia tc:

=

- ln ca gia tc:

= 2 + 2

- Phng v ca gia tc:

tan() =

V d 4: Mt chic xe hi trn ng chy th nghim n khc cua th chy vi biu thc ng

i l = 20 + 0.23; = 20 22 vi v tnh theo metre () v l giy ().

a) V th ng cong vi 0 8.

b) Tnh gia tc ca xe khi = 0.3.

Tr li v d 4 cu a):

- th:

Tr li v d 4 cu b):

- Gia tc:

- Thnh phn ngang:

= 20 + 0.23

=

= 0.62

=2

2= 1.2

- Vi = 3.0; = 3.6.

- Thnh phn dc:

= 20 22

=

= 20 4



61

=2

2= 4

- Vi = 3; = 4.

- By gi:

= 2 + 2 = 5.38

- V:

= arctan () = 312, (gc phn t th 4)

- Vy gia tc ca xe c ln l 5.38 2 v c phng v 312 hp vi trc theo chiu dng.

III. Vy nu nh v khng phi l phng trnh theo tham s th gii quyt nh th

no?

V d 5: Mt ht di chuyn theo ng = 2 + 4 + 2 tnh theo . Cho vn tc ngang =

3 , xc nh ln v phng v ca vn tc ti im (1;1).

Tr li v d 5:

- y l cch gii quyt khc cho v d. Ln ny ta c theo v khng c biu thc no cha

tham s na.

- c th xc nh ln cng nh phng v ca vn tc, ta cn bit:

=

- V,

=

- Nhng trong cu hi cho ta:

=

= 3

- Vy ta cn tm

.

- tm c, ta tnh vi phn phng trnh cho theo bng cch s dng cc k thut ta

nghin cu trong bi vi phn hm n:

= 2 + 4 + 2

= 2

+ 4

- V

= 3 v ta mun bit vn tc ti = 1 nn ta c:

= = 6

- Vy ln vn tc l:

= 2 + 2 = 6.708 2

- Vy vn tc l 6.708 2 vi phng v 63.4.

V d 6:

- Mt qu tn la c bn theo qu o (tnh theo ): = 3

90.

- Nu vn tc ngang c cho bi () = , xc nh ln v phng v ca vn tc khi qu tn



62

la chm t (coi nh a hnh bng phng) vi thi gian tnh theo pht.

Tr li v d 6:

- Hy nhn vo th chuyn ng hiu r hn vn t ra:

- Ta thy rng qu tn la chm t gn v tr = 9.5. Ti im ny vn tc ngang c gi tr

dng (qu tn la i t tri sang phi) v vn tc dc c gi tr m (qu tn la i xung).

- () = c ngha l khi tng, vn tc ngang cng tng vi cng mt gi tr (ng nhin l

khc n v o). Vy vi v d ny, vi = 2 th tc ngang l 2 pht , v vi = 7,

tc ngang l 7 pht v c th.

- tnh ln vn tc khi tn la chm t, ta cn xc nh cc thnh phn ngang v dc ca vn

tc thi im .

(i) Vn tc ngang: Ta cn gii phng trnh sau tm chnh xc im va chm ca tn la vi mt

t:

3

90= 0

- Rt nhn t, ta c:

(1 2

90) = 0

[ = 310 = 0

= 310

- Ta ch cn gi tr = 310 9.4868 (gi tr ny ph hp vi th trn).

- Vy tc ngang khi tn la chm mt t l 9.4868 pht (v () = ).

(ii) Vn tc dc: By gi ta cn dng vi phn hm n theo (ch khng phi theo ) tm vn tc

dc:

= 3

90

=

1

302

- Nhng ta bit

v c nh hng vi nhau, vy ta ch cn thay gi tr tm c phn (i),

kt qu s ra s m ng nh ta d on ban u:

18.973 665 96



63

- By gi ta tnh ln vn tc, bao gm vn tc ngang v vn tc dc:

(

)2

+ (

)2

21.213 203 44

- Vn tc c ln v phng v. By gi ta tnh phng v (tc gc ca chuyn ng).

