Bài tập toán cao cấp part 9 pdf

9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 127
hay l`a
f(x +∆x, y +∆y) ≈ f(x, y)+
∂f
∂x
(M)∆x +
∂f
∂y
(M)∆y
(9.8)
Cˆong th´u
.
c (9.8) l`a co
.
so
.
’
d
ˆe
’
´ap du
.
ng vi phˆan t´ınh gˆa
`
nd´ung. Dˆo

´
i
v´o
.
i h`am c´o sˆo
´
biˆe
´
n nhiˆe
`
uho
.
n2tac˜ung c´o cˆong th´u
.
ctu
.
o
.
ng tu
.
.
.
9.2.3 C´ac t´ınh chˆa
´
tcu
’
a vi phˆan
Dˆo
´
iv´o

.
i c´ac h`am kha
’
vi f v`a g ta c´o:
(i) d(f ± g)=df ±dg;
(ii) d(fg)=fdg + gdf, d(αf)=αdf, α ∈ R;
(iii) d

f
g

=
gdf − fdg
g
2
, g =0;
(iv) Vi phˆan cˆa
´
p1cu
’
a h`am hai biˆe
´
n f(x, y)bˆa
´
tbiˆe
´
nvˆe
`
da
.

ng bˆa
´
t
luˆa
.
n x v`a y l`a biˆe
´
ndˆo
.
clˆa
.
p hay l`a h`am cu
’
a c´ac biˆe
´
ndˆo
.
clˆa
.
p kh´ac.
9.2.4 Vi phˆan cˆa
´
p cao
Gia
’
su
.
’
h`am w = f(x, y) kha
’

vi trong miˆe
`
n D. Khi d
´o vi phˆan cˆa
´
p1
cu
’
a n´o ta
.
idiˆe
’
m(x, y) ∈ D tu
.
o
.
ng ´u
.
ng v´o
.
i c´ac sˆo
´
gia dx v`a dy cu
’
a c´ac
biˆe
´
ndˆo
.
clˆa

.
pdu
.
o
.
.
cbiˆe
’
udiˆe
˜
nbo
.
’
i cˆong th´u
.
c
df =
∂f
∂x
dx +
∂f
∂y
dy. (9.9)
O
.
’
d
ˆa y , dx =∆x, dy =∆y l`a nh ˜u
.
ng sˆo

´
gia t`uy ´y cu
’
abiˆe
´
ndˆo
.
clˆa
.
p, d´o
l`a nh˜u
.
ng sˆo
´
khˆong phu
.
thuˆo
.
c v`ao x v`a y.Nhu
.
vˆa
.
y, khi cˆo
´
di
.
nh dx v`a
dy vi phˆan df l`a h`am cu
’
a x v`a y.

Theo di
.
nh ngh˜ıa: Vi phˆan th ´u
.
hai d
2
f (hay vi phˆan cˆa
´
p 2) cu
’
a
h`am f(x, y)ta
.
idiˆe
’
m M(x, y)du
.
o
.
.
cdi
.
nh ngh˜ıa nhu
.
l`a vi phˆan cu
’
avi
phˆan th ´u
.
nhˆa

´
tta
.
idiˆe
’
m M v´o
.
i c´ac diˆe
`
ukiˆe
.
n sau dˆay:
(1) Vi phˆan df l`a h`am chı
’
cu
’
a c´ac biˆe
´
ndˆo
.
clˆa
.
p x v`a y.
128 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe

´
n
(2) Sˆo
´
gia cu
’
a c´ac biˆe
´
ndˆo
.
clˆa
.
p x v`a y xuˆa
´
thiˆe
.
n khi t´ınh vi phˆan
cu
’
a f

x
v`a f

y
du
.
o
.
.

c xem l`a b˘a
`
ng sˆo
´
gia d
ˆa
`
u tiˆen, t ´u
.
cl`ab˘a
`
ng dx v`a dy.
T`u
.
d´o
d
2
f(M)=
∂
2
f(M)
∂x
2
dx
2
+2
∂
2
f
∂x∂y

(M)dxdy +
∂
2
f
∂y
2
(M)dy
2
(9.10)
trong d´o dx
2
=(dx)
2
, dy
2
=(dy)
2
v`a ta xem c´ac da
.
o h`am riˆeng hˆo
˜
n
ho
.
.
pb˘a
`
ng nhau.
Mˆo
.

t c´ach h`ınh th´u
.
c d
˘a
’
ng th ´u
.
c (9.10) c´o thˆe
’
viˆe
´
tdu
.
´o
.
ida
.
ng
d
2
f =

∂
∂x
dx +
∂
∂y
dy

2

f(x, y)
t´u
.
c l`a sau khi thu
.
.
chiˆe
.
n ph´ep “b`ınh phu
.
o
.
ng” ta cˆa
`
ndiˆe
`
n f(x, y) v`ao
“ˆo trˆo
´
ng”.
Tu
.
o
.
ng tu
.
.
d
3
f =


