пособие 6 по матану (2 курс)

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Ǒà ªâ¨ç¥áª¨¥ § -ïâ¨ï ¬ ⥬ â¨ç¥áª®¬ã - «¨§ã. — áâì 6: — áâ-

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•¥ª ¬¥-¤ãîâ ¯¥ç

ª 䥤à

⥠ਨ äã-ªæ¨©

¯à ¡«¨ ¥-¨©

‘ à ⮢᪮£® £®áã-¨¢¥àáâ¨:â¥â

“„Š 517.262(075.3 )

ISBN 978-5-91879-125-7

••Š 22.161ï722

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— áâ-ë¥

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.

¢¨á¨¬®©- -äã¨å-ªæ¨ï¬®¯¥à¥¥¬â¥=¨§¬--®©(¥-;ïâìá). ï,’ 浪¤àãª

£ ï¨ á®åà- -¥§-ïâ좨ᨬë᢮¥¥§¯-¥à祥¬-¥¨--¥.

„ëǑãáâ쥤¨¬, â®--®¤¥§-¯à®¨§¢®¤

z

f x y

x y

- §¬¥ ë¬. ’®£¤

x ¯à¨à é¥-¨¥ x, á®åà -ïï §- ç¥-¨¥ y

¯à¨à é¥-¨¥¬

z ¯®«ãç¨â ¯à¨à é¥-¨¥, ª®â®à®¥ - §ë¢ ¥âáï ç áâ-ë¬

z

¯®

¥âá xz

. ˆâ ª,

x ¨ ®¡®§-

Ǒ®«-®¥ ¯à¨à é¥-¨¥ y z = f (x; y + y) − f (x; y).

, â®

€- «®£¨ç-® ¯®«ãç ¥¬xçz =áâ-f®(x¥ +¯à¨àx;éy)¥−-¨f¥(x; y).

®-

z ¯® y:

z äã-ªæ¨¨ z ®¯à¥¤¥«ï¥âáï à ¢¥-á⢮¬

Ž¯à¥¤¥«¥-¨¥ 1. …᫨z =áãéf (x¥áâ¢ã+ x¥;ây +¯à¥¤y¥)«− flim(x; y).

f (x+ x;y)−f (x;y)

- §ë¢ ¥âáï ç áâ-®© ¯à®¨§¢®¤-®© (¯¥à¢®£® ¯®à浪 ) äã-ªæx ¨

x→0

â®çª¥ Œ(

z = f (x; y) ¢

x; y) ¯® ¯¥à¥¬¥--®© x ¨ ®¡®§- ç ¥âáï ®¤-¨¬ ¨§ ᨬ¢®«®¢:

′

∂z

′

∂f

€-®©- «®£¨ç-®¡à®¯à¥¤¥«ï¥ ᮯàz¨¥®¡®§,¤¥«ï-¥âç,á ¥âáïf

,ç áâ- .ï ¯à®¨§¢®¤- ï ¯® ¯¥à¥¬¥--

x

∂x

x

∂z

ле¡®«ми’¯ª¨¬.¥а¥¬) ¥¯--¥ал姮¬,.¥¬¥--¯а¨Ǒзленбвгб«®- г¢лз¨б«¨¨¯а®¨§¢®¤бв¯®бв®л¥¥¨п¯а®¨§¢®¤--ªбв¢пªдг¯а®¨§¢®¤--ªж¨¨-лз¥-дг¨©--¥-бªп®бвªж¨¨®«мª¨едг-ªж¨¨-ле(¤¢ге,®¥§-®©¢¨б¨ва¥е§

íâ¨åy

¥âá⢥-- -

-®©- 寥¤ïâà¥

¯à®¨§¢®¤-ëå äã-ªæ¨¨( ®; )

®®â¢¬

¥à

í⮬¨¯à ᢨ«

®à¬ã«(¯à¨©

Ǒਬ--ä

f x y

¯®

- ©).

1.

• ©â¨ ç

-ë¥ ¯à®¨§¢®¤x ¨«¨-ë¥yäãáç¨â-ªæ¨¨¥âáï ¯®áâ®ï--®© ¢¥«¨ç¨-

•¥è¥-¨¥:

y + ex2−2 + 1)′

z = 2y + ¥x2−y + 1.

z′

y)′

ex −y )′

′

=x0+=(2

2

x = (2

x + (

x + (1)x =

x2−y

· (x

′

x2−y

x2

−y ;

e

− y x + 0 = e

· (2x − 0) = 2x · e

zy′

= 2 + ex2

−y · (−1).