= arctan(

) 1.107 148 718

- Tnh theo th iu ny tng ng:

= 1.107148718 57.25578 = 63.3907

- Ta c th thy p n trn hp l bng cch phng to mt phn ca th ni tn la chm t.

- Vy ta kt lun vn tc tn la khi chm t l 21.2 pht , phng v 6.4 hp vi phng

nm ngang.



64

- Nu ta c 2 i lng ph thuc theo thi gian v gia chng c s lin quan vi nhau, ta c th

biu th tc thay i ca i lng ny theo i lng khc. Khi ta cn vi phn c hai bn

theo thi gian, tc l ta s tm

ca hm () no .

V d 1: Mt cy thang cao 20 da vo tng, nh thang trt xung vi tc 4 . Hi tc

di chuyn ca im gia thang lc cch mt t 16 l bao nhiu?

Cc bc thc hin v d 1:

(i) V hnh minh ha cho bi ton.

(ii) Xc nh hng s v s lng cc bin.

(iii) Thit lp mi quan h gia cc i lng ny.

(iv) Vi phn theo thi gian.

(v) Xc nh gi tr im gia thang.

Tr li v d 1:

- By gi mi quan h gia v l 2 + 2 = 202.

- Vi phn hon ton theo thi gian (v v u ph thuc theo thi gian ):

2 +

2 = 0

2

+ 2

= 0

- Tc l:

+

= 0

- By gi, ta bit

= 4 v ta cn tm vn tc ngang

khi = 16.

- Mt i lng cha xc nh gi tr l , s dng nh l Pythagoras:

= 202 162 = 12

- Vy:

16

+ 12 (4) = 0

= 3

V d 2: Mt hn ri vo mt ci ao to thnh nhng gn sng hnh trn ng tm loang rng ra

xung quanh. Hi tc din tch mt trong cc hnh trn ny gia tng l bao nhiu vo thi im



65

bn knh hnh trn tng thnh 4 v loang rng vi vn tc 4 ?

Tr li v d 2:

- Mi quan h: = 2.

- Vi phn theo thi gian, sau thay vo cc gi tr c sn:

=

(2) = 2

= 2 4 0.5 12.562

V d 3: Mt v tinh di chuyn theo qu o c phng trnh:

2

72.5+2

71.5= 1

- vi v tnh theo n v nghn km.

- Nu

= 12,900 khi = 3,200 v > 0, xc nh gi tr

.

Tr li v d 3:

- Ta vi phn biu thc theo :

2

72.5+2

71.5= 1

2

72.5+2

71.5= 0

- Ta tm bng cch thay = 3.2 vo phng trnh gi thit cho lc u:

3.22

72.5+2

71.5= 1

7.836

- Da vo cu hi nn ta ly gi tr dng, thay vo phng trnh vi phn, ta c:

2 3.2 12.9

72.5+2 7.836

71.5= 0

5.195

- iu ny c ngha vn tc theo chiu trc l 5.195 .

V d 4: Mt my iu hng in t c tn s thay i t l nghch vi cn bc hai ca t in

trong mch. Nu = 920 khi = 3.5, hi tc tn s thay i nhanh th no nu nh

= 0.3 ?

Tr li v d 4:



66

- Ta c:

=

- Thay th cc gi tr c, ta c:

920,000 =

3.5 1012

= 1.721

- Vy = 1.721.

- Ta cn tm

:

=

1.721

2

32

- M = 3.5 v

= 0.3 .

- Vy:

=

1.721

2 (3.5 1012)

32 (0.3 1012) = 39,424.8

- Vy tn s ca my iu hng in t gim vi tc 39.4 .



67

Lu :

- Hin c rt nhiu phn mm v th (Mathcad, Scientific Notebook, graphic calculator,...).

Trong bi vit ny iu cn lu l bn phi nm bt c hnh dng c bn ca ng cong bn

gp. Vic hiu c nt t nhin ca nhng hm s rt quan trng cho vic nghin cu ca bn sau

ny. a s cc m hnh ton hc u bt u t th.

- Bn cn phi v th, a ra nhng v tr c bit, trnh vic v hp v chm cc im.

- Ta s dng vi tch phn tm ra cc im c bit.

