Ukwehluka - yintoni na le nto? Indlela ukufumana umahluko lo msebenzi?

Kunye yezabelo imisebenzi yazo ukwehluka - it ezinye nezimaphambili ezingundoqo ye calculus umahluko, kwisiqendu esiyintloko uhlalutyo zezibalo. Njengoko ezingenakwahlulwa, bobabini kwiinkulungwane eziliqela zazisetyenziswa ekusombululeni phantse zonke iingxaki eyayikho ikhondo lomsebenzi lwenzululwazi nobugcisa.

Ukuvela isigama umehluko

Kuba okokuqala wakwenza kwacaca ukuba umahluko onjalo, omnye wabaseki (kunye Isaakom Nyutonom) ezohlukileyo calculus edumileyo yezibalo German Gotfrid Vilgelm Leybnits. Ngaphambi kokuba zezibalo kwenkulungwane ye-17. kusetyenziswa mbono ayicacanga kakhulu kwaye ngokungacacanga ezinye "epheleleyo 'omncane nawuphi na umsebenzi eyaziwa, emele ixabiso elincinane kakhulu rhoqo kodwa akalingani zero, ngaphantsi apho kuza kuba lo msebenzi nje. Ngoko ke inyathelo elinye kuphela intshayelelo kwengcinga yokunyuka omncane umsebenzi iimpikiswano kunye nokunyuswa bazo imisebenzi anokuthi luboniswe ngokwemiqathango avela kule yokugqibela. Kwaye eli nyathelo sathathwa phantse ngaxeshanye ngasentla oosonzululwazi ezimbini ezinkulu.

Ngokusekelwe kwi imfuno ukujongana ubucukubhede iingxaki engxamisekileyo ezenziwayo ezijongene inzululwazi ekuphuhliseni shishino kunye nobuchwepheshe ngokukhawuleza, Newton Leibniz wadala iindlela eziqhelekileyo lokufumana imisebenzi izinga tshintsho (ingakumbi ngokumalunga isantya ngomatshini umzimba isikhokelo eyaziwa), nto leyo eyakhokelela ekubeni kwintshayelelo kweekhonsepthi ezinjalo, njengoko umsebenzi esukela kunye umahluko, kwaye wafumana izisombululo ingxaki algorithm yokuguquleka eyaziwa nganye se (variable) ngesantya onqamlezwe ukufumana indlela eye yabangela ukuba ingqikelelo ebalulekile Ala.

Xa imisebenzi Leibniz kunye Newton nomqondo kuqala kwabonakala ukuba nomahluko - kuba ngokomlinganiselo ukwenyuselwa wempikiswano ezisisiseko Δh songezo imisebenzi Δu enokusetyenziswa ngempumelelo ukubala ixabiso yokugqibela. Ngamanye amazwi, baye bafumanisa ukuba umsebenzi okukodwa kungenzeka nayiphi na indawo (kombuso yayo kwenkcazelo) sikubonisa yayo yemvelaphi Δu = y '(x) Δh + αΔh zombini apho α Δh - intsalela, ukulima zero Δh → 0, ngokukhawuleza kakhulu kunokuba eyona i Δh.

Ngokutsho wabaseki uhlalutyo zezibalo, nomahluko - oku kanye kwikota yokuqala ngokwemizuzu nayiphi na eminye imisebenzi. Nokuba ngaphandle kokuba ulandelelwano ingqiqo umda ichazwe ngokucacileyo ziyaqondwa ungaziva ukuba ixabiso mahluko ukuba ithe kudla ukusebenza xa Δh → 0 - Δu / Δh → y '(x).

Ngokungafaniyo Newton, ngubani ngokuyintloko yemvelo kunye nezixhobo mathematical ingqalelo njengesixhobo abangabancedani ukuba isifundo iingxaki emzimbeni, Leibniz yakholwa ngakumbi kule izixhobo, kubandakanywa inkqubo weesimboli ezibonwayo kunye ukuyiqonda amaxabiso zemathematika. Yaba nguye owoyiswayo ubhalo umgangatho ukohluka umsebenzi dy = y '(x) DX, DX, kwaye yemvelaphi umsebenzi argument njengoko y ulwalamano lwabo' (x) = Dy / DX.

