Bài tập toán cao cấp part 9 - Pdf 18

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 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
.

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

´
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.

´
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

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

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



∂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
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
´

+
∂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


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

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


)+
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

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
.

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

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
`

F

w
(·)
,w

y
= −
F

y
(·)
F

w
(·)
,
rˆo
`
ithˆe
´
v`ao biˆe

uth´u
.
c
dw =
∂w
∂x
dx +


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
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

∂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

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

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.

(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)

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 +

∂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


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
+ xyf

vv
+ f

v
,

2
f
∂y
2
= f

tt
+ f

tv
x + f

tv
x + f

vv
x
2
= 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

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 =



tt
+2yf

tv
+ y
2
f

vv

dx
2
+

f

tt
+2xf

tv
+ x
2
f

vv

dy
2
+2

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

´
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


+∆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

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


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
(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)

`
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

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


−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


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
.

.
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

(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

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

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
.

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


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
.

.
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

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

(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
.

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

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



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

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

(DS. df


M
=(f

t
+ f

v
)dx +(f

v
− f

t
)dy,
df


M
0
=

f

t
(2, 0) + f

v

(DS. df


M
=

yf

t
+
1
y
f

v

dx +

xf

t

x
y
2
f

v

dy,

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



,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

2
)
2

(2x
2
f

tt
−x
2
f

t
+ y
2
f

t
)dx
2
+(4xyf

tt
− 4xyf

t
)dxdy +(x
2
f

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


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 +


4x
y
2
f

+
2x
2
y
3
f


dxdy −

2x
2
y
3
f


x
3
y
4
f



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

(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



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

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


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