∂
∂x
dx +
∂
∂y
dy

3
f(x, y)
=
∂
3
f
∂x
3
dx
3
+3
∂
3
f
∂x
2
∂y
dx
2
dy +3
∂
3

f
∂x∂y
2
dxdy
2
+
∂
3
f
∂y
3
dy
3
,
v.v Mˆo
.
t c´ach quy na
.
p ta c´o
d
n
f(x, y)=
n

k=0
C
k
n
∂
n

f
∂x
n−k
∂y
k
dx
n−k
dy
k
. (9.11)
Trong tru
.
`o
.
ng ho
.
.
pnˆe
´
u
w = f(t, v),t= ϕ(x, y),v= ψ(x, y)
th`ı
dw =
∂f
∂t
dt +
∂f
∂v
dx (t´ınh bˆa
´

tbiˆe
´
nvˆe
`
da
.
ng !)
d
2
w =
∂
2
f
∂t
2
dt
2
+2
∂
2
f
∂t∂v
dtdy +
∂
2
f
∂v
2
dv
2

+
∂f
∂t
d
2
t +
∂f
∂v
d
2
v. (9.12)
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 129
9.2.5 Cˆong th´u
.
c Taylor
Nˆe
´
u h`am f(x, y)l`an +1 lˆa
`
n kha
’
vi trong ε-lˆan cˆa
.
n V cu

’
adiˆe
’
m
M
0
(x
0
,y
0
)th`ıdˆo
´
iv´o
.
idiˆe
’
mbˆa
´
tk`yM(x, y) ∈Vta c´o cˆong th´u
.
c Taylor
f(x, y)=f(x
0
,y
0
)+
1
1!

f


x
(x
0
,y
0
)(x − x
0
)+f

y
(x
0
,y
0
)(y −y
0
)

+
1
2!

f

xx
(x
0
,y
0

)(x − x
0
)
2
+2f

xy
(x
0
,y
0
)(x −x
0
)(y − y
0
)
+ f

yy
(x
0
,y
0
)(y −y
0
)

+ ···+
1
n!

m

i=0
C
i
n
∂
n
f(x
0
,y
0
)
∂x
n−i
∂y
i
(x −x
0
)
n−i
(y − y
0
)
i
+
1
(n + 1)!
n


i=0
∂
n+1
f(ξ,η)
∂x
n−i
∂y
i
(x − x
0
)
n−i
(y −y
0
), (9.13)
trong d
´o ξ = x
0
+ θ(x − x
0
), η = y
0
+ θ(y −y
0
), 0 <θ<1.
hay l`a
f(x, y)=f(x
0
,y
0

)+
1
1!
df (x
0
,y
0
)+
1
2!
d
2
f(x
0
,y
0
)+
+
1
n!
d
n
f(x
0
,y
0
)+R
n+1
,
= P

n
(x, y)+R
n+1
(9.14)
trong d´o P
n
(x, y)go
.
il`adath´u
.
c Taylor bˆa
.
c n cu
’
a hai biˆe
´
n x v`a y,
R
n+1
l`a sˆo
´
ha
.
ng du
.
.Nˆe
´
ud
˘a
.

t
ρ =

∆x
2
+∆y
2
th`ı (9.14) c´o thˆe
’
viˆe
´
tdu
.
´o
.
ida
.
ng
f(x, y)=P
n
(x, y)+0(ρ),ρ→ 0,
o
.
’
dˆay R
n+1
= o(ρ) l`a phˆa
`
ndu
.

da
.
ng Peano.
130 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
9.2.6 Vi phˆan cu
’
a h`am ˆa
’
n
Theo di
.
nh ngh˜ıa: biˆe
´
n w du
.
o
.
.
cgo
.
i l`a h`am ˆa
’

ncu
’
a c´ac biˆe
´
nd
ˆo
.
clˆa
.
p
x, y, , t nˆe
´
un´od
u
.
o
.
.
cchobo
.
’
iphu
.
o
.
ng tr`ınh
F (x,y, ,w)=0
khˆong gia
’
id

u
.
o
.
.
cd
ˆo
´
iv´o
.
i w.
D
ˆe
’
t´ınh vi phˆan cu
’
a h`am ˆa
’
n w ta lˆa
´
y vi phˆan ca
’
hai vˆe
´
cu
’
aphu
.
o
.

ng
tr`ınh (xem nhu
.
dˆo
`
ng nhˆa
´
tth´u
.
c) rˆo
`
it`u
.
d´ot`ımdw.Dˆe
’
t´ınh d
2
w ta cˆa
`
n
lˆa
´
y vi phˆan cu
’
a dw v´o
.
ilu
.
u´yr˘a
`

ng dx v`a dy l`a h˘a
`
ng sˆo
´
, c`on dw l`a vi
phˆan cu
’
a h`am.
Ta c˜ung c´o thˆe
’
thu d
u
.
o
.
.
c vi phˆan dw b˘a
`
ng c´ach t´ınh c´ac da
.
o h`am
riˆeng:
w

x
= −
F

x
(·)