- Nhng iu ta s tm trong bi vit ny l:

Giao im

trc

S dng = 0

Lu : Trong nhiu trng hp, tm giao im trc

khng d, khi hy b qua bc ny.

Giao im

trc S dng = 0

Cc i

a phng

S dng

.

Du ca o hm u tin thay i + sang .

Cc tiu

a phng

S dng

.

Du ca o hm u tin thay i sang +.

im un S dng 2

2, v du ca

2

2 thay i.

I. Tm cc i v cc tiu:

1. Cc i a phng:

- Cc i a phng xut hin khi phng trnh = 0 c nghim v du ca thay i t dng

sang m khi i t tri sang phi:

2. Cc tiu a phng:

- Cc tiu a phng xut hin khi phng trnh = 0 c nghim v du ca thay i t m

sang dng khi i t tri sang phi.



68

II. o hm bc :

- o hm bc 2 cho ta bit hnh dng ca ng cong bt k im no.

1. Lm xung:

- Nu 2

2> 0 th ng cong s c hnh kiu cc tiu gi l lm xung (hay lm).

V d 1:

- ng cong = 2 + 3 2 c

= 2 + 3.

- By gi 2

2= 2 v ng nhin gi tr ny > 0 vi mi nn c hnh lm xung vi mi gi tr

.

2. Lm ln:

- Nu 2

2< 0 th ng cong s c hnh kiu cc i gi l lm ln (hay li).

V d 2:

- ng cong = 3 2 + 5 c

= 32 2. o hm bc hai l

2

2= 6 < 0, < 0. V

vy ng cong lm xung vi mi gi tr < 0 (v lm ln vi mi gi tr > 0).

III. Tm im un:

- im un l im m hnh dng ca ng cong thay i t hnh kiu cc i 2

2< 0 sang hnh

kiu cc tiu 2

2> 0.

- R rng, im un s xut hin khi 2

2> 0 v c s i du (t + sang hoc t sang +).

V d 3: V th sau bng cch tm giao im hai trc ta , im un, (cc) im cc i, cc

tiu: = 3 9.



69

Tr li v d 3:

- Hnh dng c bn ca phng trnh bc 3 l:

Hy nh hnh ny. S gip ch cho bn

trong qu trnh v hnh y.

(i) Giao im vi trc :

= 3 9 = ( + 3)( 3)

Nn,

= 0 [ = 0 = 3 = 3

(ii) Giao im vi trc :

- Khi = 0 th = 0.

(iii) Cc im cc i v cc tiu:

= 32 9 = 3( + 3)( 3)

Nn,

= 0 [ = 3

= 3

- Vy ta c cc im cc i, cc tiu vi cc gi tr xp x (1.7; 10.4) v (1.7;10.4).

- Ta c th kim tra iu ny bng cch kim tra mt vi im gn 1.7 v 1.7 kim chng xem

du thay i nh th no. Nhng ta cn phi tm o hm cp 2 xc nh im un, v vy ta s

tn dng iu ny kim chng cc i, cc tiu.

(iv) o hm bc 2:

2

2= 6

- V = 1.7 th < 0 nn (1.7; 10.4) l cc i a phng.

- V = 1.7 th > 0 nn (1.7; 10.4) l cc tiu a phng.

(v) im un:

2

2= 6

Nn,



70

2

2= 0 = 0

- V 2

2 i du t m (lm xung) sang dng (lm ln) khi i qua 0.

(vi) th:

- By gi ta d liu v th:

IV. Cc dng hnh tng qut:

- Nu ta bit c nhng hnh dng c bn ny th vic v th s d dng hn rt nhiu. ng

nhin nhng hnh di y l hnh l tng, v cn rt nhiu dng hnh khc v trng hp khc.

Nhng t ra nhng hnh ny chnh l nn tng cc bn nghin cu nhng hnh phc tp hn.

Hm bc Hm bc

M cao nht ca : 2 M cao nht ca : 3

1 cc tiu.

0 cc i.

(nu hm s c h s trc 2 l dng)

1 cc tiu.

1 cc i.

0 im un. 1 im un.

