Title: Gii tch 2 cho khi k thut
URL Source: blob://pdf/417dec02-70c2-42b0-a33d-f2981bef3554
Markdown Content:
## Gii tch 2 cho khi k thut
Ngy 21 thng 3 nm 2022
Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 1 / 42 Gii thiu chung
Gii thiu chung v mn Gii tch 2
S tn ch: 3
im tng kt gm: im qu trnh (40 %) v im thi (60 %), trong
im qu trnh gm: im im danh, im bi tp nhm v im kim tra 1,2
Hnh thc thi:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 2 / 42
## Gii thiu chung
Gii thiu chung v mn Gii tch 2
S tn ch: 3
im tng kt gm: im qu trnh (40 %) v im thi (60 %), trong
im qu trnh gm: im im danh, im bi tp nhm v im kim tra 1,2
Hnh thc thi:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 2 / 42
## Gii thiu chung
Gii thiu chung v mn Gii tch 2
S tn ch: 3
im tng kt gm: im qu trnh (40 %) v im thi (60 %), trong
im qu trnh gm: im im danh, im bi tp nhm v im kim tra 1,2
Hnh thc thi:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 2 / 42
## Gii thiu chung
Gii thiu chung v mn Gii tch 2
S tn ch: 3
im tng kt gm: im qu trnh (40 %) v im thi (60 %), trong
im qu trnh gm: im im danh, im bi tp nhm v im kim tra 1,2
Hnh thc thi:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 2 / 42
## Gii thiu chung
Ti liu tham kho
Gio trnh Gii tch 2 - HGTVT
Ton cao cp A3 - NXB Gio dc
Gii tch ton hc (Cc VD v bi tp)-Tp 2- Liasko
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 3 / 42
## Gii thiu chung
Ti liu tham kho
Gio trnh Gii tch 2 - HGTVT
Ton cao cp A3 - NXB Gio dc
Gii tch ton hc (Cc VD v bi tp)-Tp 2- Liasko
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 3 / 42
## Gii thiu chung
Ti liu tham kho
Gio trnh Gii tch 2 - HGTVT
Ton cao cp A3 - NXB Gio dc
Gii tch ton hc (Cc VD v bi tp)-Tp 2- Liasko
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 3 / 42
## Chng 1. Hm s nhiu bin
Chng 1. Hm s nhiu bin
1.1. Mt s khi nim chung v hm s nhiu bin 1.1.1. Hm s 2 bin
xy dng l thuyt cho hm s thc nhiu bin s ta bt u t cc khi nim i vi hm s 2 bin, vic pht biu cho l thuyt ca hm s nhiu hn 2 bin c tin hnh tng t.
a) Cc khi nim
Cho D l mt min trong mt phng Oxy , ta ni rng trn D xc nh mt hm s f nu vi mi im M(x, y ) D tn ti duy nht mt gi tr z = f (x, y ) R. y ta ni x v y l cc bin s, ta coi z hay f gi l hm ca cc bin (x, y ),v D c gi l min xc nh ca hm f .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 4 / 42
## Chng 1. Hm s nhiu bin
Chng 1. Hm s nhiu bin
1.1. Mt s khi nim chung v hm s nhiu bin 1.1.1. Hm s 2 bin
xy dng l thuyt cho hm s thc nhiu bin s ta bt u t cc khi nim i vi hm s 2 bin, vic pht biu cho l thuyt ca hm s nhiu hn 2 bin c tin hnh tng t.
a) Cc khi nim
Cho D l mt min trong mt phng Oxy , ta ni rng trn D xc nh mt hm s f nu vi mi im M(x, y ) D tn ti duy nht mt gi tr z = f (x, y ) R. y ta ni x v y l cc bin s, ta coi z hay f gi l hm ca cc bin (x, y ),v D c gi l min xc nh ca hm f .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 4 / 42
## Chng 1. Hm s nhiu bin
Chng 1. Hm s nhiu bin
1.1. Mt s khi nim chung v hm s nhiu bin 1.1.1. Hm s 2 bin
xy dng l thuyt cho hm s thc nhiu bin s ta bt u t cc khi nim i vi hm s 2 bin, vic pht biu cho l thuyt ca hm s nhiu hn 2 bin c tin hnh tng t.
a) Cc khi nim
Cho D l mt min trong mt phng Oxy , ta ni rng trn D xc nh mt hm s f nu vi mi im M(x, y ) D tn ti duy nht mt gi tr z = f (x, y ) R. y ta ni x v y l cc bin s, ta coi z hay f gi l hm ca cc bin (x, y ),v D c gi l min xc nh ca hm f .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 4 / 42
## Chng 1. Hm s nhiu bin
Chng 1. Hm s nhiu bin
1.1. Mt s khi nim chung v hm s nhiu bin 1.1.1. Hm s 2 bin
xy dng l thuyt cho hm s thc nhiu bin s ta bt u t cc khi nim i vi hm s 2 bin, vic pht biu cho l thuyt ca hm s nhiu hn 2 bin c tin hnh tng t.
a) Cc khi nim
Cho D l mt min trong mt phng Oxy , ta ni rng trn D xc nh mt hm s f nu vi mi im M(x, y ) D tn ti duy nht mt gi tr z = f (x, y ) R. y ta ni x v y l cc bin s, ta coi z hay f gi l hm ca cc bin (x, y ),v D c gi l min xc nh ca hm f .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 4 / 42
## Chng 1. Hm s nhiu bin
Chng 1. Hm s nhiu bin
1.1. Mt s khi nim chung v hm s nhiu bin 1.1.1. Hm s 2 bin
xy dng l thuyt cho hm s thc nhiu bin s ta bt u t cc khi nim i vi hm s 2 bin, vic pht biu cho l thuyt ca hm s nhiu hn 2 bin c tin hnh tng t.
a) Cc khi nim
Cho D l mt min trong mt phng Oxy , ta ni rng trn D xc nh mt hm s f nu vi mi im M(x, y ) D tn ti duy nht mt gi tr z = f (x, y ) R. y ta ni x v y l cc bin s, ta coi z hay f gi l hm ca cc bin (x, y ),v D c gi l min xc nh ca hm f .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 4 / 42
## Chng 1. Hm s nhiu bin
- Ta gi tp G = {(x, y , f (x, y )) |(x, y ) D} R3 l th ca hm s f .- V mt hnh hc th th ca hm s 2 bin f (x, y ) m t mt mt cong trong khng gian, v min xc nh D l hnh chiu ca tp G xung mt phng Oxy .
b) Nhn xt
- Thng thng vi hm s 2 bin ta hay coi z l hm cn (x, y ) l bin v vit
z = z(x, y ), i khi ta cng c th coi x hay y l cc hm s, tc l x = x(y , z)
hay y = y (x, z).- Khi coi x (hoc y ) l hm th min xc nh ca hm s c xt trn mt Oyz
(hoc Ozx ) tng ng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 5 / 42
## Chng 1. Hm s nhiu bin
- Ta gi tp G = {(x, y , f (x, y )) |(x, y ) D} R3 l th ca hm s f .- V mt hnh hc th th ca hm s 2 bin f (x, y ) m t mt mt cong trong khng gian, v min xc nh D l hnh chiu ca tp G xung mt phng Oxy .
b) Nhn xt
- Thng thng vi hm s 2 bin ta hay coi z l hm cn (x, y ) l bin v vit
z = z(x, y ), i khi ta cng c th coi x hay y l cc hm s, tc l x = x(y , z)
hay y = y (x, z).- Khi coi x (hoc y ) l hm th min xc nh ca hm s c xt trn mt Oyz
(hoc Ozx ) tng ng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 5 / 42
## Chng 1. Hm s nhiu bin
- Ta gi tp G = {(x, y , f (x, y )) |(x, y ) D} R3 l th ca hm s f .- V mt hnh hc th th ca hm s 2 bin f (x, y ) m t mt mt cong trong khng gian, v min xc nh D l hnh chiu ca tp G xung mt phng Oxy .
b) Nhn xt
- Thng thng vi hm s 2 bin ta hay coi z l hm cn (x, y ) l bin v vit
z = z(x, y ), i khi ta cng c th coi x hay y l cc hm s, tc l x = x(y , z)
hay y = y (x, z).- Khi coi x (hoc y ) l hm th min xc nh ca hm s c xt trn mt Oyz
(hoc Ozx ) tng ng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 5 / 42
## Hm s nhiu bin
Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 6 / 42 Mt s mt bc hai thng gp
1.1.2. th ca mt s hm hai bin thng gp
a) Mt cu
x2 + y 2 + z2 = a2, a > 0
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 7 / 42
## Mt s mt bc hai thng gp
1.1.2. th ca mt s hm hai bin thng gp
b) Mt Ellipsoid
x2
a2 + y 2
b2 + z2
c2 = 1
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 8 / 42
## Mt s mt bc hai thng gp
1.1.2. th ca mt s hm hai bin thng gp
c) Mt tr
x2 + y 2 = R2, c z b
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 9 / 42
## Mt s mt bc hai thng gp
1.1.2. th ca mt s hm hai bin thng gp
d) Mt Paraboloid
z = x2 + y 2
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 10 / 42
## Mt s mt bc hai thng gp
1.1.2. th ca mt s hm hai bin thng gp
e) Mt nn
z2 = x2 + y 2
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 11 / 42
## Mt s khi nim v hm 2 bin
1.1.3. Tnh lin tc ca hm s hai bin
- Trong mt phng cho 2 im M0(x0, y0) v im M(x, y ), khong cch (Euclid) gia M v M0 c xc nh bi
(M, M0) = | ~MM 0| = (x0 x)2 + ( y0 y )2.