F

w
(·)
,w

y
= −
F

y
(·)
F

w
(·)
,
rˆo
`
ithˆe
´
v`ao biˆe
’
uth´u
.
c
dw =
∂w
∂x
dx +

∂w
∂y
dy + ···+
∂w
∂t
dt, v.v
C
´
AC V
´
IDU
.
V´ı du
.
1. T´ınh vi phˆan df nˆe
´
u
1) f(x, y)=xy
2
,2)f(x, y)=

x
2
+ y
2
.
Gia
’
i. 1) Ta c´o
f


x
=

xy
2


x
= y
2
,f

y
=

xy
2
)

y
=2xy.
Do d´o
df (x, y)=y
2
dx +2xydy.
2) Ta t´ınh c´ac da
.
o h`am riˆeng:
f


x
=
x

x
2
+ y
2
,f

y
=
y

x
2
+ y
2
·
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 131
Do d´o
df =
x


x
2
+ y
2
dx +
y

x
2
+ y
2
dy =
xdx + ydy

x
2
+ y
2
· 
V´ı du
.
2. T´ınh df (M
0
)nˆe
´
u f(x, y, z)=e
x
2
+y

2
+z
2
v`a M
0
= M
0
(0, 1, 2).
Gia
’
i. Ta c´o
df (M)=
∂f
∂x
(M)dx +
∂f
∂y
(M)dy +
∂f
∂z
(M)dz, M = M(x, y, z).
Ta t´ınh c´ac da
.
o h`am riˆeng
∂f
∂x
=2xe
x
2
+y

2
+z
2
⇒
∂f
∂x
(M
0
)=0, (v`ı x =0)
∂f
∂y
=2ye
x
2
+y
2
+z
2
⇒
∂f
∂y
(M
0
)=2e
5
,
∂f
∂z
=2ze
x

2
+y
2
+z
2
⇒
∂f
∂z
(M
0
)=4e
5
.
T`u
.
d
´o
df (M
0
)=2e
5
dy +4e
5
dz. 
V´ı d u
.
3. T´ınh dw ta
.
idiˆe
’

m M
0
(−1, 1) nˆe
´
u
w = f(x + y
2
,y+ x
2
).
Gia
’
i. C´ach 1. T´ınh c´ac da
.
o h`am riˆeng cu
’
a h`am f(x, y) theo x v`a
theo y rˆo
`
i´apdu
.
ng cˆong th´u
.
c (9.9). T`u
.
v´ıdu
.
4, mu
.
c 9.1 ta c´o

∂f
∂x
(M
0
)=f

t
(0, 2) − 2f

v
(0, 2)
∂f
∂y
(M
0
)=2f

t
(0, 2) + f

v
(0, 2)
t = x + y
2
,v= y + x
2
v`a do d´o
df (M
0
)=


f

t
(0, 2) − 2f

v
(0, 2)

dx +2

2f

t
(0, 2) + f

v
(0, 2)

dy.
132 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
C´ach 2.

´
Ap du
.
ng t´ınh bˆa
´
tbiˆe
´
nvˆe
`
da
.
ng cu
’
a vi phˆan cˆa
´
p1.
Ta c´o
t = x + y
2
⇒ dt = dx +2ydy,
v = y + x
2
⇒ dv =2xdx + dy.
Do d
´o
df (M
0
)=
∂f
∂t

(0, 2)dt +
∂f
∂v
(0, 2)dv
= f

t
(0, 2)[dx +2ydy]+f

v
(0, 2)[2xdx + dy]
=

f

t
(0, 2) − 2f

v
(0, 2)

dx +

2f

t
(0, 2) + f

v
(0, 2)


dy. 
V´ı d u
.
4. 1) Cho h`am f(x, y)=x
y
. H˜ay t`ım vi phˆan cˆa
´
p hai cu
’
a f
nˆe
´
u x v`a y l`a biˆe
´
ndˆo
.
clˆa
.
p.
2) T`ım vi phˆan cˆa
´
p hai cu
’
a h`am f(x + y, xy)nˆe
´
u x v`a y l`a biˆe
´
n
d

ˆo
.
clˆa
.
p.
Gia
’
i. 1) T`u
.
v´ıdu
.
2, 1) v`a cˆong th ´u
.
c (9.10) ta c´o
d
2
f =
∂
2
f
∂x
2
dx
2
+2
∂
2
f
∂x∂y
dxdy +