Ž¯à¥¤¥«¥-¨¥ 2. — áâ-ë¥ ¯à®¨§¢®¤-ë¥ ¯¥à¢®£® ¯®à浪

∂f (x;y) ¨ ∂f (x;y)

¬® ® «áᥤãî騬âਢ âì ª ª äã-ªæ¨¨ ®â ¯¥à¥¬¥ -ëå

∂x

∂y

∂x µ ∂x ¶

=

∂x2 = zxx = fx2 (x; y);

∂

∂z

∂2z

′′

′′

∂

∂z

∂ z

′′

′′

µ

¶

zxy

fxy x y

∂x

∂y

∂y∂x

∂y µ

∂x ¶

∂x∂y = zyx = fyx(x; y);

∂

∂z

=

∂ z

′′

′′

Ǒав.®¢л¤§«¨з—€. -¯®абв¨«®--п¤ªл¬п¢«повбп,£¨з®¢

∂y µ ∂y ¶

=

∂y2

3

zyy = fy2

x y .

∂

∂z

∂2z

′′

′′

¯-¥à®¨§¢®.®¯à६¥--¥¤ë¥-¯а¨«¬,повбп¬-¢â®à®¥§ë¢çà, £®á¥ââᨫ-ïë¨á¬¥=¯à®¨§¢®¤¡®«¥è¥¥--¢ë᮪®®©-ç(ë¥;áâ£)®à-¥®©¯®à浪â쥯ந§¢®¤£®, ç¥, ⢢§ïâ¥-àâ®®©.ï’¯®£® ¨

′′

∂

z

′′′

ªæ¨ ¬¥à 2. • ©â¨ ᬥè --ë¥zç ,áâ

-ë¥

¯à®¨§¢®¤, z .

-ë¥ ¢â®à®£® ¯®à浪 äã--

xy

∂x∂y2

xyz

•¥è¥z-=¨¥x:4’−ª2xª2yª3 +′

y5 + 1:

′

zx = 4x3 − 4xy3 ¨ zy = −6x y2 + 5y4,

′′

3

− 4xy

3)′

2

zxy = (4x

2

y = −12xy

;

′′

y

2

+ 5y

4)′

2

.

Žª § «®áì, çâ®

zyz = (−6x

x =

−12xy

’॥¬®à:¥¬

′′

′′

1 (˜¢ zàæ).

zǑà¥. ¤¯®«®•â®â २¬,§ã«ìçâ®-1)¥ á«ãç ¥-. ˆ¬¥¥â ¬¥á⮠⥮

xy = yx

â¨

f (x; y) ®¯à¥¤¥«¥-

¢ ®¡«

D, ¢â®àë2) í⮩¥

®¡« á⨠áãé¥áâ¢ãîâ ¯¥à¢ë¥ ¯à®¨§¢®¤-ë¥ f ′

f ′

ª ¥

ᬥè --ë¥ ¯à®¨§¢®¤-ë¥

x íâ¨y ,

á«¥¤-¨¥ ¯à®¨§¢ ¤-ë¥

f

′′

f ′′

¯®-

xy ¨ yx, ¨, - ª® ¥æ, 3)

â®çª¥ (

fxy′′

fyx′′

, ª ª äã-ªæ¨¨ x ¨ y, -¥¯à¥àë¢-ë ¢ -¥ª®â®à®©

x0, y0) ®¡« á⨠D. ’®£¤

í⮩ â®çª¥

f ′′

x0; y0) = f

′′

x0; y0)

xy (

yx(

®ªàŽ¯à¥á⥤-¥®á⨫¥-¨¥â®çª¨3. Ǒãáâì äã-ªæ¨ï z

= f (x; y) ®¯à ¤¥«¥-

¢ -¥ª®â®à®©

M (x; y). …᫨ ¯®«-®¥ ¯à¨à é¥-¨¥ äã-ªæ¨¨ z =

f (x + x; y + y) − f (x; y) ¬® -® § ¯¨á âì ¢ ¢¨¤¥

£¤¥

z = A · x + B · y + α x + β y,

(1)

α =ïα( x, y(1),→ 0 ¤¨äβ = β( x, y) → 0 ¯à¨ x → 0, y → 0, â®

-

äã¬левбп-ªæ¨¢á¨¬¢®«®à¨,¢-¥--¢¨á¨¬ëå§ë¢á⢧뢥¥âáï¥âáï¯à®«¥¤áâ-ë¬ä¥¢«ïî館ääà¥-æ¨àã¥à¥ï¥-¬®©,æ¨á®¡®©«®¬£á㬬í⮩« ¢-ãîäã¯ç¥-à¢ëåªæ¨¨áâ줢ãå¯à¨à®¡®§á«é-¥£-ç¨ï¥

¥

f

dz:

„«ï -¥§

¯¥à¥¬dz¥--=¯ëå¥A෥寨áx +ã B¯®«· y£. îâ å =

(2)

Ǒ®í⮬ã à ¢¥-á⢮ (2) ¬® -®

âì ¢ ¢¨¤¥:

dx ¨ ã = dy.