V ta ni im M tin ti M0 (M M0) khi x x0, y y0 nu (M, M0) 0- Ta ni rng hm s f (x, y ) c gii hn bng A khi (x, y ) tin ti (x0, y0) v vit
lim
> (x,y)(x0,y0)
f (x, y ) = A
nu vi mi dy (xn, yn) (x0, y0) th dy f (xn, yn) A.- Ta ni hm s f (x, y ) lin tc ti (x0, y0) nu lim
> (x,y)(x0,y0)
f (x, y ) = f (x0, y0).
Nhn xt : Cc hm s s cp lun lin tc trn min xc nh ca n.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 12 / 42
## Mt s khi nim v hm 2 bin
1.1.3. Tnh lin tc ca hm s hai bin
- Trong mt phng cho 2 im M0(x0, y0) v im M(x, y ), khong cch (Euclid) gia M v M0 c xc nh bi
(M, M0) = | ~MM 0| = (x0 x)2 + ( y0 y )2.
V ta ni im M tin ti M0 (M M0) khi x x0, y y0 nu (M, M0) 0- Ta ni rng hm s f (x, y ) c gii hn bng A khi (x, y ) tin ti (x0, y0) v vit
lim
> (x,y)(x0,y0)
f (x, y ) = A
nu vi mi dy (xn, yn) (x0, y0) th dy f (xn, yn) A.- Ta ni hm s f (x, y ) lin tc ti (x0, y0) nu lim
> (x,y)(x0,y0)
f (x, y ) = f (x0, y0).
Nhn xt : Cc hm s s cp lun lin tc trn min xc nh ca n.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 12 / 42
## Mt s khi nim v hm 2 bin
1.1.3. Tnh lin tc ca hm s hai bin
- Trong mt phng cho 2 im M0(x0, y0) v im M(x, y ), khong cch (Euclid) gia M v M0 c xc nh bi
(M, M0) = | ~MM 0| = (x0 x)2 + ( y0 y )2.
V ta ni im M tin ti M0 (M M0) khi x x0, y y0 nu (M, M0) 0- Ta ni rng hm s f (x, y ) c gii hn bng A khi (x, y ) tin ti (x0, y0) v vit
lim
> (x,y)(x0,y0)
f (x, y ) = A
nu vi mi dy (xn, yn) (x0, y0) th dy f (xn, yn) A.- Ta ni hm s f (x, y ) lin tc ti (x0, y0) nu lim
> (x,y)(x0,y0)
f (x, y ) = f (x0, y0).
Nhn xt : Cc hm s s cp lun lin tc trn min xc nh ca n.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 12 / 42
## Mt s khi nim v hm 2 bin
1.1.3. Tnh lin tc ca hm s hai bin
- Trong mt phng cho 2 im M0(x0, y0) v im M(x, y ), khong cch (Euclid) gia M v M0 c xc nh bi
(M, M0) = | ~MM 0| = (x0 x)2 + ( y0 y )2.
V ta ni im M tin ti M0 (M M0) khi x x0, y y0 nu (M, M0) 0- Ta ni rng hm s f (x, y ) c gii hn bng A khi (x, y ) tin ti (x0, y0) v vit
lim
> (x,y)(x0,y0)
f (x, y ) = A
nu vi mi dy (xn, yn) (x0, y0) th dy f (xn, yn) A.- Ta ni hm s f (x, y ) lin tc ti (x0, y0) nu lim
> (x,y)(x0,y0)
f (x, y ) = f (x0, y0).
Nhn xt : Cc hm s s cp lun lin tc trn min xc nh ca n.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 12 / 42
## Mt s khi nim v hm 2 bin
1.1.3. Tnh lin tc ca hm s hai bin
- Trong mt phng cho 2 im M0(x0, y0) v im M(x, y ), khong cch (Euclid) gia M v M0 c xc nh bi
(M, M0) = | ~MM 0| = (x0 x)2 + ( y0 y )2.
V ta ni im M tin ti M0 (M M0) khi x x0, y y0 nu (M, M0) 0- Ta ni rng hm s f (x, y ) c gii hn bng A khi (x, y ) tin ti (x0, y0) v vit
lim
> (x,y)(x0,y0)
f (x, y ) = A
nu vi mi dy (xn, yn) (x0, y0) th dy f (xn, yn) A.- Ta ni hm s f (x, y ) lin tc ti (x0, y0) nu lim
> (x,y)(x0,y0)
f (x, y ) = f (x0, y0).
Nhn xt : Cc hm s s cp lun lin tc trn min xc nh ca n.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 12 / 42
## o hm v vi phn ca hm nhiu bin
1.2. o hm ring, vi phn ton phn
1.2.1. o hm ring
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca im M0, khi ta c nh bin y = y0 th hm f (x, y0) = g (x) l hm s ch ph thuc vo bin x.Trong trng hp ny nu tn ti o hm g (x0) th n c gi l o hm ring theo bin x ca hm s f ti im M0, k hiu l f
> x
(M0) hoc f
x (M0).- Theo biu din o hm ca hm 1 bin i vi hm g (x), ta c
g (x0) = lim
> x0
g (x0 + x) g (x0)x
Khi ta c th vit (trong trng hp gii hn v phi tn ti)
f
> x
(x0, y0) = lim
> x0
f (x0 + x, y0) f (x0, y0)x
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 13 / 42
## o hm v vi phn ca hm nhiu bin
1.2. o hm ring, vi phn ton phn
1.2.1. o hm ring
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca im M0, khi ta c nh bin y = y0 th hm f (x, y0) = g (x) l hm s ch ph thuc vo bin x.Trong trng hp ny nu tn ti o hm g (x0) th n c gi l o hm ring theo bin x ca hm s f ti im M0, k hiu l f
> x
(M0) hoc f
x (M0).- Theo biu din o hm ca hm 1 bin i vi hm g (x), ta c
g (x0) = lim
> x0
g (x0 + x) g (x0)x
Khi ta c th vit (trong trng hp gii hn v phi tn ti)
f
> x
(x0, y0) = lim
> x0
f (x0 + x, y0) f (x0, y0)x
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 13 / 42
## o hm v vi phn ca hm nhiu bin
1.2. o hm ring, vi phn ton phn
1.2.1. o hm ring
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca im M0, khi ta c nh bin y = y0 th hm f (x, y0) = g (x) l hm s ch ph thuc vo bin x.Trong trng hp ny nu tn ti o hm g (x0) th n c gi l o hm ring theo bin x ca hm s f ti im M0, k hiu l f
> x
(M0) hoc f
x (M0).- Theo biu din o hm ca hm 1 bin i vi hm g (x), ta c
g (x0) = lim
> x0
g (x0 + x) g (x0)x
Khi ta c th vit (trong trng hp gii hn v phi tn ti)
f
> x
(x0, y0) = lim
> x0
f (x0 + x, y0) f (x0, y0)x
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 13 / 42
## o hm v vi phn ca hm nhiu bin
Ni mt cch khc
f
> x
= lim
> x0
f (x + x, y ) f (x, y )x
v
f
> y
= lim
> y0
f (x, y + y ) f (x, y )y .
b) Nhn xt:
+ Khi tnh o hm ring ca hm s theo mt bin no ta coi bin cn li nh l hng s. + o hm ring c tnh cht tng t nh o hm ca hm s 1 bin s.