∂
2
f
∂y
2
dy
2
,
trong d´o
∂
2
f
∂x
2
= y(y − 1)x
y−2
,
∂
2
f
∂y
2
= x
y
(lnx)
2
,
∂
2
f

∂x∂y
= x
y−1
(1 + ylnx)
v`a do d
´o
d
2
f = y(y − 1)x
y−2
dx
2
+ x
y−1
(1 + ylnx)dxdy + x
y
(lnx)
2
dy
2
.
2) Ta viˆe
´
t h`am d˜a cho du
.
´o
.
ida
.
ng u = f(t, v), trong d

´o t = x + y,
v = xy.
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 133
1
+
C´ach I. T´ınh c´ac da
.
o h`am riˆeng rˆo
`
i ´ap du
.
ng (9.10). Ta c´o:
∂f
∂x
= f

t
(x + y,xy)+f

v
(x + y,xy) · y,
∂f
∂y
= f


t
(x + y,xy)+f

v
(x + y,xy) · x,
∂
2
f
∂x
2
= f

tt
+ f

tv
y + f

tv
y + f

vv
y
2
= f

tt
+2yf


tv
+ y
2
f

vv
,
∂
2
f
∂x∂y
= f

tt
+ f

tv
x + f

tv
y + f

vv
xy + f

v
= f

tt
+(x + y)f


tv
+ xyf

vv
+ f

v
,
∂
2
f
∂y
2
= f

tt
+ f

tv
x + f

tv
x + f

vv
x
2
= f


tt
+2xf

tv
+ x
2
f

vv
.
Thˆe
´
c´ac d
a
.
o h`am riˆeng t`ım du
.
o
.
.
c v`ao (9.10) ta thu du
.
o
.
.
c
d
2
f =(f


tt
+2yf

tv
+ y
2
f

vv
)dx
2
+2(f

tt
+(x + y)f

tv
+ xyf

vv
+ f

v
)dxdy
+(f

tt
+2xf

tv

+ x
2
f

vv
)dy
2
.
2
+
C´ach II. Ta c´o thˆe
’
thu du
.
o
.
.
ckˆe
´
t qua
’
n`ay nˆe
´
ulu
.
u´yr˘a
`
ng v´o
.
i

t = x + y ⇒ dt = dx + dy v`a v = xy → dv = xdy + ydx v`a t`u
.
d
´o
d
2
t = d(dx + dy)=d
2
x + d
2
y =0
(v`ı x v`a y l`a biˆe
´
nd
ˆo
.
clˆa
.
p) v`a
d
2
v = d(xdy + ydx)=dxdy + dxdy =2dxdy.
´
Ap du
.
ng (9.12) ta c´o
d
2
f =
∂

2
f
∂t
2
(dx + dy)
2
+2
∂
2
f
∂t∂v
(dx + dy)(xdy + ydx)
+
∂
2
f
∂v
2
(xdy + ydx)
2
+
∂f
∂t
· 0+
∂f
∂v
(2dxdy)
=

f


tt
+2yf

tv
+ y
2
f

vv

dx
2
+

f

tt
+2xf

tv
+ x
2
f

vv

dy
2
+2


f

tt
+(x + y)f

tv
+ xyf

vv
+ f

v

dxdy. 
134 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
V´ı du
.
5.
´
Ap du
.

ng vi phˆan dˆe
’
t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
:
1) a =(1,04)
2,03
2) b = arctg

1, 97
1, 02
− 1

3) c =

(1, 04)
1,99
+ ln(1, 02)
4) d =
sin 1, 49 · arctg0, 07
2
2,95
.
Gia
’
i. Dˆe
’
´ap du

.
ng vi phˆan v`ao t´ınh gˆa
`
nd´ung ta cˆa
`
n thu
.
.
chiˆe
.
n c´ac
bu
.
´o
.
c sau dˆay:
Th´u
.
nhˆa
´
t l`a chı
’
r˜o biˆe
’
uth´u
.
c gia
’
it´ıchd
ˆo

´
iv´o
.
i h`am m`a gi´a tri
.
gˆa
`
n
d
´ung cu
’
a n´o cˆa
`
n pha
’
i t´ınh.
Th´u
.
hai l`a cho
.
ndiˆe
’
mdˆa
`
u M
0
sao cho gi´a tri
.
cu
’

a h`am v`a cu
’
a c´ac
da
.
o h`am riˆeng cu
’
a n´o ta
.
idiˆe
’
mˆa
´
y c´o thˆe
’
t´ınh m`a khˆong cˆa
`
nd`ung
ba
’
ng.
Cuˆo
´
ic`ung ta ´ap du
.
ng cˆong th´u
.
c
f(x
0

