Dao Ham Va Bai Toan Tiep Tuyen Cho Lop 11Day Du

T Ton _ Tin trng thpt lc ngn s 2

I. Kin thc c bn.1. Bng o hm cc hm s c bn.

Hm s(y = f(x))o hm(y = f(x))Hm so hm

y = c0y = tanx

y = x1y = cotx

y = xnnxn-1y = exex

y = 1/x

y = ax ax. lna

y = lnx1/x

y = sinxcosxy = logax

y = cosx-sinx

2. o hm ca hm hp.

Ta xt hm s y = f(u(x)). Ta tnh o hm ca hm s cho theo x nh sau

Bng o hm ca hm s hp

Hm s

o hm

Hm so hm

y = unn.un-1.uy = tanu. u

y = 1/u

y = cotu. u

y = euu.eu

y = sinuu.cosuy = au u.au. lna

y = cosu- u.sinuy = lnu

y = logau

Ch : Khi p dng tnh o hm ca hm hp ta ch ban u tnh o hm ca hm s theo bin u ri nhn vi o hm ca hm s u theo bin x.

3. Cc php ton o hm.

Cho hai hm s y = u(x), y = v(x). Khi

*) (u + v) = u + v

*) (u - v) = u v

*) (uv) = uv + vu

*) (ku) = k.u ( k l hng s)

*)

4. o hm bc cao ca hm s.

o hm bc n ca hm s y = f(x) l o hm bc 1 ca o hm bc n 1 ca hm s y = f(x) ( n > 1).II. Cc dng ton c bn.1. Dng 1. Tnh o hm ca hm s.Phng php. Ta vn dng cc quy tc v php tnh o hm, c bit l o hm ca hm hp. Nu yu cu tnh o hm ti mt im ta cn tnh o hm ri thay vo e c kt qu.V d 1. Tnh o hm cc hm s sau

a)

b)

c)

d)

Giia) Ta c

b) Ta c

c) Ta c

d) Ta c

V d 2. Tnh o hm cc hm s sau ti cc im tng ng.

a) ti x0 = -1.

b) ti .

c) ti x0 = 2 .

Gii

a) Ta c

suy ra

b) Ta c

suy ra

c) Ta c

suy ra

V d 3. Tnh o hm cc hm s sau

a)

b)

c)

d)

e)

f)

g)

Gii

a) Ta c

b) Ta c

c) Ta c

d) Ta c

e) Ta c

f) Ta c

g) Ta c

2. Dng 2. Gii phng trnh y = 0.

Phng php. Ta tnh y sau gii phng trnh y = 0.

V d 1. Gii phng trnh y = 0 bit.

a)

b)

c)

d)

e)

f)

g)

h)

i)

Gii

a) Ta c

suy ra

Vy phng trnh y = 0 c hai nghim phn bit x = 0 v x = 2.

b) Ta c

suy ra

Vy phng trnh y = 0 c hai nghim phn bit x = 0 v x = 2.

c) Ta c

Suy ra

Vy phng trnh y = 0 c hai nghim phn bit

d) Ta c

suy ra

Vy phng trnh y = 0 c hai nghim phn bit x = 0 v x = -2.

e) Ta c

suy ra

Vy phng trnh y = 0 c hai nghim phn bit x = 0 v x = -2.

f) Ta c

Suy ra

Vy phng trnh y = 0 c ba nghim phn bit .

g) Ta c

Suy ra

Vy phng trnh y = 0 c nghim duy nht x = 0.h) Ta c

Suy ra

Vy phng trnh y = 0 c hai nghim phn bit x = -1 v x = 3.

i) Ta c

Suy ra

Vy phng trnh y = 0 c hai nghim phn bit

3. Dng 3: Chng minh ng thc v o hm.Phng php: Tnh o hm v s dng cc php bin i c bit l v hm lng gic.

V d 1. Chng minh rng

a) y y2 -1 = 0 vi y = tanx.

b) y + 2y2 + 2 = 0 vi y = cot2x.

c) y2 + 4y2 = 4 vi y = sin2x.