V d: Tnh o hm ring ca cc hm s sau:
(1)z = x3 3y 2 + 2x2y + x + 2y ; (2) z = xy ; (3) z = ln (x + x2 + y 2)(4) z = arctan yx ; (5) f (x, y , z) = xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 14 / 42
## o hm v vi phn ca hm nhiu bin
Ni mt cch khc
f
> x
= lim
> x0
f (x + x, y ) f (x, y )x
v
f
> y
= lim
> y0
f (x, y + y ) f (x, y )y .
b) Nhn xt:
+ Khi tnh o hm ring ca hm s theo mt bin no ta coi bin cn li nh l hng s. + o hm ring c tnh cht tng t nh o hm ca hm s 1 bin s.
V d: Tnh o hm ring ca cc hm s sau:
(1)z = x3 3y 2 + 2x2y + x + 2y ; (2) z = xy ; (3) z = ln (x + x2 + y 2)(4) z = arctan yx ; (5) f (x, y , z) = xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 14 / 42
## o hm v vi phn ca hm nhiu bin
Ni mt cch khc
f
> x
= lim
> x0
f (x + x, y ) f (x, y )x
v
f
> y
= lim
> y0
f (x, y + y ) f (x, y )y .
b) Nhn xt:
+ Khi tnh o hm ring ca hm s theo mt bin no ta coi bin cn li nh l hng s. + o hm ring c tnh cht tng t nh o hm ca hm s 1 bin s.
V d: Tnh o hm ring ca cc hm s sau:
(1)z = x3 3y 2 + 2x2y + x + 2y ; (2) z = xy ; (3) z = ln (x + x2 + y 2)(4) z = arctan yx ; (5) f (x, y , z) = xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 14 / 42
## o hm v vi phn ca hm nhiu bin
Ni mt cch khc
f
> x
= lim
> x0
f (x + x, y ) f (x, y )x
v
f
> y
= lim
> y0
f (x, y + y ) f (x, y )y .
b) Nhn xt:
+ Khi tnh o hm ring ca hm s theo mt bin no ta coi bin cn li nh l hng s. + o hm ring c tnh cht tng t nh o hm ca hm s 1 bin s.
V d: Tnh o hm ring ca cc hm s sau:
(1)z = x3 3y 2 + 2x2y + x + 2y ; (2) z = xy ; (3) z = ln (x + x2 + y 2)(4) z = arctan yx ; (5) f (x, y , z) = xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 14 / 42
## o hm v vi phn ca hm nhiu bin
Ni mt cch khc
f
> x
= lim
> x0
f (x + x, y ) f (x, y )x
v
f
> y
= lim
> y0
f (x, y + y ) f (x, y )y .
b) Nhn xt:
+ Khi tnh o hm ring ca hm s theo mt bin no ta coi bin cn li nh l hng s. + o hm ring c tnh cht tng t nh o hm ca hm s 1 bin s.
V d: Tnh o hm ring ca cc hm s sau:
(1)z = x3 3y 2 + 2x2y + x + 2y ; (2) z = xy ; (3) z = ln (x + x2 + y 2)(4) z = arctan yx ; (5) f (x, y , z) = xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 14 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) z = x3 3y 2 + 2x2y + x + 2yz
x = 3x2 + 4xy + 1; z
y = 6y + 2x2 + 2(2) z = xy z
x = yx y 1, z
y = xy lnx
(3) z = ln (x + x2 + y 2)
z
x = (x + x2 + y 2)
x
x + x2 + y 2 =
1 + xx2+y 2
x + x2 + y 2 = 1
x2 + y 2 .
z
y = (x + x2 + y 2)
y
x + x2 + y 2 =
yx2+y 2
x + x2 + y 2 = y
x2 + y 2(x + x2 + y 2) .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 15 / 42
## o hm v vi phn ca hm nhiu bin
(4) z = arctan yxz
x = ( yx )
x
1 + ( yx )2
= yx2
1 + y 2
x2
= yx2 + y 2 .
z
y = ( yx )
y
1 + ( yx )2
=
1
x
1 + y 2
x2
= xx2 + y 2 .
(5) f (x, y , z) = xyz
f
x = yzx yz 1,
f
y = zx yz lnx ,
f
z = yx yz lnx .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 16 / 42
## o hm v vi phn ca hm nhiu bin
(4) z = arctan yxz
x = ( yx )
x
1 + ( yx )2
= yx2
1 + y 2
x2
= yx2 + y 2 .
z
y = ( yx )
y
1 + ( yx )2
=
1
x
1 + y 2
x2
= xx2 + y 2 .
(5) f (x, y , z) = xyz
f
x = yzx yz 1,
f
y = zx yz lnx ,
f
z = yx yz lnx .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 16 / 42
## o hm v vi phn ca hm nhiu bin
(4) z = arctan yxz
x = ( yx )
x
1 + ( yx )2
= yx2
1 + y 2
x2
= yx2 + y 2 .
z
y = ( yx )
y
1 + ( yx )2
=
1
x
1 + y 2
x2
= xx2 + y 2 .
(5) f (x, y , z) = xyz
f
x = yzx yz 1,
f
y = zx yz lnx ,
f
z = yx yz lnx .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 16 / 42
## o hm v vi phn ca hm nhiu bin
1.2.2. Vi phn ton phn
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0, gi cc s gia ca x, y ti x0, y0 tng ng l x, y . Khi s gia ca f (x, y ) ti M0 l
f (x0, y0) = f (x0 + x, y0 + y ) f (x0, y0).
Nu ta c th biu din
f (x0, y0) = A.x + B.y + . x + . y
trong , A, B khng ph thuc vo x, y , cn 0, 0 khi
x 0, y 0, th ta ni rng hm s f (x, y ) kh vi ti M0 v i lng
(A.x + B.y ) c gi l vi phn ton phn ca hm s f (x, y ) ti M0, k hiu l df (M0).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 17 / 42
## o hm v vi phn ca hm nhiu bin
1.2.2. Vi phn ton phn
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0, gi cc s gia ca x, y ti x0, y0 tng ng l x, y . Khi s gia ca f (x, y ) ti M0 l
f (x0, y0) = f (x0 + x, y0 + y ) f (x0, y0).
Nu ta c th biu din
f (x0, y0) = A.x + B.y + . x + . y
trong , A, B khng ph thuc vo x, y , cn 0, 0 khi
x 0, y 0, th ta ni rng hm s f (x, y ) kh vi ti M0 v i lng
(A.x + B.y ) c gi l vi phn ton phn ca hm s f (x, y ) ti M0, k hiu l df (M0).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 17 / 42
## o hm v vi phn ca hm nhiu bin
* Mt s nhn xt cho tnh kh vi ca hm s nhiu bin s
- Cc hm s s cp lun kh vi trn min xc nh ca n. - Nu hm s f (x, y ) kh vi ti M0(x0, y0) th lin tc ti M0, tuy nhin iu ngc li khng ng. - Nu hm s f (x, y ) c cc o hm ring lin tc ti M0 th kh vi ti M0 v ngc li. Khi ta c th biu din
df (M0) = f
> x
(M0) x + f
> y
(M0) y .
- Khng mt tng qut ta c th biu din cng thc vi phn ton phn ca mt hm s theo cng thc:
df (x, y ) = f
> x
dx + f
> y
dy ,
df (x, y , z) = f
> x
dx + f
> y
dy + f
> z
dz , ...
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 18 / 42
## o hm v vi phn ca hm nhiu bin
* Mt s nhn xt cho tnh kh vi ca hm s nhiu bin s
- Cc hm s s cp lun kh vi trn min xc nh ca n. - Nu hm s f (x, y ) kh vi ti M0(x0, y0) th lin tc ti M0, tuy nhin iu ngc li khng ng. - Nu hm s f (x, y ) c cc o hm ring lin tc ti M0 th kh vi ti M0 v ngc li. Khi ta c th biu din
df (M0) = f
> x
(M0) x + f
> y
(M0) y .
- Khng mt tng qut ta c th biu din cng thc vi phn ton phn ca mt hm s theo cng thc:
df (x, y ) = f
> x
dx + f
> y
dy ,
df (x, y , z) = f
> x
dx + f
> y
dy + f
> z
dz , ...
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 18 / 42
## o hm v vi phn ca hm nhiu bin
* Mt s nhn xt cho tnh kh vi ca hm s nhiu bin s
- Cc hm s s cp lun kh vi trn min xc nh ca n. - Nu hm s f (x, y ) kh vi ti M0(x0, y0) th lin tc ti M0, tuy nhin iu ngc li khng ng. - Nu hm s f (x, y ) c cc o hm ring lin tc ti M0 th kh vi ti M0 v ngc li. Khi ta c th biu din
df (M0) = f
> x
(M0) x + f
> y
(M0) y .