+∆x, y
0
+∆y)=f(x
0
,y
0
)+f

x
(x
0
,y
0
)∆x + f

y
(x
0
,y
0
)∆y.
1) T´ınh a =(1, 04)
2,03
. Ta x´et h`am f(x, y)=x
y
.Sˆo
´
a cˆa
`
n t´ınh l`a

gi´a tri
.
cu
’
a h`am khi x =1,04 v`a y =2, 03.
Ta lˆa
´
y M
0
= M
0
(1, 2). Khi d´o∆x =0, 04, ∆y =0, 03.
Tiˆe
´
p theo ta c´o
∂f
∂x
= yx
y−1
⇒
∂f
∂x


M
0
=2
∂f
∂y
= x

y
lnx ⇒
∂f
∂y


M
0
=1·ln1 = 0.
Bˆay gi`o
.
´ap du
.
ng cˆong th´u
.
cv`u
.
anˆeuo
.
’
trˆen ta c´o:
a = f(1, 04; 2, 03) = (1, 04)
2,03
≈ f(1, 2) + 2 · 0, 04 = 1 + 0, 08 = 1, 08.
2) Ta nhˆa
.
nx´etr˘a
`
ng arctg


1, 97
1, 02
− 1

l`a gi´a tri
.
cu
’
a h`am
f(x, y) = arctg

x
y
− 1

9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 135
ta
.
idiˆe
’
m M(1, 97; 1, 02).
Ta cho
.
n M

0
= M
0
(2, 1) v`a c´o
∆x =1, 97 − 2=−0, 03,
∆y =1, 02 − 1=0, 02.
Tiˆe
´
pd
ˆe
´
n ta c´o
∂f
∂x
=
1
y
1+

x
y
− 1

2
=
y
y
2
+(x − y)
2

∂f
∂y
= −
x
y
2
+(x −y)
2
·
T`u
.
d´o
∂f
∂x
(M
0
)=f

x
(2, 1) =
1
1
2
+(2− 1)
2
=0, 5
∂f
∂y
(M
0

)=f

y
(2, 1) = −1.
Do d´o
arctg

1, 97
1, 02
− 1

= arctg

2
1
−1

+(0, 5) · (−0, 03) + 1 · (0, 02)
=
π
4
−0, 015 − 0, 02=0, 785 −0, 035
=0, 75.
3) Ta thˆa
´
yr˘a
`
ng c =

(1, 04)

1,99
+ ln(1, 02) l`a gi´a tri
.
cu
’
a h`am
u = f(x, y, z)=
√
x
y
+lnz ta
.
idiˆe
’
m M(1, 04; 1, 99; 1, 02).
Ta cho
.
n M
0
= M
0
(1, 2, 1). Khi d´o
∆x =1, 04 − 1=0, 04
∆y =1, 99 − 2=−0, 01
∆z =1, 02 − 1=0, 02.
136 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe

`
ubiˆe
´
n
Bˆay gi`o
.
ta t´ınh gi´a tri
.
c´ac d
a
.
o h`am riˆeng ta
.
idiˆe
’
m M
0
.Tac´o
∂f
∂x
=
yx
y−1
2
√
x
y
+lnz
⇒
∂f

∂x
(M
0
)=
2 · 1
2
√
1 + ln1
=1,
∂f
∂y
=
x
y
lnx
2
√
x
y
+lnz
⇒
∂f
∂y
(M
0
)=0,
∂f
∂z
=
1

2z
√
x
y
+lnz
⇒
∂f
∂z
(M
0
)=
1
2
·
T`u
.
d´o suy ra

(1, 04)
1,99
+ ln(1, 02) ≈
√
1+ln1+1· (0, 04) + 0 · (−0, 01)
+(1/2) · 0, 02=1,05.
4) Ta thˆa
´
y d l`a gi´a tri
.
cu
’

a h`am f(x, y, z)=2
x
sin y arctgx ta
.
idiˆe
’
m
M(−2, 95; 1, 49; 0, 07)
Ta lˆa
´
y M
0
= M
0

− 3,
π
2
, 0

. Khi d´o
∆x = −2, 95 − (−3)=0, 05
∆y =1,49 − 1, 57 = −0, 08
∆z =0, 07.
Tiˆe
´
p theo ta c´o
f(M
0
)=2

−3
sin(π/2) arctg0 = 0,
f

x
(M
0
)=2
x
ln2 · sin y arctgz


M
0
=0,
f

y
(M
0
)=2
x
cos y arctgz


M
0
=0,
f


z
(M
0
)=
2
x
sin y
1+z
2


M
0
=2
−3
.
T`u
.
d´o ta thu du
.
o
.
.
c
sin 1, 49 arctg0, 07
2
2,95
≈ 2
−3
· 0, 07 ≈ 0, 01. 

V´ı du
.
6. Khai triˆe
’
n h`am f(x, y)=x
y
theo cˆong th´u
.
c Taylor ta
.
i lˆan
cˆa
.
nd
iˆe
’
m(1, 1) v´o
.
i n =3.
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 137
Gia
’
i. Trong tru
.

