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Basic theorems of differential calculus and their application

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This article gives you some easy ways to solve common. Basic theorems of differential calculus and their application


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Basic theorems of differential calculus and their application

Yarmetova Mukaddas Ruzimovna-UrSU teacher

Saidova Aziza Yuldash qizi-UrSU student

Annotation. This article gives you some easy ways to solve common. Basic theorems of differential calculus and their application

Key words: vector, inequality, angle, identity

We can often use theorems on derivative functions to solve some problems. These theorems play an important role in checking functions.

Theorem 1 (Farm theorem).)(xf

functionRX  given in the package.Xx 0 for the circumference of the point)0(),()( 000 +−=

XxxxU The following conditions must be met:

1))( 0xUx

 da)),()(()()( 00 xfxfxfxf 

2))( 0xf 

be available and limited. Then0)( 0 = xf is being..

Let's say,)( 0xUx

 in)()( 0xfxf  let it beObviously, in this case0)()( 0 − xfxf

will be.Conditionally)(xf function0x limited in point)( 0xf  yield.Then0

lim

lim

lim)( 000 xx

xfxf

xx

xfxf

xx

xfxf

xf xxxxxx −

= −→+→→ Will be.At the moment ,0xx  will be,0)('

lim0

)()( 0

=



+→ xf

xx

xfxf

xx

xfxf

xx0xx 

will be0)('

lim0

)()( 0

=



−→ xf

xx

xfxf

xx

xfxf

xx

from0)(' 0 =xf It

turns out that.

Theorem 2 (Roll theorem).Suppose,)(xf function],[ ba to meet the following conditions:

1)],,[)( baCxf 

2)),( bax  in)(xf  available and limited,

3))()( bfaf = let it be.),(0 bax 0)( 0 = xf

Conditionally],[)( baCxf  . According to Weierstrass's second theorem)(xf function],[ ba at its maximum and minimum values,21 , cс points]),[,( 21 bacc  found,]},,[|)(max{)( 1 baxxfcf =]},[|)(min{)( 2 baxxfcf = Has been.

If)()( 21 cfcf = been, Then],[ ba inconstxf =)( is being,),(0 bax  at0)(' 0 =xf

. If)()( 21 cfcf  be, thats)()( bfaf = because)(xf function)( 1cf and)( 2cf to at least one of the values],[ ba the interior of the segment)( 00 bxax  .reach the point According to the farm theorem0)( 0 = xf will be. ►

3-theorem (Lagranj by theorem).Suppose,)(xf function],[ ba at will be,fulfill the following conditions:

1)],[)( baCxf  ,

2)),( bax  at)(xf  the product is available and limitedIn that case it is so),( baс

poind found ,))(()()( abcfafbf −=− will be.

This)(

)()()( ax

ab

afxfxF −

−−= (1) Let's look at the function. This function satisfies all the conditions of the Roll theorem. At the same time, its a productab

xfxF −

−= )()( will be.According to the roll theorem, so)),(( baсc  the point is found,0)(' =cF

(2) Will be.

(1) and (2) from equations0

)( =

− ab

,that is))(()()( abcfafbf −=−

1-result. Let's say,)(xf function),( ba at)(xf  ,having a product),( bax  at0)( = xf

being to. Then),( bax  atconstxf =)( will be.),(, 0 baxx 

take, edgesx and0x in the segment)(xf using Lagrange's theorem on the functionconstxfxf == )()( 0 being found. ►

2-result.)(xf and)(xg function),( ba at)(xf  ,)(xg ,products),( bax  in)()( xgxf =

been. Then),( bax  inconstxgxf += )()( will be.

This is proof of the result)()( xgxf − by applying result 1 to the function.Theorem 4 (Cauchy Theorem). Let, and let the functions fulfill the following conditions.

1)],[)( baCxf  ,],[)( baCxg  ,

2)),( bax  da)(xf  va)(xg crops are available and limited;

3)),( bax  da0)(' xg will be. Then),( baс ,the point is found)( Willbe. First of all)()( agbg  We emphasize that because)()( agbg = if so, then according to Roll's theorem),( baс the point would be found0)( = cg would be This is contrary to condition 3) The following]),[()]()([

)()()( baxagxg

afxfxФ −

−−=),( baс

found point ,

7150)( = cФ

will be (3) Obviouly,)(

)()( xg

xfxФ 

−=

(4) (3) and (4) relationship0)(

)( =

− cg

that is)(

Example 1.Rxx  '',' for|'''||''sin'sin| xxxx −− prove the inequality.

Let's say,''' xx  will be.xxf sin)( = in]'','[ xx We apply Lagrange's theorem.That's it)'','( xxc  the point is that,)'''(|cos||''sin'sin| xxcxx −=− willbe.IfRt  at1|cos| t Given that, then from the above relationship,)'','(|'''||''sin'sin| Rxxxxxx −−

Being.

References

1.Mirzaahmedov.M , Sotiboldiyev T “Matematikadanolimpiadamasalalari” Toshkent-2003 y

2.Гелфанд.F “методы доказательства неравенства” Москва-1972г

3.Сивяшинский.В “доказательство тригонометрических неравенств” Москва-1985г

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