- Khng mt tng qut ta c th biu din cng thc vi phn ton phn ca mt hm s theo cng thc:
df (x, y ) = f
> x
dx + f
> y
dy ,
df (x, y , z) = f
> x
dx + f
> y
dy + f
> z
dz , ...
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 18 / 42
## o hm v vi phn ca hm nhiu bin
* Mt s nhn xt cho tnh kh vi ca hm s nhiu bin s
- Cc hm s s cp lun kh vi trn min xc nh ca n. - Nu hm s f (x, y ) kh vi ti M0(x0, y0) th lin tc ti M0, tuy nhin iu ngc li khng ng. - Nu hm s f (x, y ) c cc o hm ring lin tc ti M0 th kh vi ti M0 v ngc li. Khi ta c th biu din
df (M0) = f
> x
(M0) x + f
> y
(M0) y .
- Khng mt tng qut ta c th biu din cng thc vi phn ton phn ca mt hm s theo cng thc:
df (x, y ) = f
> x
dx + f
> y
dy ,
df (x, y , z) = f
> x
dx + f
> y
dy + f
> z
dz , ...
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 18 / 42
## o hm v vi phn ca hm nhiu bin
* Mt s nhn xt cho tnh kh vi ca hm s nhiu bin s
- Cc hm s s cp lun kh vi trn min xc nh ca n. - Nu hm s f (x, y ) kh vi ti M0(x0, y0) th lin tc ti M0, tuy nhin iu ngc li khng ng. - Nu hm s f (x, y ) c cc o hm ring lin tc ti M0 th kh vi ti M0 v ngc li. Khi ta c th biu din
df (M0) = f
> x
(M0) x + f
> y
(M0) y .
- Khng mt tng qut ta c th biu din cng thc vi phn ton phn ca mt hm s theo cng thc:
df (x, y ) = f
> x
dx + f
> y
dy ,
df (x, y , z) = f
> x
dx + f
> y
dy + f
> z
dz , ...
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 18 / 42
## o hm v vi phn ca hm nhiu bin
Cng thc tnh gn ng theo vi phn
Gi s f (x, y ) l mt hm s kh vi ti M0(x0, y0), khi vi x v y kh b ta c th xp x
f (x0 + x, y0 + y ) f (M0) + f
> x
(M0) x + f
> y
(M0) y
Tng t cho hm s 3 bin vi M0(x0, y0, z0) v vi x, y , z kh b ta c
f (x0 + x, y0 + y , z0 + z) f (M0) + f
> x
(M0) x + f
> y
(M0) y + f
> z
(M0) z
V d: Dng vi phn tnh gn ng gi tr ca biu thc (1) A = 1, 98 4 + 3, 03 2
(2) B = ln (1, 03 + 3
0, 99 1)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 19 / 42
## o hm v vi phn ca hm nhiu bin
Cng thc tnh gn ng theo vi phn
Gi s f (x, y ) l mt hm s kh vi ti M0(x0, y0), khi vi x v y kh b ta c th xp x
f (x0 + x, y0 + y ) f (M0) + f
> x
(M0) x + f
> y
(M0) y
Tng t cho hm s 3 bin vi M0(x0, y0, z0) v vi x, y , z kh b ta c
f (x0 + x, y0 + y , z0 + z) f (M0) + f
> x
(M0) x + f
> y
(M0) y + f
> z
(M0) z
V d: Dng vi phn tnh gn ng gi tr ca biu thc (1) A = 1, 98 4 + 3, 03 2
(2) B = ln (1, 03 + 3
0, 99 1)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 19 / 42
## o hm v vi phn ca hm nhiu bin
Cng thc tnh gn ng theo vi phn
Gi s f (x, y ) l mt hm s kh vi ti M0(x0, y0), khi vi x v y kh b ta c th xp x
f (x0 + x, y0 + y ) f (M0) + f
> x
(M0) x + f
> y
(M0) y
Tng t cho hm s 3 bin vi M0(x0, y0, z0) v vi x, y , z kh b ta c
f (x0 + x, y0 + y , z0 + z) f (M0) + f
> x
(M0) x + f
> y
(M0) y + f
> z
(M0) z
V d: Dng vi phn tnh gn ng gi tr ca biu thc (1) A = 1, 98 4 + 3, 03 2
(2) B = ln (1, 03 + 3
0, 99 1)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 19 / 42
## o hm v vi phn ca hm nhiu bin
Gii.
(1) Xt hm s f (x, y ) = x4 + y 2. Chn M0(2, 3), khi
x = 0, 02 , y = 0, 03, f (M0) = 24 + 32 = 5.
f
x = 2x3
x4 + y 2 , f
x (M0) = 16 5
f
y = y
x4 + y 2 , f
y (M0) = 35Khi A 5 + 16 5 .(0, 02 ) + 35 .0, 03 = 4, 954 (2) Xt hm s f (x, y ) = ln (x + 3
y 1). Chn M0(1, 1), khi
x = 0, 03 , y = 0, 01, f (M0) = ln (1 + 3
1 1) = 0 (sv t lm tip).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 20 / 42
## o hm v vi phn ca hm nhiu bin
Gii.
(1) Xt hm s f (x, y ) = x4 + y 2. Chn M0(2, 3), khi
x = 0, 02 , y = 0, 03, f (M0) = 24 + 32 = 5.
f
x = 2x3
x4 + y 2 , f
x (M0) = 16 5
f
y = y
x4 + y 2 , f
y (M0) = 35Khi A 5 + 16 5 .(0, 02 ) + 35 .0, 03 = 4, 954 (2) Xt hm s f (x, y ) = ln (x + 3
y 1). Chn M0(1, 1), khi
x = 0, 03 , y = 0, 01, f (M0) = ln (1 + 3
1 1) = 0 (sv t lm tip).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 20 / 42
## o hm v vi phn ca hm nhiu bin
1.2.3. o hm ring ca hm hp
a) Gi s z = f (u), vi u = u(x, y ). Khi
z
x = f (u).u
x
z
y = f (u).u
y .
b) Gi s z = f (x, y ), vi x = x(t), y = y (t). Khi
z(t) = f
x .x(t) + f
y .y (t).
c) Gi s z = f (u, v ), vi u = u(x, y ), v = v (x, y ). Khi
z
x = f
u .u
x + f
v .v
x
z
y = f
u .u
y + f
v .v
y .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 21 / 42
## o hm v vi phn ca hm nhiu bin
1.2.3. o hm ring ca hm hp
a) Gi s z = f (u), vi u = u(x, y ). Khi
z
x = f (u).u
x
z
y = f (u).u
y .
b) Gi s z = f (x, y ), vi x = x(t), y = y (t). Khi
z(t) = f
x .x(t) + f
y .y (t).
c) Gi s z = f (u, v ), vi u = u(x, y ), v = v (x, y ). Khi
z
x = f
u .u
x + f
v .v
x
z
y = f
u .u
y + f
v .v
y .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 21 / 42
## o hm v vi phn ca hm nhiu bin
1.2.3. o hm ring ca hm hp
a) Gi s z = f (u), vi u = u(x, y ). Khi
z
x = f (u).u
x
z
y = f (u).u
y .
b) Gi s z = f (x, y ), vi x = x(t), y = y (t). Khi
z(t) = f
x .x(t) + f
y .y (t).
c) Gi s z = f (u, v ), vi u = u(x, y ), v = v (x, y ). Khi
z
x = f
u .u
x + f
v .v
x
z
y = f
u .u
y + f
v .v
y .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 21 / 42
## o hm v vi phn ca hm nhiu bin
V d 1. Cho z = f (x2 y 2), vi f l hm s kh vi. Hy rt gn biu thc
A = y .z
x + x.z
y .
Gii. t u = x2 y 2. Khi ta c z = f (u), v
z
x = f (u).u
x = f (u).2xz
y = f (u).u
y = f (u).(2y ),
thay vo biu thc ta c
A = y .f (u).2x + x.f (u).(2y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 22 / 42
## o hm v vi phn ca hm nhiu bin
V d 1. Cho z = f (x2 y 2), vi f l hm s kh vi. Hy rt gn biu thc
A = y .z
x + x.z
y .