`o
.
ng ho
.
.
p n`ay cˆong th´u
.
c Taylor c´o da
.
ng sau d
ˆa y
f(x, y)=f(1, 1) +
df (1, 1)
1!
+
d
2
f(1, 1)
2!
+
d
2
f(1, 1)
3!
+ R
3
. (*)
1
+
T´ınh mo

.
ida
.
o h`am riˆeng cu
’
a h`am cho dˆe
´
nxˆa
´
p 3. Ta c´o
f

x
= yx
y−1
,f

y
= x
y
lnx, f

x
2
= y(y − 1)x
y−2
,
f

xy

= x
y−1
+ yx
y−1
lnx, f

y
2
= x
y
(lnx)
2
,
f
(3)
x
3
= y(y − 1)(y −2)x
y−3
,f
(3)
x
2
y
=(2y − 1)x
y−2
+ y(y − 1)x
y−2
lnx,
f

(3)
xy
2
=2x
y−1
lnx + yx
y−1
(lnx)
2
,f
(3)
y
3
= x
y
(lnx)
3
.
2+ T´ınh gi´a tri
.
cu
’
a c´ac d
a
.
o h`am riˆeng ta
.
idiˆe
’
m(1, 1). Ta c´o

f(1, 1) = 1,f

x
(1, 1)=1,f

y
(1, 1)=0,f

x
2
(1, 1) = 0,
f

xy
(1, 1)=1,f

y
2
(1, 1) = 0,f
(3)
x
3
(1, 1)=0,f
(3)
x
2
y
(1, 1)=1,
f
(3)

xy
2
(1, 1) = 0,f
(3)
y
3
(1, 1)=0.
3
+
Thˆe
´
v`ao cˆong th´u
.
c (*) ta c´o
df (1, 1) = f

x
(1, 1)∆x + f

y
(1, 1)∆y =∆x,
d
2
f(1, 1) = f

x
2
(1, 1)∆x
2
+2f


xy
(1, 1)∆x∆y + f

y
2
(1, 1)∆y
2
=2∆x∆y,
d
3
f(1, 1) = 3∆x
2
∆y
v`a do d
´o
x
y
=1+∆x +∆x∆y +
1
2
∆x
2
∆y + R
3
. 
V´ı du
.
7. T´ınh vi phˆan cu
’

a h`am ˆa
’
n w(x, y)d
u
.
o
.
.
cchobo
.
’
iphu
.
o
.
ng
tr`ınh
w
3
+3x
2
y + xw + y
2
w
2
+ y − 2x =0.
Gia
’
i. Ta xem phu
.

o
.
ng tr`ınh d˜a cho nhu
.
mˆo
.
tdˆo
`
ng nhˆa
´
t v`a lˆa
´
yvi
phˆan cu
’
avˆe
´
tr´ai v`a vˆe
´
pha
’
i:
3w
2
dw +6xydx +3x
2
dy + wdx + xdw +2y ·w
2
dy
+2y

2
wdw − 2dx + dy =0
138 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
v`a t `u
.
d
´or´ut ra dw.Tac´o
(6xy + w − 2)dx +(3x
2
+2yw
2
+1)dy +(3w
2
+ x +2y
2
w)dw =0
v`a do d
´o
dw =
2 − 6xy − w
3w
2

+ x +2y
2
w
dx −
3x
2
+2yw
2
+1
3w
2
+ x +2y
2
w
dy. 
V´ı du
.
8. T´ınh dw v`a d
2
w cu
’
a h`am ˆa
’
n w(x, y)du
.
o
.
.
cchobo
.

’
iphu
.
o
.
ng
tr`ınh
x
2
2
+
y
2
6
+
w
2
8
=1.
Gia
’
i. D
ˆa
`
utiˆent`ım dw.Tu
.
o
.
ng tu
.

.
nhu
.
trong v´ıdu
.
7 ta c´o
xdx +
ydy
3
+
wdw
4
=0⇒ dw = −
4x
w
dx −
4y
3w
dy. (*)
La
.
ilˆa
´
y vi phˆan to`an phˆa
`
nd˘a
’
ng th´u
.
cthudu

.
o
.
.
cv´o
.
ilu
.
u´yl`adx, dy l`a
h˘a
`
ng sˆo
´
; dw l`a vi phˆan cu
’
a h`am.
Ta c´o
d
2
w = −4
wdx − xdw
w
2
dx −
4
3
·
wdy − ydw
w
2

dy
hay l`a
d
2
w =4

1
w
dx
2
−
x
2
w
2
dxdw +
1
3w
dy
2
−
y
3w
2
dydw

(**)
D
ˆe
’

c´o biˆe
’
uth´u
.
c d
2
w qua x, y, w, dx v`a dy ta cˆa
`
nthˆe
´
dw t`u
.
(*) v`ao
(**). 
V´ı du
.
9. C´ac h`am ˆa
’
n u(x, y)v`av(x, y)du
.
o
.
.
c x´ac d
i
.
nh bo
.
’
ihˆe

.
xy + uv =1,
xv − yu =3.
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 139
T´ınh du(1, −1), d
2
u(1, −1); dv(1, −1), d
2
v(1, −1) nˆe
´
u u(1, −1) = 1,
v(1, −1) = 2.
Gia
’
i. Lˆa
´
y vi phˆan hˆe
.
d˜a cho hai lˆa
`
n ta c´o

ydx + xdy + udv + vdu =0,
xdv + vdx − ydu −udy =0.