Gii

a) Ta c

Khi

Vy ta c iu cn chng minh.

b) Ta c

Khi


c) Ta cy = 2cos2x

Khi

Vy ta c iu cn chng minh.III. Bi tp t luyn.Bi 1. Tnh o hm cc hm s sau

a)

b)

c)

d)

e)

f)

Bi 2. Tnh o hm cc hm s sau

a)

b)

c)

d)

e)

f)

Bi 3. Tnh o hm cc hm s sau ti cc im tng ng

a) ti im x0 = -1

b) ti im x0 = 2

c) ti im .Bi 4. Gii phng trnh y = 0 trong cc trng hp sau

a)

b)

c)

d)

e)

f)

I. Kin thc c bn.1. Tip tuyn ti mt im: Cho hm s y= f(x) (C), x0 l mt im thuc vo TX ca hm s trn v tn ti o hm ti . Khi ta c tip tuyn vi (C) ti im

(x0; f(x0)) c phng trnh l y = y/(x0)(x-x0) + f(x0)

Nhn xt: trn ta c y/(x0) l h s gc ca tip tuyn. Ta cn tm c h s gc v tip im trong trng hp ny nu mun vit phng trnh tip tuyn vi ng cong no . Cc bi tp hay gp trong phn ny: Cho honh tip im; tung tip im; hay ti giao im ca th hm s vi ng thng no .2. iu kin tip xc ca hai th.

Cho hai hm s y = f(x) (C1), y = g(x) (C2).

Khi (C1) tip xc vi (C2) khi v ch khi h phng trnh c nghim.

Ch :

+ Nu hai th (C1) v (C2) l hai ng cong th chng tip xc vi nhau ti hai im khi h trn c hai nghim phn bit.

+ Nu mt trong hai ng l ng thng th c hai tip tuyn ta cn h trn c hai nghim phn bit.II. Dng ton c bn.1. Dng 1. Vit phng trnh tip tuyn ti mt im.

Phng php: Ta cn tm c to tip im da vo cc d kin bi ton cho.Nhn xt: Trong dng ny ta thng gp cc trng hp sau

+ Cho bit ta ca tip im.

+ Cho bit honh ca tip im hoc iu kin no tm c honh tip im.

+ Bit tung tip im hoc iu kin no tm c tung tip im.

+ Tip im l giao im ca th vi mt th khc. Khi ta cn gii h phng trnh tm to ca tip im.

2. Dng 2. Tip tuyn i qua mt im: Cho hm s y= f(x) (C) vit phng trnh tip tuyn vi (C) i qua im M(xM; yM)

Phng php:

Cch 1: Tm tip im

Gi s tip tuyn vi (C) cn tm c tip im l M0(x0; y0). Khi tip tuyn cn tm c phng trnh y = f/(x0)(x-x0) + f(x0).

M tip tuyn i qua im M(xM; yM) suy ra yM = f/(x0)(xM-x0) + f(x0) gii phng trnh ny ta tm c honh tip im sau tm y0 = f(x0) ri vit phng trnh tip tuyn cn tm theo dng 1.Cch 2: S dng iu kin tip xc

Gi s ng thng qua M(xM; yM) c h s gc k khi n c phng trnh

y = k(x-xM) + yMTa c ng thng y = k(x-xM) + yM l tip tuyn ca ng cong (C) gii h ny ta tm c honh ca tip im sau vit phng trnh tip tuyn tng ng.

Nhn xt: trn c bao nhiu nghim x ta c by nhiu tip tuyn i qua im M.

3. Dng 3. Tip tuyn cho trc h s gc:Phng php.

Cch 1. Tm tip im

Gi s tip tuyn cn tm c tip im l M0(x0; y0). Khi tip tuyn cn tm c phng trnh y = f/(x0)(x-x0) + f(x0).

Khi theo gii thit ta c f/(x0) = k. Gii phng trnh ny ta tm c honh tip im sau tm y0 = f(x0) ri vit phng trnh tip tuyn cn tm theo dng 1.Nhn xt: Trong dng ny ta c th gp cc bi tp nh sau:

*) Tip tuyn c h s gc k khi ta tm tip im M0(x0; y0) bng cch gii phng trnh f/(x0) = k sau vit phng trnh tip tuyn tng ng.

*) Tip tuyn vung gc vi ng thng y = ax + b khi tip tuyn c h s gc l k = sau tm tip im M0(x0; y0) bng cch gii phng trnh f/(x0) = k v vit phng trnh tip tuyn tng ng.

*) Tip tuyn song song vi ng thng y = ax+ b khi tip tuyn c h s gc l k= a sau tm tip im M0(x0; y0) bng cch gii phng trnh f/(x0) = k v vit phng trnh tip tuyn tng ng.

*) Tip tuyn to vi chiu dng trc honh gc khi h s gc ca tip tuyn l k = tan sau tm tip im M0(x0; y0) bng cch gii phng trnh f/(x0) = k v vit phng trnh tip tuyn tng ng.