Gii. t u = x2 y 2. Khi ta c z = f (u), v
z
x = f (u).u
x = f (u).2xz
y = f (u).u
y = f (u).(2y ),
thay vo biu thc ta c
A = y .f (u).2x + x.f (u).(2y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 22 / 42
## o hm v vi phn ca hm nhiu bin
V d 1. Cho z = f (x2 y 2), vi f l hm s kh vi. Hy rt gn biu thc
A = y .z
x + x.z
y .
Gii. t u = x2 y 2. Khi ta c z = f (u), v
z
x = f (u).u
x = f (u).2xz
y = f (u).u
y = f (u).(2y ),
thay vo biu thc ta c
A = y .f (u).2x + x.f (u).(2y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 22 / 42
## o hm v vi phn ca hm nhiu bin
V d 2. Cho z = ln (3x + 2y + 1), vi x = et , y = sint . Hy tnh z
x , z
y , dz dt .
Gii. Ta c:
z
x = (3x + 2y + 1)
x
3x + 2y + 1 = 33x + 2y + 1
z
y = 23x + 2y + 1
dz dt = ( ln (3et + 2sint + 1)) = 3et + 2cost
3et + 2sint + 1 .
V d 3. Cho z = ln (u2 + v 2), vi u = xy , v = ex+y . Hy tnh z
x , z
y .(Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 23 / 42
## o hm v vi phn ca hm nhiu bin
V d 2. Cho z = ln (3x + 2y + 1), vi x = et , y = sint . Hy tnh z
x , z
y , dz dt .
Gii. Ta c:
z
x = (3x + 2y + 1)
x
3x + 2y + 1 = 33x + 2y + 1
z
y = 23x + 2y + 1
dz dt = ( ln (3et + 2sint + 1)) = 3et + 2cost
3et + 2sint + 1 .
V d 3. Cho z = ln (u2 + v 2), vi u = xy , v = ex+y . Hy tnh z
x , z
y .(Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 23 / 42
## o hm v vi phn ca hm nhiu bin
V d 2. Cho z = ln (3x + 2y + 1), vi x = et , y = sint . Hy tnh z
x , z
y , dz dt .
Gii. Ta c:
z
x = (3x + 2y + 1)
x
3x + 2y + 1 = 33x + 2y + 1
z
y = 23x + 2y + 1
dz dt = ( ln (3et + 2sint + 1)) = 3et + 2cost
3et + 2sint + 1 .
V d 3. Cho z = ln (u2 + v 2), vi u = xy , v = ex+y . Hy tnh z
x , z
y .(Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 23 / 42
## o hm v vi phn ca hm nhiu bin
1.2.4. o hm ring ca hm n
a) Gi s y = y (x) l hm s (n) c cho bi phng trnh
F (x, y ) = 0. ()
By gi ta o hm 2 v ca phng trnh () theo bin x ta c
F
x .1 + F
y .y (x) = 0.
Khi
y (x) = F
x
F
y
.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 24 / 42
## o hm v vi phn ca hm nhiu bin
1.2.4. o hm ring ca hm n
a) Gi s y = y (x) l hm s (n) c cho bi phng trnh
F (x, y ) = 0. ()
By gi ta o hm 2 v ca phng trnh () theo bin x ta c
F
x .1 + F
y .y (x) = 0.
Khi
y (x) = F
x
F
y
.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 24 / 42
## o hm v vi phn ca hm nhiu bin
1.2.4. o hm ring ca hm n
a) Gi s y = y (x) l hm s (n) c cho bi phng trnh
F (x, y ) = 0. ()
By gi ta o hm 2 v ca phng trnh () theo bin x ta c
F
x .1 + F
y .y (x) = 0.
Khi
y (x) = F
x
F
y
.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 24 / 42
## o hm v vi phn ca hm nhiu bin
b) Gi s z = z(x, y ) l hm s xc nh bi phng trnh
F (x, y , z) = 0bng cch lm tng t ta thu c
z
> x
= F
> x
F
> z
, z
> y
= F
> y
F
> z
.
V d.
(1) Tnh y (x) bit rng y = y (x) l hm xc nh bi phng trnh
xe y + ye x = 1. T tnh y (0) bit rng y (0) = 1. (2) Tnh dz bit rng z = z(x, y ) xc nh bi phng trnh
x3 + y 3 + z3 = 3xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 25 / 42
## o hm v vi phn ca hm nhiu bin
b) Gi s z = z(x, y ) l hm s xc nh bi phng trnh
F (x, y , z) = 0bng cch lm tng t ta thu c
z
> x
= F
> x
F
> z
, z
> y
= F
> y
F
> z
.
V d.
(1) Tnh y (x) bit rng y = y (x) l hm xc nh bi phng trnh
xe y + ye x = 1. T tnh y (0) bit rng y (0) = 1. (2) Tnh dz bit rng z = z(x, y ) xc nh bi phng trnh
x3 + y 3 + z3 = 3xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 25 / 42
## o hm v vi phn ca hm nhiu bin
b) Gi s z = z(x, y ) l hm s xc nh bi phng trnh
F (x, y , z) = 0bng cch lm tng t ta thu c
z
> x
= F
> x
F
> z
, z
> y
= F
> y
F
> z
.
V d.
(1) Tnh y (x) bit rng y = y (x) l hm xc nh bi phng trnh
xe y + ye x = 1. T tnh y (0) bit rng y (0) = 1. (2) Tnh dz bit rng z = z(x, y ) xc nh bi phng trnh
x3 + y 3 + z3 = 3xyz .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 25 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) Xt hm s F (x, y ) = xe y + ye x 1 = 0. Ta c:
F
x = ey + ye x , F
y = xe y + ex .
Khi
y (x) = F
x
F
y
= ey + ye x
xe y + ex
v
y (0) = e + 11 = (e + 1).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 26 / 42
## o hm v vi phn ca hm nhiu bin
Gii:
(1) Xt hm s F (x, y ) = xe y + ye x 1 = 0. Ta c:
F
x = ey + ye x , F
y = xe y + ex .
Khi
y (x) = F
x
F
y
= ey + ye x
xe y + ex
v
y (0) = e + 11 = (e + 1).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 26 / 42
## o hm v vi phn ca hm nhiu bin
(2) Xt hm s F (x, y , z) = x3 + y 3 + z3 3xyz = 0. Ta c:
F
x = 3x2 3yz , F
y = 3y 2 3xz , F
z = 3z2 3xy .
Khi
z
x = F
x
F
z
= 3x2 3yz
3z2 3xy , z
y = F
y
F
z
= 3y 2 3xz
3z2 3xy dz = z
x dx + z
y dy = 3yz 3x2
3z2 3xy dx + 3xz 3y 2
3z2 3xy dy .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 27 / 42
## o hm v vi phn ca hm nhiu bin
(2) Xt hm s F (x, y , z) = x3 + y 3 + z3 3xyz = 0. Ta c:
F
x = 3x2 3yz , F
y = 3y 2 3xz , F
z = 3z2 3xy .
Khi
z
x = F
x
F
z
= 3x2 3yz
3z2 3xy , z
y = F
y
F
z
= 3y 2 3xz
3z2 3xy dz = z
x dx + z
y dy = 3yz 3x2
3z2 3xy dx + 3xz 3y 2
3z2 3xy dy .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 27 / 42
## o hm v vi phn ca hm nhiu bin
(2) Xt hm s F (x, y , z) = x3 + y 3 + z3 3xyz = 0. Ta c:
F
x = 3x2 3yz , F
y = 3y 2 3xz , F
z = 3z2 3xy .