(I)

2dxdy +2dudv + ud
2
v + vd
2
u =0,
2dxdv −2dudv + xd
2
v − yd
2
u =0.
(I I)
Thˆe
´
v`ao (I) gi´a tri
.
x =1,y = −1, u =1,v =2tac´o

−dx + dy + dv +2du =0
2dx − dy + dv + du =0
⇒
du =3dx − 2dy
dv = −5dx +3dy
(I II)
T`u
.
(I II) ta c˜ung thu d
u
.

o
.
.
c u

x
=3,u

v
= −2; v

x
= −5, v

y
=3.
Thay v`ao (II) c´ac gi´a tri
.
x =1,y = −1, u =1,v =2v`adu, dv t`u
.
(I II) ta c´o:
d
2
v +2d
2
u = −2dxdy −2(3dx −2dy)(3dy −5dx)
d
2
v + d
2

u =2dy(3dx − 2dy) − 2dx(3dy −5dx)
v`a do d
´o
d
2
u = 4(5dx
2
− 10dxdy +4dy
2
),
d
2
v = 10(−dx
2
+4dxdy − 2dy
2
). 
B
`
AI T
ˆ
A
.
P
T´ınh vi phˆan dw cu
’
a c´ac h`am sau
1. w = x
2
y − y

2
x + 3. (DS. dw =(2xy − y
2
)dx +(x
2
−2xy)dy)
2. w =(x
2
+ y
2
)
3
.(DS. 6(x
2
+ y
2
)
2
(xdx + ydy))
3. w = x − 3 sin y.(D
S. dw = dx − 3 cos ydy)
140 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n

4. w = ln(x
2
+ y). (DS.
2xdx
x
2
+ y
+
dy
x
2
+ y
)
5. w =

y
x

x
.(DS.

y
x

x
ln
y
x
−


y
x

x

dx +

y
x

x−1
dy)
6. w = ln tg
y
x
.(D
S. −
2ydx
x
2
sin
2y
x
+
2dy
x sin
2y
x
).
T´ınh dw(M

0
)cu
’
a c´ac h`am ta
.
idiˆe
’
m M
0
d˜a cho (7-14)
7. w = e
−
y
x
, M
0
(1, 0). (DS. dw(1, 0) = −dy)
8. w = y
3
√
x, M
0
(1, 1). (DS. dw(1, 1) =
1
3
dx + dy)
9. f( x, y)=
yz
x
, M

0
(1, 2, 3). (DS. df


M
0
= −6dx +3dy +2dz)
10. f( x, y, z)=cos(xy + xz), M
0

1,
π
6
,
π
6

.
(DS. df


M
0
= −
√
3
2

π
3

dx + dy + dz

)
11. f( x, y)=e
xy
, M
0
(0, 0). (DS. df


M
0
=0)
12. f( x, y)=x
y
, M
0
(2, 3). (DS. df


M
0
=12dx + 8ln2dy)
13. f( x, y)=xln(xy), M
0
(1, 1). (DS. df


M
0

= dx + dy)
14. f( x, y) = arctg
x
y
, M
)
(1, 2). (DS. df


M
0
=
1
5
(2dx − dy)).
T`ım vi phˆan cu
’
a c´ac h`am ho
.
.
psaudˆay ta
.
i c´ac diˆe
’
md˜a c h ı
’
ra (15-18)
15. f( x, y)=f(x − y,x+ y), M(x, y), M
0
(1, −1).

(DS. df


M
=(f

t
+ f

v
)dx +(f

v
− f

t
)dy,
df


M
0
=

f

t
(2, 0) + f

v

(2, 0)

dx +

f

v
(2, 0) − f

t
(2, 0)

dy,
t = x − y, v = x + y)
9.2. Vi phˆan cu
’
a h`am nhiˆe
`
ubiˆe
´
n 141
16. f(x, y)=f

xy,
x
y

, M(x, y), M
0
(0, 1).