*) Tip tuyn to vi ng thng y = ax +b mt gc khi h s hc ca tip tuyn l k tho mn hoc chng ta dng tch v hng ca hai vct php tuyn tm h s gc k sau tm tip im M0(x0; y0) bng cch gii phng trnh f/(x0) = k v vit phng trnh tip tuyn tng ng.

III. V d.V d 1: Cho hm s . Vit phng trnh tip tuyn vi (C) bit

a) Honh tip im ln lt l -1; 3;

b) Tung tip im ln lt l -4.

c) Tip im l giao ca (C) vi trc honh.

Gii

TX:

Ta c

a) Vi honh tip im x0 = -1 ta c y0 = f(x0) = f(-1) = - 4; suy ra tip tuyn vi (C) khi c phng trnh y = f/(-1)(x+1) 4 hay y = - 4

Vi honh tip im x0 = 3 ta c y0 = f(x0) = f(3) = 44; suy ra tip tuyn vi (C) khi c phng trnh y = f/(3)(x-3) + 44 hay y = 40x 76

b) Vi tung tip im y0 = - 4 ta c x0 = -1 hoc x0 = 0

Vi honh tip im x0 = -1 ta c suy ra tip tuyn vi (C) khi c phng trnh y = f/(-1)(x+1) 4 hay y = - 4

Vi x0 = 0 ta c suy ra tip tuyn vi (C) khi c phng trnh y = f/(0)(x+1) 4 hay y = x 3.

c) Giao im ca (C) vi trc honh c honh l nghim ca phng trnh

Khi suy ra tip tuyn vi (C) khi c phng trnh y = f/(1)(x-1) hay y = 8x 8.

V d 2: Cho hm s (Cm). Vit phng trnh tip tuyn ca (Cm) ti giao im ca n vi Oy, tm m tip tuyn trn chn trn hai trc to ra mt tam gic c din tch bng 8.

Gii

TX:

Ta c (Cm) giao vi Oy ti im A(0; 1 -m)

. Khi tip tuyn cn tm l y = y/(0)x +1 m hay y =-mx +1-m

Tip tuyn trn ct trc honh ti im suy ra

Vi m = 0 th th hm s cho khng ct trc honh suy ra khng tn ti tam gic OAB. Vy vi th tip tuyn cn tm ct hai trc ta to ra tam gic c din tch bng 8.

V d 3: Cho hm s vit phng trnh tip tuyn vi (C) bit

a) Tip tuyn c h s gc k = 9

b) Tip tuyn vung gc vi ng thng

Gii

TX: . Ta c

a) Gi A(xA; yA) l tip im ca tip tuyn cn tm khi ta c

Vi ta c khi tip tuyn vi (C) cn tm l y = 9(x+1) 4 hay

y=9x+5.

Vi xA = 3 ta c yA = 0 khi tip tuyn vi (C ) cn tm l y =9(x-3) hay y= 9x 27

Vy c hai tip tuyn vi (C) c h s gc l k = 9 l

y=9x+5 v y= 9x 27.

b) Gi M(xM ;yM) l tip im ca tip tuyn cn tm.

Tip tuyn cn tm vung gc vi ng thng suy ra h s gc ca n l

k = -3 (Lm tng t nh phn a)

V d 4: Cho hm s (C). Vit phng trnh tip tuyn vi (C) trong cc trng hp sau

a) Tip tuyn song song vi ng thng y = 6x 4.

b) Tip tuyn to vi ng thng mt gc 450.Gii

TX: . Ta c

a) V tip tuyn song song vi ng thng y = 6x 4 suy ra h s gc ca tip tuyn l k = 6.

Gi M0(x0; y0) l tip im ca tip tuyn cn tm. Khi ta c

Vi ta c khi tip tuyn cn tm l

Vi ta c khi tip tuyn cn tm l

b) V tip tuyn cn tm to vi ng thng mt gc 450 suy ra h s gc ca tip tuyn l k tho mn

sau lm tng t nh phn a (Tm tip im).

V d 5: Vit phng trnh tip tuyn vi (C) : i qua im .

Gii

Gi s ng thng i qua c h s gc k, khi n c dng

(d)

Ta c (d) tip xc vi (C) khi v ch khi h phng trnh sau c nghm

Thay (2) vo (1) ta c

Vy c ba tip tuyn vi (C) i qua im ( T vit phng trnh tip tuyn).V d 6. Cho hm s

a) CMR: Khng tn ti hai im no trn (C ) sao cho tip tuyn ti hai im vung gc vi nhau.

b) Tm k sao cho trn (C) c t nht mt im sao cho tip tuyn ti vung gc vi ng thng y = kx + m.