Khi
z
x = F
x
F
z
= 3x2 3yz
3z2 3xy , z
y = F
y
F
z
= 3y 2 3xz
3z2 3xy dz = z
x dx + z
y dy = 3yz 3x2
3z2 3xy dx + 3xz 3y 2
3z2 3xy dy .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 27 / 42
## o hm ring v vi phn cp cao
1.3. o hm ring v vi phn cp cao 1.3.1. o hm ring cp cao
a) Cc khi nim
Gi s hm s f (x, y ) c cc o hm ring f
> x
, f
> y
, cc o hm ring ca f
> x
v f
> y
(nu c) c gi l cc o hm ring cp 2 ca hm s f (x, y ), k hiu v c tnh tng ng nh sau:
f
> xx
= ( f
> x
)
> x
, f
> xy
= ( f
> x
)
> y
f
> yx
= ( f
> y
)
> x
, f
> yy
= ( f
> y
)
> y
b) Nhn xt (nh l Schwaz) . Nu hm s f (x, y ) c cc o hm ring cp hai
f
> xy
v f
> yx
lin tc ti M0(x0, y0) th
f
> xy
(M0) = f
> yx
(M0).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 28 / 42
## o hm ring v vi phn cp cao
1.3. o hm ring v vi phn cp cao 1.3.1. o hm ring cp cao
a) Cc khi nim
Gi s hm s f (x, y ) c cc o hm ring f
> x
, f
> y
, cc o hm ring ca f
> x
v f
> y
(nu c) c gi l cc o hm ring cp 2 ca hm s f (x, y ), k hiu v c tnh tng ng nh sau:
f
> xx
= ( f
> x
)
> x
, f
> xy
= ( f
> x
)
> y
f
> yx
= ( f
> y
)
> x
, f
> yy
= ( f
> y
)
> y
b) Nhn xt (nh l Schwaz) . Nu hm s f (x, y ) c cc o hm ring cp hai
f
> xy
v f
> yx
lin tc ti M0(x0, y0) th
f
> xy
(M0) = f
> yx
(M0).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 28 / 42
## o hm ring v vi phn cp cao
1.3. o hm ring v vi phn cp cao 1.3.1. o hm ring cp cao
a) Cc khi nim
Gi s hm s f (x, y ) c cc o hm ring f
> x
, f
> y
, cc o hm ring ca f
> x
v f
> y
(nu c) c gi l cc o hm ring cp 2 ca hm s f (x, y ), k hiu v c tnh tng ng nh sau:
f
> xx
= ( f
> x
)
> x
, f
> xy
= ( f
> x
)
> y
f
> yx
= ( f
> y
)
> x
, f
> yy
= ( f
> y
)
> y
b) Nhn xt (nh l Schwaz) . Nu hm s f (x, y ) c cc o hm ring cp hai
f
> xy
v f
> yx
lin tc ti M0(x0, y0) th
f
> xy
(M0) = f
> yx
(M0).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 28 / 42
## o hm ring v vi phn cp cao
T nhn xt trn ta thy, vic tnh o hm ring cp cao ca hm s nhiu bin khng ph thuc vo th t ly o hm ring ca hm s theo bin s, iu ny gip chng ta gim bt rt nhiu trong vic tnh o hm ring cp cao ca hm s c nhiu bin s.
V d.
(1) Tnh cc o hm ring cp 2 ca hm s sau
f (x, y ) = xsin (x2 + 3y ) + ln (x + 2y )
(2) Cho hm s
u = x2 + y 2 + z2
Chng minh rng u
> xx
+ u
> yy
+ u
> zz
= 2
u .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 29 / 42
## o hm ring v vi phn cp cao
T nhn xt trn ta thy, vic tnh o hm ring cp cao ca hm s nhiu bin khng ph thuc vo th t ly o hm ring ca hm s theo bin s, iu ny gip chng ta gim bt rt nhiu trong vic tnh o hm ring cp cao ca hm s c nhiu bin s.
V d.
(1) Tnh cc o hm ring cp 2 ca hm s sau
f (x, y ) = xsin (x2 + 3y ) + ln (x + 2y )
(2) Cho hm s
u = x2 + y 2 + z2
Chng minh rng u
> xx
+ u
> yy
+ u
> zz
= 2
u .
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 29 / 42
## o hm ring v vi phn cp cao
Gii:
(1) Ta c f (x, y ) = xsin (x2 + 3y ) + ln (x + 2y ). Vy
f
x = sin (x2 + 3y ) + 2x2cos (x2 + 3y ) + 1
x + 2y , f
y = 3xcos (x2 + 3y ) + 2
x + 2y
v
f
xx = 6xcos (x2 + 3y ) 4x3sin (x2 + 3y ) 1
(x + 2y )2
f
xy = f
yx = 3cos (x2 + 3y ) 6x2sin (x2 + 3y ) 2
(x + 2y )2
f
yy = 9xsin (x2 + 3y ) 4
(x + 2y )2 .
(2) (Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 30 / 42
## o hm ring v vi phn cp cao
Gii:
(1) Ta c f (x, y ) = xsin (x2 + 3y ) + ln (x + 2y ). Vy
f
x = sin (x2 + 3y ) + 2x2cos (x2 + 3y ) + 1
x + 2y , f
y = 3xcos (x2 + 3y ) + 2
x + 2y
v
f
xx = 6xcos (x2 + 3y ) 4x3sin (x2 + 3y ) 1
(x + 2y )2
f
xy = f
yx = 3cos (x2 + 3y ) 6x2sin (x2 + 3y ) 2
(x + 2y )2
f
yy = 9xsin (x2 + 3y ) 4
(x + 2y )2 .
(2) (Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 30 / 42
## o hm ring v vi phn cp cao
Gii:
(1) Ta c f (x, y ) = xsin (x2 + 3y ) + ln (x + 2y ). Vy
f
x = sin (x2 + 3y ) + 2x2cos (x2 + 3y ) + 1
x + 2y , f
y = 3xcos (x2 + 3y ) + 2
x + 2y
v
f
xx = 6xcos (x2 + 3y ) 4x3sin (x2 + 3y ) 1
(x + 2y )2
f
xy = f
yx = 3cos (x2 + 3y ) 6x2sin (x2 + 3y ) 2
(x + 2y )2
f
yy = 9xsin (x2 + 3y ) 4
(x + 2y )2 .
(2) (Sinh vin t gii)
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 30 / 42
## o hm ring v vi phn cp cao
1.3.2. Vi phn cp cao
i vi hm s hai bin f (x, y ) ta c th biu din cng thc vi phn cp cao theo quy np nh sau:
df = f
x dx + f
y dy = f
x dx + f
y dy = ( x dx + y dy )1 fd2f = f
xx dx 2 + 2f
xy dxdy + f
yy dy 2 = ( x dx + y dy )2 f
dnf = ( x dx + y dy )n f
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 31 / 42
## o hm ring v vi phn cp cao
1.3.2. Vi phn cp cao
i vi hm s hai bin f (x, y ) ta c th biu din cng thc vi phn cp cao theo quy np nh sau:
df = f
x dx + f
y dy = f
x dx + f
y dy = ( x dx + y dy )1 fd2f = f
xx dx 2 + 2f
xy dxdy + f
yy dy 2 = ( x dx + y dy )2 f
dnf = ( x dx + y dy )n f
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 31 / 42
## o hm ring v vi phn cp cao
1.3.2. Vi phn cp cao
i vi hm s hai bin f (x, y ) ta c th biu din cng thc vi phn cp cao theo quy np nh sau:
df = f
x dx + f
y dy = f
x dx + f
y dy = ( x dx + y dy )1 fd2f = f
xx dx 2 + 2f
xy dxdy + f
yy dy 2 = ( x dx + y dy )2 f
dnf = ( x dx + y dy )n f
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 31 / 42
## o hm ring v vi phn cp cao
1.3.2. Vi phn cp cao
i vi hm s hai bin f (x, y ) ta c th biu din cng thc vi phn cp cao theo quy np nh sau:
df = f
x dx + f
y dy = f
x dx + f
y dy = ( x dx + y dy )1 fd2f = f
xx dx 2 + 2f
xy dxdy + f
yy dy 2 = ( x dx + y dy )2 f
dnf = ( x dx + y dy )n f
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 31 / 42
## o hm ring v vi phn cp cao
V d. Tm d2z bit
z = sin (x + 2y )
Gii: Ta c
z
x = cos (x + 2y ), z
y = 2cos (x + 2y )
z
xx = sin (x + 2y ), z
xy = 2sin (x + 2y ), z
yy = 4sin (x + 2y ).