Hm bc Hm bc

M cao nht ca : 4 M cao nht ca : 5



71

2 cc tiu.

0 cc i.

(nu hm s c h s trc 2 l dng)

2 cc tiu.

2 cc i.

2 im un. 3 im un.

V d 4:

- V th v th hin giao im vi cc trc th, cc i, cc tiu v im un:

= 4 62

Tr li v d 4:


= 4 62 = 2( + 6)( 6)

Nn,

= 0 [

= 0

= 6

= 6


- Khi = 0 th = 0.


= 43 12 = 4( + 3)( 3)

Nn,

= 0 [

= 0

= 3

= 3

- Vy ta c cc im cc i, cc tiu l (0; 0) v (3;9) v (3;9).

(iv) o hm bc 2:

2

2= 122 12

- V = 3 th > 0 nn (3; 9) l cc tiu a phng.

- V = 3 th > 0 nn (3; 9) l cc tiu a phng.

- V = 0 th < 0 nn (0; 0) l cc i a phng.

(v) im un:

2

2= 122 12 = 12( + 1)( 1)

Nn,



72

2

2= 0 [

= 1 = 1

- Nu < 1 th > 0.

- Nu 1 < < 1 th < 0.

- Nu 1 < th > 0.

- Vy du ca thay i khi = 1. Vy (1;5) l im un.


(vi) th:


V d 5: V th v th hin giao im vi cc trc th, cc i, cc tiu v im un:

= 5 54

Tr li v d 5:

- Lu : Cu hi ny kh l phc tp v y cng l v d cho mt vi rc ri bn c th gp. Nu

bn vn cha hon ton hiu, ng lo lng qu nh!


= 5 54 = 4( 5)

Nn,

= 0 [ = 0 = 5


- Khi = 0 th = 0.


= 54 203 = 53( 4)

Nn,

= 0 [

= 0 = 4

- Vy ta c cc im cc i, cc tiu l (0; 0) v (4;256).

(iv) o hm bc 2:



73

2

2= 203 602

- V = 0 th = 0 nn (0; 0) khng l cc tr.

- V = 4 th > 0 nn (4;256) l cc tiu a phng.

(v) im un:

2

2= 203 602 = 202( 3)

Nn,

2

2= 0 [

= 0 = 3

- Nu < 0 th < 0.

- Nu 0 < < 3 th < 0.

- Nu 3 < th > 0.

- Vy du ca khng thay i khi = 0. Vy (0; 0) khng l im un.


Ch : Tht ra ti = 0 ta c im nhn. y khng phi cc i a phng mc d nhn hnh v

th c v ging cc i a phng

(vi) th:


- By gi ta s nghin cu cch v nhng ng cong khng biu din cho a thc. Nhng ng

cong ny c th khng lin tc hay c nhng hnh dng c bit. Do tnh ng dng ca cc hnh ny

trong cuc sng rt ln nn ta cn nm r cc dng th, ng thi khi s dng my tnh v, ta

cn xc nh c nhng li sai hay hnh dng khc thng ca th.

- Ta s dng nhng k thut v th c bn, kim tra c im ca hm s khi:

+

bn tri im khng lin tc

bn phi im khng lin tc

V. i xng: Ta s dng tnh i xng qua trc v ng cong.



74

VI. Tp xc nh v tp gi tr:

- Tp xc nh (tt c cc gi tr c th) v tp gi tr (gi tr tng ng vi ) rt quan trng

v hnh theo mt s yu cu (chng hn cn bc hai).

VII. Quy trnh thc hin:

- Khi v hnh ta cn xc nh:

(i) Giao im ca th vi trc .

(ii) Giao im ca th vi trc .

(iii) Gii hn khi tin n .

(iv) Tp xc nh v tp gi tr.

(v) Cc i v cc tiu.

(vi) o hm bc hai.

(vii) Trng thi th gn im khng lin tc.

V d 6: V th:

= +4

2

Tr li v d 6:


+4

2= 0

3 + 4 = 0

= 43

1.6


- Ta khng xc nh c hm s ti = 0 nn khng c giao im ca th vi trc , t s

c tim cn ti = 0.