Giia) Gi s trn (C) c hai im M1(x1; y1) v M2(x2; y2) m tip tuyn vi (C) ti vung gc vi nhau.

Ta c y = 3x2 + 6x + 3 = 3(x+1)2.Khi ta c

v l

Suy ra gi s l sai hay ta c iu cn chng minh.

b)

V d 7. Cho hm s y = x3 - x2 c th (C)

Vit phng trnh tip tuyn vi th (C) i qua im A(3; 0).Gii

ng thng () i qua A(3; 0) v c h s gc k c dng: y = k(x - 3)

+) () l tp tuyn vi (C)

H c nghim.

Th (1) vo (2):

2x3 -12x2 + 18x = 0

+) Vi x1 = 0 k1 = 0 PTT2: y = 0

+) Vi x2 = 3 k2 = 3 PTT2: y = 3x - 9.

Vy c hai tip tuyn vi th hm s cho tho mn yu cu bi ton y = 0 v y = 3x 9.V d 8. Tm a th hm s (C) tip xc vi (P): y = x2 + a.Giiiu kin tip xc ca th (C) vi (P)

H c nghim

Gii (1) x = 0 Th vo (2) a = - 1

Vy vi a = -1 th (1) tip xc vi (P).V d 9. Cho ng cong (C)

Tm cc im trn Ox t k c hai tip tuyn vi (C) m hai tip tuyn ny vung gc vi nhau.

Gii:

Gi M(a; 0) ( Ox; l ng thng qua M c h s gc k: y = k(x - a)

() l tip tuyn ca (C)

EMBED Equation.DSMT4 H c nghim.

EMBED Equation.DSMT4 (2) - (1)

Kt hp (3) v (1) ta c:

(4) k2(1 - a)2 + 4k - 4 = 0

T M k c hai tip tuyn vung gc vi nhau ti (C) H trn c hai nghim phn bit k1, k2 v k1.k2 = -1.

EMBED Equation.DSMT4 Vy cc im cn tm l (-1; 0); (3; 0)Nhn xt: T h (I) ta phi bin i thnh h tng ng m ch c a v k. Nhn thy nu tnh c theo a v k thay vo phng trnh (1) th c mt h mi tng ng trong c mt phng trnh ch cha a v k t ta c php bin i nh trn v cch gii ny l ngn gn. V d 10. Cho ng cong (C)

Tm cc im trn mt phng to m t k c hai tip tuyn gc vi (C), hai tip tuyn ny vung gc vi nhau.

Gii:

() l ng thng i qua M(a; b) v c h s gc k nn PT (): y = k(x - a) + b.



EMBED Equation.DSMT4 Ly (4) - (3)

EMBED Equation.DSMT4


EMBED Equation.DSMT4 (5)

Kt hp (5) v (1) ta c h

( k 1 v t (1) nu k = 1 th x, h v nghim.)

EMBED Equation.DSMT4 V t M k c hai tip tuyn vung gc vi nhau ti (C) h trn c hai nghim phn bit k1, k2 v k1.k2 = - 1


EMBED Equation.DSMT4 Th (10) vo (9): 2[(1 - a)b + 2] 0 (1 - a)b + 2 0

T (10) (1 - a)2 + b2 + 2(1 - a)b = 4 + 2(1 - a)b

(1 - a + b)2 = 2(2 + (1 - a)b)

V 2+ (1 - a)b 0 1 - a + b 0.

Vy ta c tp hp cc im M cn tm l ng trn tm I(1; 0) bn knh R = 2, b i 4 im l giao cc ng thng x = 1 v - x + y + 1 = 0 vi ng trn l cc im (1; 2); (); ().V d 11. Cho ng cong: (C)

Tm tt c cc im trn ng thng y = 7 m t k c hai tip tuyn vi ng cong (C) m hai tip tuyn hp vi nhau gc = 450.Gii:

Gi M ( t: y = 7 M(a; 7).

Phng trnh ng thng () qua M c h s gc k: y = k(x - a) + 7.



EMBED Equation.DSMT4 Ly (4) - (3):

EMBED Equation.DSMT4 (5)

Kt hp (5) v (1)


T M k hai tip tuyn hp vi nhau gc = 450.