(DS. df


M
=

yf

t
+
1
y
f

v

dx +

xf

t
−
x
y
2
f

v

dy,

df


M
0
=

f

t
(0, 0) + f

v
(0, 0)

dx, t = xy, v =
x
y
)
17. f(x, y, z)=f(x
2
− y
2
,y
2
− z
2
,z
2
− x

2
), M(x, y, z), M
0
(1, 1, 1).
(DS. df


M
=2(xf

t
− xf

w
)dx +2y(f

v
−f

t
)dy +2z(f

w
− f

v
)dz,
df



M
0
=2(f

t
(0, 0, 0) − f

w
(0, 0, 0))dx +2(f

v
(0, 0, 0) − f

t
(0, 0, 0))dy
+2(f

w
(0, 0, 0) − f

v
(0, 0, 0))dz,
t = x
2
− y
2
,v= y
2
− z
2

,w= z
2
− x
2
)
18. f(x, y, z)=f(sin x+sin y,cos x−cos z), M(x, y, z)v`aM
0
(0, 0, 0).
(D
S. df


M
=(f

t
cos x − f

v
sin x)dx + f

t
cos ydy + f

v
sin zdz,
df


M

0
= f

t
(0, 0)dx + f

v
(0, 0)dy,
t = sin x + sin y, v = cos x − cos z).
T´ınh vi phˆan dw v`a d
2
w ta
.
idiˆe
’
m M(x, y) (19-22) nˆe
´
u:
19. w = f(lnz), z = x
2
+ y
2
.
(DS. d
2
w =
2
(x
2
+ y

2
)
2

(2x
2
f

tt
−x
2
f

t
+ y
2
f

t
)dx
2
+(4xyf

tt
− 4xyf

t
)dxdy +(x
2
f


t
− yf

t
+2yf

t
2
)dy
2
)
20. w = f(α,β,γ), α = ax, β = by, γ = cz; a, b, c-h˘a
`
ng sˆo
´
.
(D
S. dw = af

α
dx + bf

β
dy + cf

γ
dz;
d
2

w = a
2
f

α
2
dx
2
+ b
2
f

β
2
dy
2
+ c
2
f

γ
2
dz
2
+2(f

αβ
abdxdy + f

βγ

bcdydz + f

αγ
acdxdz))
21. w = f(x + y,x − y). (D
S. x + y = u, x − y = v;
d
2
w =(f

u
2
+2f

uv
+ f

v
2
)dx
2
+(f

u
2
− 2f

v
2
)dxdy +(f


u
2
− 2f

uv
+ f

v
2
)dy
2
)
142 Chu
.
o
.
ng 9. Ph´ep t´ınh vi phˆan h`am nhiˆe
`
ubiˆe
´
n
22. w = xf

x
y

.(DS. dw =

f +

x
y
f


dx −
x
2
y
2
f

dy,
d
2
w =

2
y
f

+
x
y
2
f

)dx
2
−


4x
y
2
f

+
2x
2
y
3
f


dxdy −

2x
2
y
3
f

−
x
3
y
4
f



dy
2
)
T´ınh vi phˆan cˆa
´
p hai cu
’
a c´ac h`am sau dˆay ta
.
ic´acdiˆe
’
m M(x, y)
v`a M
0
(x
0
,y
0
)nˆe
´
u f l`a h`am hai lˆa
`
n kha
’
vi v`a x, y, z l`a biˆe
´
ndˆo
.
clˆa
.

p
(23-25)
23. u = f(x − y, x + y), M(x, y), M
0
(1, 1) .
(DS. d
2
u


M
= f

tt
(dx − dy)
2
+2f

tv
(dx
2
− dy
2
)+f

vv
(dx + dy)
2
,
d

2
u


M
0
= f

tt
(0, 2)dx(dx − dy)
2
+2f

tv
(0, 2)(dx
2
− dy
2
)
+ f

vv
(0, 2)(dx + dy)
2
)
24. u = f(x + y, z
2
), M(x, y, z), M
0
(−1, −1, 0).

(DS. d
2
u


M
= f

tt
(dx + dy)
2
+4zf

tv
dz(dx + dy)
+4z
2
f

vv
dz
2
+2f

v
d
2
z,
d
2

u


M
0
= f

tt
(0, 0)(dx + dy)
2
+2f

v
(0, 0)dz
2
,
t = x + y, v = z
2
)
25. u = f(xy, x
2
+ y
2
), M(x, y), M
0
(0, 0).
(D
S. d
2
u



M
= f

tt
(ydx + xdy)
2
+4f

tv
(ydz + xdy)(xdx + ydy)
+4f

vv
(xdx + ydy)
2
+2f

t
dxdy +2f

v
(dx
2
+ dy
2
),
d
2

u


M
0
=2f

t
(0, 0)dxdy +2f

v
(0, 0)(dx
2
+ dy
2
),
t = xy, v = x
2
+ y
2
)
T´ınh vi phˆan d
n
w (26-27) nˆe
´
u:
26. w = f(ax + by + cz).
(DS. d
n
w = f

(n)
(ax + by + cz)(adx + bdy + cdz)
n
)