ªæ¨ï’¥®à¥¬

2 (-áã饮¡å¥á⢮¢¤¨¬®¥æ¨àãá«®¢¨dz¥= Adx¥ ¤¨ää+ Bdy¥à¥-æ¨à㥬®áâ¨). …᫨ äã(3)--

í⮩ zâ®çª= f¥(,x;¨¬y)¥¤¨ää¥â ¥à¥¥©- ç áâ¬-륢¯à®¨§¢®¤â®çª¥ Œ(-ëx¥; y), â® -

-¥¯à¥àë¢- ¢

z′

zx′

zy′ , ¯à¨ç¥¬ zx′ = €,

-®ª,¥-¨ïãâ¢-¥¯àç¥àá«¥à뢥-¤¤á⢨ëå¥-- ¯à®¨§¢®¤ï¥-äã⥢-®à¥ªæ¨ïà¥-¬ë-®, åâ.-¥.᫨§¥¤ã- ¯àâ ¤¨ää¥àë¢-¥®áâ¨à¥--

æ¨àãäãy Žâ¬=-ªæ¨¨¥‚¬®áâì.¥â¨¬,¨«¨äãç-ªæ¨¨®¡à. ’

¢ëç¨á«à¥-æ¨àã¥-¨ï ¢¯®«

z

p

x

y

â®çª-®£®(0;0)¤¨ää. Š¥à¥ª-æ¨

:

¯®«ãç= 2¥¬+ä®à¬ã«ã2 -

¤«ï

∂z

∂z

ªæ¨ï’¥®à¥¬

3 (¤®áâ â®ç-®æ¨à㥠dzá«®¢¥= ¨¥dx¤¨ä+ ä¥dy.à¥-æ à㥬®áâ¨). …᫨ äã(4)--

∂x

∂y

(å; ã) ¨¬¥¥â -¥¯à¥àë¢-ë¥ ç áâ-ë¥ ¯à®¨§¢®¤-ë¥

′

′

z =âf¥®áï-¨-ä®à¬ã«®©¥ 4¤¨ää. „¨ää¥à¥(4)-¥à¥.-樬

¬í⮩ â 窥 ¨ ¥¥ ¯®«-ë©z¤¨ääz ¥à¢¥â®çª-æ¨ «¥

Ž¯à¢ëàŒ(å;¥ã),¤¥«

x

y

k-£® ¯®à浪 - §ë¢ ¥âáï ¢ëà ¥- ¥

k

k−

d

йz ¥=-¨пеd(d 1z), £¤¥ ¢б¥ ¤¨дд¥а¥-ж¨ «л ¡¥агвбп ¯а¨ ®¤-¨е ¨ в¥е ¥ ¯а¨-

í⮩ äãk--ªæ¨¨£® ¯®à浪

¢ª«îç¨â¢¨¤:

¥«ì-®, â® ¤¨ää¥à¥-æ¨

k-

¯®à浪 ¤«ï

᪮¡ª®«ãç•â®¥å,--¢ëàä®à¬ë¥ ç«¥-«ì-¨ë¯®á«¥-d"ã¬á«z

µ ∂x1 dx

...

∂xn dxn¶

k

· z.

∂

∂

k¢®§¢®¤¥¥¤ã-=í⮥⣨âᯮîâá® -館ï"¢1áâ-+쥯¥+-ì:

¯®á-

¯àç ¢¨«¬-¬

«£¥¡àë,- áâ®ï騩§â¥¬ ¢á¥

®¨§¢®

z (ª®â®à¤¨ä䤮¯¨á뢥ॠ¥âáï ¢ ç¨á«¨â¥«¨

¯à• ¯à¨¬),¤-ëå⮫쪮¨¤«ï¤¨¥à,

ää-¥ªæ¨¨à¥-樤¢ãå«®¢¥¬.¯¥á¨¬¢®«à¥¬¥--ë嬢®§¢à é ¥-âáïæ¨ ¨å«ë§¡ã¤ãâ- ç¥-¨¨¬¥ ª¥âìª

¢¨¤:∂k

3

d2z =

3

∂2z

dx + 2

∂2z

dxdy +

∂2z

dy,

∂x2

∂x∂y

∂y2

∂3z 3

∂3z

2

∂3z

2

∂3z

⮬¤¨ä䢨¤¥¥àdâ¥-zª®¢:æ¨=

« 3-

£dx® ¯®à浪+ 3

¤«ïdxäãdy-ªæ¨¨+ 3

âà¥ådxdy¯¥à¥¬+¥--ëådy¢ ,à §¢¥à-ã-

d3u =

∂x3

∂x2∂y

∂x∂y2

dx2dy

∂y3

∂3u

dx3+

∂3u

dy3 + +

∂3u

dz3 + 3

∂3u

∂x3

∂y

∂z3

∂x2∂y

∂ u

dxdy2

∂3u

dx2dz + 3

∂3u

dxdz2

+ 3

∂x∂y2

∂x∂z2

2

∂x2∂z

∂ u

∂3u

2

∂3u

Ǒਬ¥à 3. • ©â¨ ¤¨ä䥥à॥-dxæ¨dy« 2+-£3® ¯®à浪dydzäã+-ªæ¨¨3

dxdydz.

∂y2∂z

∂y∂z2

∂x∂y∂z

•¥è¥-¨¥: ‚â®à®©

-

« äã-ªæ¨¨

f (x; y) = exy .

ä®à¬ã«¥

2

f ¬® ¥â ¡ëâì ¢ëç¨á«¥- ¯®

∂2f

2

∂2f 2

∂2f

‚ëç¨á«¨¬ d¯¥fà¢ë(x;¥y,) =§

⥬(x¢â®àë; y)dx¥ +ç áâ-ëdy¥ ¯à®¨+ 2 §¢®¤-(ëx;¥y: )dxdy

∂x2

∂y2

∂x∂y

∂f

= yexy ,

∂f

= xexy ;

∂x

∂y

∂2f

∂2f

2

xy

∂2f

xy

xy ;

’ ª¨¬ ®¡à §®¬ ¯®«ãç2 xy¥¬:

= y e ,

= x

e ,

= e

+ xye

∂x2

∂y2

∂x∂y

d2f (x; y) = y2exy dx2

+ x2exy dy2 + (exy + xyexy )dxdy.

• ©â¨ ç áâ-ë¥ ¯à®¨§¢®¤-ë¥ ¯¥à¢®£® ¨ ¢â®à®£® ¯®à浪®¢ ®â á«¥¤ãîé¨å

äã-ªæ¨©:

(1) uŽâ¢=¥â:x4 + y4 − 4x2y2;

∂2u

∂2u

∂u

= 4x3 − 8xy2,

∂u

= 4y3 − 8x2y,

= 12x2 − 8y3,

= −16xy,

u =

∂x

∂y

∂x2

∂x∂y

;

(2)

∂2u

= 12y2 − 8x3.

∂y2

u = xy + x ;

1 , ∂u

1 , ∂2u = 2x

Žâ¢¥â:

y

∂u

x , ∂2u = 0, ∂2u

x

;

= y + y

= x −

= 1 −

.

(3)

u =

∂x

∂y

y2

∂x2

∂x∂y

y2

∂y2

y3

1 , ∂u

2 , ∂2u = 6x

Žâ¢¥â:y2

∂u

2x , ∂2u = 0, ∂2u

=

= −

= −

.

,

(4)

∂xx y

∂y

y3

∂x2

∂x∂y

y3

∂y2

y4

p

∂u = y

y2

, ∂u

xy

, ∂2u

3xy

Žâ¢¥â: x2 +

2

= −

= −

∂x

x + y23/

∂y

x2 + y 3/2

∂x2

x2 + y25/2

∂2u

=

y 2x2 − y2)

,

∂2u

=

x(x2 − 2y2)

.

(5)

∂x∂y

x2 + y25/2

∂y2

− x2 + y25/2

uŽâ¢=¥â:x sin(x + y);

∂u

∂u

∂2u

= sin(x + y) + x os(x + y),

= x os(x + y),

= 2 os(x + y) −

∂x

∂y

∂x2

∂2u

∂2u

äã•-(6)ªæ¨©:©â¨sin(¤¨ää+ ),¥à

¥-æ¨=«ëos(¯¥+ࢮ) £®

®

¯®à浪®¢= sin( ¤«ï+

).á«¥¤ãîé¨å

x

x

y

∂x∂y

x

y

− x ¨sin(¢â®à®x + y),£ ∂y2

−x

x

y

u =

os(x2)

;

y