Khng mt tnh cht tng qut

Ta gi s:


k1 - k1.k2 = 1 + k2 (7)

V (6) phi c hai nghim phn bit m

EMBED Equation.DSMT4 c mt nghim bng 0 v mt nghim khc 0.

Vy t (6)

EMBED Equation.DSMT4 hoc (8)

Kt hp (8) v (7) ta c: hoc

Nu k1 = 1, t (6): .

Nu k2 = -1 , t (8):

Vy cc im tm c l: M1;2 (; 7); M3;4(; 7)

V d 12. Vit phng trnh tip tuyn chung ca hai (P) sau:

y = x2 - 3x + 2 (1) v y = - x2 + 7x - 11 (2)

Gii:

Gi tip tuyn chung l: y = ax + b. Gi M0(x0; y0) v l tip im ca tip tuyn vi Parabol (1) v (2)

Theo iu kin tip xc ca hai ng ta c h sau:

H c nghim.

T (1) v (2) (5)

T (3) v (4)

Gii ra tm c

Kt lun: Tip tuyn chung l: y = 3x - 7 v y = x 2.V d 13. Tm tip tuyn c nh ca h ng cong c phng trnh:

Gii:

Gi ng thng: y = ax + b l tip tuyn c nh ca h ng cong

H phng trnh sau c nghim (m 0

EMBED Equation.DSMT4 Ly (3) - (4):

Kt hp (2) v (5) ta c:

(a + 1)2m2 + 2(a - 1)(b + 1)m + (b + 1)2 = 0

Phng trnh ny tha mn (m 0

Kt lun: Vy h ng cong c mt tip tuyn c nh l: y = - x - 1IV. Bi tp t luyn.Bi 1. Cho . Tm m ct ng thng y = -x + 1 ti ba im A(0; 1), B, C sao cho tip tuyn vi ti B v C vung gc vi nhau.

Bi 2. Tm cc im trn th hm s m tip tuyn ti vung gc vi ng thng .

Bi 3. Cho hm s . CMR: Trn (C) c v s cp im m tip tuyn ti tng cp im song song vi nhau ng thi cc ng thng ni cc cp im ny ng quy ti mt im c nh.

Bi 4. Cho . Tm tip tuyn vi (C) c h s gc nh nht.

Bi 5. Cho Vit phng trnh tip tuyn vi hai th trn ti giao im ca chng.

Bi 6. Vit phng trnh tip tuyn vi ti giao im ca n vi trc Oy. Tm k tip tuyn to vi hai trc ta mt tam gic c din tch bng 8.

Bi 7. Cho hm s . Vit phng trnh tip tuyn vi (C) trong cc trng hp sau

a) C h s gc k = - 2.

b) Tip tuyn to vi chiu dng trc honh gc 600.

c) Tip tuyn to vi chiu dng trc honh gc 150.

d) Tip tuyn to vi chiu dng trc honh gc 750.

e) Tip tuyn to song song vi ng thng y = - x + 2.

f) Tip tuyn vung gc vi ng thng y = 2x 3.

g) Tip tuyn to vi ng thng y= 3x + 7 gc 450.

Bi 8. Cho hm s

a) Vit phng trnh tip tuyn vi (C) i qua im .

b) Tm trn ng thng y = - 2 nhng im k c hai tip tuyn ti (C) vung gc vi nhau.

Bi 9. Cho hm s . Tm trn trc honh nhng im k c ba tip tuyn vi (C).

(H SPHN2- KB-1999)

Bi 10. Cho hm s . Vit phng trnh tip tuyn vi (C) i qua im A(2; 0).

(H THHN- 1994).

Bi 11. Cho hm s . Vit phng trnh tip tuyn vi (C) to vi trc honh gc 450.

Bi 12. Cho hm s . Vit phng trnh tip tuyn vi (C) to vi ng thng y = 3x gc 450.

Bi 13. Tm trn Oy nhng im k c ng mt tip tuyn vi .

Bi 14. Cho hm s . Tm M trn (C) sao cho tip tuyn vi (C) ti M ct hai trc Ox, Oy ti A, B to ra tam gic OAB vung cn. (HVBCVTHN - 1997).

Bi 15. Cho hm s . CMR: Tip tuyn vi (C) ti mi im M ty lun to vi hai tim cn mt tam gic c din tch khng i.

Bi 16. Tm cc im trn th m tip tuyn ti vung gc vi ng thng . (H Ngoi Ng H Ni 2001)Bi 17. Tm tip tuyn c h s gc nh nht vi th .