Do
d2z = z
xx dx 2 + 2z
xy dxdy + z
yy dy 2
= sin (x + 2y )dx 2 4sin (x + 2y )dxdy 4sin (x + 2y )dy 2
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 32 / 42
1.4. Cc tr ca hm nhiu bin s 1.4.1. Cc tr khng rng buc (Cc tr t do)
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0.Ta ni f (x, y ) t cc i ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ta ni f (x, y ) t cc tiu ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ti nhng im M0 m hm s f (x, y ) t cc i hay cc tiu ti c gi chung l nhng im cc tr ca hm s. - Nhng im M0 m ti cc o hm ring ca f (x, y ) trit tiu, tc
{
f
> x
(M0) = 0
f
> y
(M0) = 0hoc t nht 1 trong cc o hm ring ca hm s khng tn ti c gi l im dng (im ti hn) ca hm s.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 33 / 42
1.4. Cc tr ca hm nhiu bin s 1.4.1. Cc tr khng rng buc (Cc tr t do)
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0.Ta ni f (x, y ) t cc i ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ta ni f (x, y ) t cc tiu ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ti nhng im M0 m hm s f (x, y ) t cc i hay cc tiu ti c gi chung l nhng im cc tr ca hm s. - Nhng im M0 m ti cc o hm ring ca f (x, y ) trit tiu, tc
{
f
> x
(M0) = 0
f
> y
(M0) = 0hoc t nht 1 trong cc o hm ring ca hm s khng tn ti c gi l im dng (im ti hn) ca hm s.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 33 / 42
1.4. Cc tr ca hm nhiu bin s 1.4.1. Cc tr khng rng buc (Cc tr t do)
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0.Ta ni f (x, y ) t cc i ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ta ni f (x, y ) t cc tiu ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ti nhng im M0 m hm s f (x, y ) t cc i hay cc tiu ti c gi chung l nhng im cc tr ca hm s. - Nhng im M0 m ti cc o hm ring ca f (x, y ) trit tiu, tc
{
f
> x
(M0) = 0
f
> y
(M0) = 0hoc t nht 1 trong cc o hm ring ca hm s khng tn ti c gi l im dng (im ti hn) ca hm s.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 33 / 42
1.4. Cc tr ca hm nhiu bin s 1.4.1. Cc tr khng rng buc (Cc tr t do)
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0.Ta ni f (x, y ) t cc i ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ta ni f (x, y ) t cc tiu ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ti nhng im M0 m hm s f (x, y ) t cc i hay cc tiu ti c gi chung l nhng im cc tr ca hm s. - Nhng im M0 m ti cc o hm ring ca f (x, y ) trit tiu, tc
{
f
> x
(M0) = 0
f
> y
(M0) = 0hoc t nht 1 trong cc o hm ring ca hm s khng tn ti c gi l im dng (im ti hn) ca hm s.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 33 / 42
1.4. Cc tr ca hm nhiu bin s 1.4.1. Cc tr khng rng buc (Cc tr t do)
a) Cc khi nim
- Cho hm s f (x, y ) xc nh ti im M0(x0, y0) v ln cn ca M0.Ta ni f (x, y ) t cc i ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ta ni f (x, y ) t cc tiu ti M0 nu tn ti mt ln cn U ca M0 sao cho
f (x, y ) f (x0, y0), (x, y ) U.Ti nhng im M0 m hm s f (x, y ) t cc i hay cc tiu ti c gi chung l nhng im cc tr ca hm s. - Nhng im M0 m ti cc o hm ring ca f (x, y ) trit tiu, tc
{
f
> x
(M0) = 0
f
> y
(M0) = 0hoc t nht 1 trong cc o hm ring ca hm s khng tn ti c gi l im dng (im ti hn) ca hm s.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 33 / 42
## Cc tr ca hm s nhiu bin s
Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 34 / 42 Cc tr ca hm s nhiu bin s
b) iu kin cn hm s c cc tr nh l : Nu hm s f (x, y ) t cc tr ti im M0(x0, y0) th im M0 phi l im dng ca hm s.
Nhn xt :- iu ngc li ca nh l khng ng - Cc im cc tr ca hm s ch c th nm trong cc im dng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 35 / 42
## Cc tr ca hm s nhiu bin s
b) iu kin cn hm s c cc tr nh l : Nu hm s f (x, y ) t cc tr ti im M0(x0, y0) th im M0 phi l im dng ca hm s.
Nhn xt :- iu ngc li ca nh l khng ng - Cc im cc tr ca hm s ch c th nm trong cc im dng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 35 / 42
## Cc tr ca hm s nhiu bin s
b) iu kin cn hm s c cc tr nh l : Nu hm s f (x, y ) t cc tr ti im M0(x0, y0) th im M0 phi l im dng ca hm s.
Nhn xt :- iu ngc li ca nh l khng ng - Cc im cc tr ca hm s ch c th nm trong cc im dng.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 35 / 42
## Cc tr ca hm s nhiu bin s
c) iu kin hm s c cc tr
Gi s ti im M0(x0, y0) ta c
{
f
> x
(M0) = 0
f
> y
(M0) = 0.
t A = f
> xx
(M0), B = f
> xy
(M0), C = f
> yy
(M0). Khi : - Nu
{
B2 AC < 0
A > 0 th im M0 l im t cc tiu ca hm s. - Nu
{
B2 AC < 0
A < 0 th im M0 l im t cc i ca hm s. - Nu B2 AC > 0 th M0 khng l im t cc tr ca hm s. - Nu B2 AC = 0 trong trng hp ny ta cha th kt lun, cn phi xt thm.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 36 / 42
## Cc tr ca hm s nhiu bin s
c) iu kin hm s c cc tr
Gi s ti im M0(x0, y0) ta c
{
f
> x
(M0) = 0
f
> y
(M0) = 0.
t A = f
> xx
(M0), B = f
> xy
(M0), C = f
> yy
(M0). Khi : - Nu
{
B2 AC < 0
A > 0 th im M0 l im t cc tiu ca hm s. - Nu
{
B2 AC < 0
A < 0 th im M0 l im t cc i ca hm s. - Nu B2 AC > 0 th M0 khng l im t cc tr ca hm s. - Nu B2 AC = 0 trong trng hp ny ta cha th kt lun, cn phi xt thm.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 36 / 42
## Cc tr ca hm s nhiu bin s
c) iu kin hm s c cc tr
Gi s ti im M0(x0, y0) ta c
{
f
> x
(M0) = 0
f
> y
(M0) = 0.
t A = f
> xx
(M0), B = f
> xy
(M0), C = f
> yy
(M0). Khi : - Nu
{
B2 AC < 0
A > 0 th im M0 l im t cc tiu ca hm s. - Nu
{
B2 AC < 0
A < 0 th im M0 l im t cc i ca hm s. - Nu B2 AC > 0 th M0 khng l im t cc tr ca hm s. - Nu B2 AC = 0 trong trng hp ny ta cha th kt lun, cn phi xt thm.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 36 / 42
## Cc tr ca hm s nhiu bin s
c) iu kin hm s c cc tr
Gi s ti im M0(x0, y0) ta c
{
f
> x
(M0) = 0
f
> y
(M0) = 0.
t A = f
> xx
(M0), B = f
> xy
(M0), C = f
> yy
(M0). Khi : - Nu
{
B2 AC < 0
A > 0 th im M0 l im t cc tiu ca hm s. - Nu
{
B2 AC < 0
A < 0 th im M0 l im t cc i ca hm s. - Nu B2 AC > 0 th M0 khng l im t cc tr ca hm s. - Nu B2 AC = 0 trong trng hp ny ta cha th kt lun, cn phi xt thm.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 36 / 42
## Cc tr ca hm s nhiu bin
V d 1. Tm cc tr ca hm s f (x, y ) = x3 + y 3 3xy + 2020.
Gii: Ta c
f
x = 3x2 3y , f
y = 3y 2 3x,
f
xx = 6x, f
xy = 3, f
yy = 6y .
Gii h
{
f
x = 0
f
y = 0
{
3x2 3y = 03y 2 3x = 0
{
x = 0
y = 0
{
x = 1
y = 1 .
Vy hm s c 2 im dng l: M1(0, 0), M2(1, 1).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 37 / 42
## Cc tr ca hm s nhiu bin
V d 1. Tm cc tr ca hm s f (x, y ) = x3 + y 3 3xy + 2020.
Gii: Ta c
f
x = 3x2 3y , f
y = 3y 2 3x,
f
xx = 6x, f
xy = 3, f
yy = 6y .
Gii h
{
f
x = 0
f
y = 0
{
3x2 3y = 03y 2 3x = 0
{
x = 0
y = 0
{
x = 1
y = 1 .
Vy hm s c 2 im dng l: M1(0, 0), M2(1, 1).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 37 / 42
## Cc tr ca hm s nhiu bin
V d 1. Tm cc tr ca hm s f (x, y ) = x3 + y 3 3xy + 2020.
Gii: Ta c
f
x = 3x2 3y , f
y = 3y 2 3x,
f
xx = 6x, f
xy = 3, f
yy = 6y .
Gii h
{
f
x = 0
f
y = 0
{
3x2 3y = 03y 2 3x = 0
{
x = 0
y = 0
{
x = 1
y = 1 .
Vy hm s c 2 im dng l: M1(0, 0), M2(1, 1).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 37 / 42
## Cc tr ca hm s nhiu bin
- Ti M1(0, 0), ta c A = 0, B = 3, C = 0.
B2 AC = 9 > 0, vy M1 khng l im t cc tr. - Ti M2(1, 1), ta c A = 6, B = 3, C = 6.