(iii) Cc gii hn:

- Khi th 4

2 0, vy .

- (Tht ra, ng cong s ngy cng tin gn n ng thng = phn m).

- Khi + th +.

- (Tht ra, ng cong s ngy cng tin gn n ng thng = phn dng).

(iv) Tp xc nh v tp gi tr:

- Tp xc nh: Mi gi tr thc ngoi tr 0.

- Tp gi tr: Mi gi tr thc .

(v) Cc i v cc tiu:

= +4

2

= 1 83

- Ta c 1 83 = 0 = 2.

- Vy ta s c cc i hoc cc tiu ti im (2; 3).

- By gi, khi th

1 v khi + th

1.

(vi) o hm cp 2:



75

2

2=24

4> 0,

- Vy hm lm ln vi mi gi tr , vy im (2; 3) l cc tiu.

(vii) Gn im khng lin tc:

- Khi 0 (ngha l tin gn n 0 t pha m), .

- (Bn c th kim chng bng cch th cc gi tr = 1;0.5;0.1;0.01;0.001) Khi

0+ (ngha l tin gn n 0 t pha dng), .

- (Bn c th kim chng bng cch th cc gi tr = 1; 0.5; 0.1; 0.01; 0.001).

(viii) th:

- By gi ta bt u v th:

V d 7: V th =9

2+9.

Tr li v d 7:

(i) Giao im vi trc : 9

2 + 9= 0 = 0


- Khi = 0 th = 0.

(iii) Gii hn khi tin n :

- Chia t v mu cho 2 c: 9

1 +92

- Khi th 0.

- Tng t, + th 0. (iv) Tp xc nh v tp gi tr:

-Tp xc nh: Mi s thc .

- Tp gi tr: xem phn 5.

(v) Cc i v cc tiu:



76

=9

2 + 9

=

9(2 9)

2 + 9

= 0 = 3

- Vy ta c cc i hoc cc tiu ti (3;1.5) hoc (3; 1.5).

- Mt vi miu t cho biu thc

l:

+ DNG khi 3 < < 3.

+ M khi < 3 hoc > 3.

- T ta kt lun rng (3;1.5) l cc tiu v (3; 1.5) l cc i, vy tp gi tr l 1.5

1.5, ng thi ta c dc ti = 0 l:

|=0

= 9(2 9)

2 + 9|=0

= 1

(vi) o hm bc 2:

- Trong trng hp ny, o hm bc hai c v nh phc tp tnh nhanh, rt d dn n sai st.

- By gi ta c th v th.



77

- Quy trnh tm gi tr ln nht, nh nht gi l ti u ho, l ng dng rt quan trng ca vi phn.

Quy trnh ny ging nh tm li nhun ti a ca mt cng ty, hay gim thiu gi thnh, hay s

lng vt liu t nht to ra mt sn phm no . Nhng th ny c ngha rt quan trng trong

cc ngnh cng nghip.

V d 1: Li nhun hng ngy ca mt nh my lc du tho cng thc:

= 8 0.022

- vi l s lng thng du lc c. Vi bao nhiu thng du th cng ty s t li nhun cao

nht? Gi tr li nhun y l bao nhiu?

Tr li v d 1:

- Li nhun t gi tr cao nht (thp nht) khi:

= 0.

= 8 0.04

= 0 = 200

- Liu y c phi l gi tr ln nht cha?

2

2= 0.04 < 0,

vy ta c gi tr ln nht khi = 200; = 800 ng.

- Vy nu nh nh my lc c 200 thng / ngy s t c li nhun cao nht l 800 ng.

V d 2: Mt khu vc cha hng hnh ch nht c xy st, dc theo chiu cao ca mt to nh.

Mt hng ro an ninh c thit k nm 3 mt cn li ca khu vc y. Hi nu c 800 hng

ro bao quanh hng ro bao quanh th din tch ti a ca khu vc ny l bao nhiu?

Tr li v d 2:



78

- Ta c din tch l = .

- T hnh v ta c 2 + = 800 = 800 2.

- Vy din tch l

dao ham, tich phan ung dung duoc gi

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