(H Ngoi Thng TPHCM 1998).

Bi 18. Tm tip tuyn c h s gc nh nht vi th

( Hc vin quan h quc t 2001).Bi 19. Tm im M trn th sao cho tip tuyn vi (C) tai M i qua gc ta . ( H Cng on 2001).Bi 20. Vit phng trnh tip tuyn ti cc im c nh m th . Tm qu tch giao im ca cc tip tuyn .

( H an ninh 2000_ k A).Bi 21. Cho th hm s

a) Vit phng trnh tip tuyn vi (C) i qua im .

b) Tm trn ng thng y = -2 im m t k c hai tip tuyn vi (C) v chng vung gc vi nhau.

Bi 22. Cho hm s . Tm cc im trn ng thng x = 2 k c ng ba tip tuyn vi (C).( H cn th 2000_ k A).Bi 23. Cho hm s . Vit phng trnh tip tuyn vi (C) bit tip tuyn song song vi ng thng y = -x. ( H lt 2000_ k A).Bi 24. Cho hm s . Vit phng trnh tip tuyn vi th (C) i qua im A(1; 3)

( H ty nguyn 2000_ k A).Bi 25. Cho hm s . ng thng y = 5 tip xc vi (C) ti A v ct (C ) ti im B, tm ta im B. ( H ty nguyn 2000_ k D).Bi 26. Cho hm s . Vit phng trnh tip tuyn vi (C ) i qua im A(1; 0).

( H an ninh nhn dn 2000_ k D).Bi 27. Tm cc im trn trc honh k c ng mt tip tuyn vi th

Bi 28. Cho th . CMR trn ng thng y = 7 c bn im sao cho t mi im k c hai tip tuyn ti (C) v to vi nhau mt gc 450.

Bi 29. Cho th . Tm t hp cc im trn mt phng to Oxy tho mn

a) T khng k c tip tuyn no vi th (C).

b) T k c t nht mt tip tuyn vi th (C).

c) T k c ng mt tip tuyn vi th (C).

d) T k c ng hai tip tuyn vi th (C).

e) T k c ng hai tip tuyn vi th (C) v hai tip tuyn o vung gc vi nhau.

Bi 30. Vit phng trnh tip tuyn i qua im A(1; 0) ti th .

( H dc 1999).Bi 31. Vit phng trnh tip tuyn i qua im A(-1; 0 ) ti th .

( H xy dng 1995).Bi 32. Vit phng trnh tip tuyn i qua im A(0; 5/4 ) ti th .

( Hsp vinh 1998).Bi 33. Vit phng trnh tip tuyn i qua im A(1; 1 ) ti th .

( H lt 1999).

I. Kin thc c bn.1. Khai trin nh thc Newtn.

Ta c (1)Trong :

+ a, b l hai s thc.

+ n l s nguyn dng.

Nhn xt:

+ Trong khai trin trn s m ca a gim dn t tri sang phi, ngc li s m ca b tng dn t tri sang phi. S m ca a v b trong mi s hng cng li u bng n.

+ Trong khai trin trn c n + 1 s hng.

+ S hng tng qut trong khai trin (1) l .

+ S hng thc k trong khai trin (1) l .

2. Mt vi khai trin thng dng.Ta c

Thay x = 1 vo hai v ca (2) ta c ng thc sau

Thay x = - 1 vo hai v ca (2) ta c ng thc sau

3. Mi lin h ca hai hm s bng nhau.Ta c hai hm s y = f(x) v y = g(x).

Nu f(x) = g(x) th f(x) = g(x)II. Dng ton tnh tng ca t hp lin quan ti o hm.Ta c mt vi ch khi gp tnh tng ca t hp

+ Nu trong v tnh tng khng c th ta cn dng khai trin ri o hm hai v theo x c hai v sau thay x bng mt gi tr thch hp.

+ Nu trong mt v tnh tng khng c v th ta dng khai trin ri o hm hai v theo x hai ln sau thy bng mt gi tr thch hp.

III. V d.V d 1. Chng minh rng

a)

b)

Gii

a) Ta c

o hm hai v ca (*) theo x ta c

(a)

Thay x = 1 vo ng thc (a) ta c

Vy ta c ng thc cn chng minh.

b)o hm hai v ca (*) hai ln theo x ta c

Thay x = 1 vo ng thc trn ta c


PAGE 21o hm v ng dng tp 1.

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Dao Ham Va Bai Toan Tiep Tuyen Cho Lop 11Day Du