B2 AC = 27 < 0, A = 6 > 0, vy M2 l im t cc tiu ca hm s, v
fCT = f (1, 1) = 2019.
V d 2. Tm cc tr ca hm s
f (x, y ) = x2 + 4y 2 2 ln( xy )
Gii:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 38 / 42
## Cc tr ca hm s nhiu bin
- Ti M1(0, 0), ta c A = 0, B = 3, C = 0.
B2 AC = 9 > 0, vy M1 khng l im t cc tr. - Ti M2(1, 1), ta c A = 6, B = 3, C = 6.
B2 AC = 27 < 0, A = 6 > 0, vy M2 l im t cc tiu ca hm s, v
fCT = f (1, 1) = 2019.
V d 2. Tm cc tr ca hm s
f (x, y ) = x2 + 4y 2 2 ln( xy )
Gii:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 38 / 42
## Cc tr ca hm s nhiu bin
- Ti M1(0, 0), ta c A = 0, B = 3, C = 0.
B2 AC = 9 > 0, vy M1 khng l im t cc tr. - Ti M2(1, 1), ta c A = 6, B = 3, C = 6.
B2 AC = 27 < 0, A = 6 > 0, vy M2 l im t cc tiu ca hm s, v
fCT = f (1, 1) = 2019.
V d 2. Tm cc tr ca hm s
f (x, y ) = x2 + 4y 2 2 ln( xy )
Gii:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 38 / 42
## Cc tr ca hm s nhiu bin
- Ti M1(0, 0), ta c A = 0, B = 3, C = 0.
B2 AC = 9 > 0, vy M1 khng l im t cc tr. - Ti M2(1, 1), ta c A = 6, B = 3, C = 6.
B2 AC = 27 < 0, A = 6 > 0, vy M2 l im t cc tiu ca hm s, v
fCT = f (1, 1) = 2019.
V d 2. Tm cc tr ca hm s
f (x, y ) = x2 + 4y 2 2 ln( xy )
Gii:
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 38 / 42
## Cc tr ca hm s nhiu bin s
1.4.2. Cc tr c rng buc Bi ton: Tm cc tr ca hm s f (x, y ) trong x, y tha mn iu kin:
(x, y ) = 0.
Phng php tha s Lagrange
- Lp hm b tr
F (x, y , ) = f (x, y ) + (x, y )
trong l mt hng s s xc nh sau. - Xt h
F
> x
(x, y , ) = 0
F
> y
(x, y , ) = 0
F
>
= (x, y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 39 / 42
## Cc tr ca hm s nhiu bin s
1.4.2. Cc tr c rng buc Bi ton: Tm cc tr ca hm s f (x, y ) trong x, y tha mn iu kin:
(x, y ) = 0.
Phng php tha s Lagrange
- Lp hm b tr
F (x, y , ) = f (x, y ) + (x, y )
trong l mt hng s s xc nh sau. - Xt h
F
> x
(x, y , ) = 0
F
> y
(x, y , ) = 0
F
>
= (x, y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 39 / 42
## Cc tr ca hm s nhiu bin s
1.4.2. Cc tr c rng buc Bi ton: Tm cc tr ca hm s f (x, y ) trong x, y tha mn iu kin:
(x, y ) = 0.
Phng php tha s Lagrange
- Lp hm b tr
F (x, y , ) = f (x, y ) + (x, y )
trong l mt hng s s xc nh sau. - Xt h
F
> x
(x, y , ) = 0
F
> y
(x, y , ) = 0
F
>
= (x, y ) = 0.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 39 / 42
## Cc tr ca hm s nhiu bin
Gi s h c nghim M0(x0, y0), ng vi 0, ta xt
d2F (M0) = F
xx (M0)dx 2 + 2F
xy (M0)dxdy + F
yy (M0)dy 2
Khi : - Nu d2F (M0) > 0 th M0 l im t cc tiu ca f (x, y ).- Nu d2F (M0) < 0 th M0 l im t cc i ca f (x, y ).- Nu d2F (M0) = 0 th ta phi dng phng php khc.
V d. Tm cc tr ca hm s z = x + 2y vi iu kin x2 + y 2 = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 40 / 42
## Cc tr ca hm s nhiu bin
Gi s h c nghim M0(x0, y0), ng vi 0, ta xt
d2F (M0) = F
xx (M0)dx 2 + 2F
xy (M0)dxdy + F
yy (M0)dy 2
Khi : - Nu d2F (M0) > 0 th M0 l im t cc tiu ca f (x, y ).- Nu d2F (M0) < 0 th M0 l im t cc i ca f (x, y ).- Nu d2F (M0) = 0 th ta phi dng phng php khc.
V d. Tm cc tr ca hm s z = x + 2y vi iu kin x2 + y 2 = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 40 / 42
## Cc tr ca hm s nhiu bin
Gi s h c nghim M0(x0, y0), ng vi 0, ta xt
d2F (M0) = F
xx (M0)dx 2 + 2F
xy (M0)dxdy + F
yy (M0)dy 2
Khi : - Nu d2F (M0) > 0 th M0 l im t cc tiu ca f (x, y ).- Nu d2F (M0) < 0 th M0 l im t cc i ca f (x, y ).- Nu d2F (M0) = 0 th ta phi dng phng php khc.
V d. Tm cc tr ca hm s z = x + 2y vi iu kin x2 + y 2 = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 40 / 42
## Cc tr ca hm s nhiu bin
Gi s h c nghim M0(x0, y0), ng vi 0, ta xt
d2F (M0) = F
xx (M0)dx 2 + 2F
xy (M0)dxdy + F
yy (M0)dy 2
Khi : - Nu d2F (M0) > 0 th M0 l im t cc tiu ca f (x, y ).- Nu d2F (M0) < 0 th M0 l im t cc i ca f (x, y ).- Nu d2F (M0) = 0 th ta phi dng phng php khc.
V d. Tm cc tr ca hm s z = x + 2y vi iu kin x2 + y 2 = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 40 / 42
## Cc tr ca hm s nhiu bin
Gi s h c nghim M0(x0, y0), ng vi 0, ta xt
d2F (M0) = F
xx (M0)dx 2 + 2F
xy (M0)dxdy + F
yy (M0)dy 2
Khi : - Nu d2F (M0) > 0 th M0 l im t cc tiu ca f (x, y ).- Nu d2F (M0) < 0 th M0 l im t cc i ca f (x, y ).- Nu d2F (M0) = 0 th ta phi dng phng php khc.
V d. Tm cc tr ca hm s z = x + 2y vi iu kin x2 + y 2 = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 40 / 42
## Cc tr ca hm nhiu bin s
Gii: Xt hm s F (x, y , ) = x + 2y + (x2 + y 2 5).
F
x = 1 + 2x, F
y = 2 + 2y , F
xx = 2, F
xy = 0, F
yy = 2.
F
x (x, y , ) = 0
F
y (x, y , ) = 0
F
= 0
1 + 2x = 02 + 2y = 0
x2 + y 2 = 5.
Gii h trn ta c 2 im dng: M1(1, 2) (ng vi = 12 ), v M2(1, 2) (ng vi = 12 ).
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 41 / 42
## Cc tr ca hm s nhiu bin s
Ta thy
d2F = 2(dx 2 + dy 2).
- Ti M1, d2F (M1) = ( dx 2 + dy 2) > 0, vy M1 l im t cc tiu v
zCT = z(1, 2) = 5. - Ti M2, d2F (M2) = (dx 2 + dy 2) < 0, vy M2 l im t cc i v
zC = z(1, 2) = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 42 / 42
## Cc tr ca hm s nhiu bin s
Ta thy
d2F = 2(dx 2 + dy 2).
- Ti M1, d2F (M1) = ( dx 2 + dy 2) > 0, vy M1 l im t cc tiu v
zCT = z(1, 2) = 5. - Ti M2, d2F (M2) = (dx 2 + dy 2) < 0, vy M2 l im t cc i v
zC = z(1, 2) = 5.
> Gii tch 2 cho khi k thut Ngy 21 thng 3 nm 2022 42 / 42
## Cc tr ca hm s nhiu bin s
Ta thy
d2F = 2(dx 2 + dy 2).
- Ti M1, d2F (M1) = ( dx 2 + dy 2) > 0, vy M1 l im t cc tiu v
zCT = z(1, 2) = 5. - Ti M2, d2F (M2) = (dx 2 + dy 2) < 0, vy M2 l im t cc i v
zC = z(1, 2) = 5.