Le nkcazelo yanamhlanje

Yintoni umahluko ngokwemiqathango yemathematika namhlanje? It ngokusondeleyo ukuya ingqiqo anyuswe variable. Ukuba variable y kuthatha ixabiso lokuqala y y = _1, ngoko y = y _2, umahluko y ₂ ─ y ₁ kuthiwa ixabiso ukwenyuselwa y. Le kokupha ingaba HIV. ezimbi ne zero. Igama elithi "Nciphisa" uchongwe Δ, Δu Ngokushicilela (funda 'y olwandle') ibonisa ixabiso ukwenyuselwa y. ngoko Δu = y ₂ ─ y _1.

Ukuba ixabiso Δu umsebenzi ngendlela engaqondakaliyo y = f (x) angamelwa njengoko Δu = A Δh + α, apho A kungekho uxhomekeke Δh, t. E. A = nengngqi ngenxa x enikwa, kwaye α elithi xa Δh → 0 kuyaphela ikho nokubakho ngokukhawuleza kunokuba yangempela Δh, ngoko ke lokuqala ( "inkosi") igama umlinganiselo Δh, kwaye ke (x) umehluko y = f, ezipalini zibonakaliswe dy okanye iDF (x) (funda "y de", "de EFF ukusuka X"). Ngenxa yoko ukwehluka - a yomgama 'ephambili "mayela amacandelo yokunyuka imisebenzi Δh.

incazelo mechanical

Makhe s = f (t) - umgama kumgca othe ngqo ezihamba kwindawo izinto ukusuka kwindawo yokuqala (t - ixesha lokuhamba). Nciphisa ii - yeyona ngongoma indlela ngexesha ixesha lokuphumla Δt, kwaye DS umehluko = f '(t) Δt - le ndlela, apho ubuya kubanjwa ngenxa xesha Δt, ukuba igcinwe le f isantya' (t), kufikelelwe ngexesha t . Xa omncane Δt DS indlela wentelekelelo olwahlukileyo ii actual ama ukuba umyalelo eliphezulu ngokuphathelele Δt. Ukuba isantya ngexesha t akalingani zero, i-DS sentengo unika indawo icala encinane.

incazelo lwezibalo

Vumela L line igrafu ye y = f (x). Ke Δ x = MQ, Δu = QM '(bona. Umfanekiso ngezantsi). Ilayini ezibambanayo kodwa azikweli MN breaks Δu kunqunyulwa ibe ngamaqela amabini, QN kunye NM '. Okokuqala Δh na umlinganiselo QN = MQ ∙ kubatshaba- (isibiyeli QMN) = Δh f '(x), t. E QN na umehluko Dy.

Isigaba sesibini umahluko Δu NM'daet ─ Dy, xa Δh ubude → 0 NM 'kunciphisa oluikhawuleza ngaphezu kokupha Kwempikiswano, oko iye umyalelo ubuncinane ngaphezulu Δh. Kulo mzekelo, ukuba f (x) ≠ 0 (ebaxiweyo non-parallel inkabi) ziqwempu QM'i QN alingana '; ngamanye amazwi NM awusafuni ngokukhawuleza (umyalelo ubuncinane of ephakamileyo ayo), ngaphandle kokupha iyonke Δu = QM '. Oku kubonakala Umfanekiso (esondela ingxenye M'k M NM'sostavlyaet zonke ezincinane ipesenti QM 'ingxenye).

Ngoko ke, ngokucacile ezohlukileyo umsebenzi ngendlela engaqondakaliyo lilingana sokuba le kwemisebenzi ilayini ezibambanayo kodwa azikweli.

Uvumelane kanye umehluko

Eyona kwikota yokuqala ka umsebenzi ibinzana okukodwa elingana nexabiso f yayo yemvelaphi '(x). Ngaloo ndlela, ubudlelwane ilandelayo - dy = f '(x) Δh okanye iDF (x) = f' (x) Δh.

Yinto eyaziwayo ukuba ukwenyuselwa Kwempikiswano ozimeleyo ilingana umehluko yayo Δh = DX. Ngako oko, siya kubhala: f '(x) DX = Dy.

Ukufumana (ngamanye amaxesha kuthiwa i "Isigqibo") nomahluko lwenziwa yi imigaqo ngokufanayo yezabelo. Uluhlu kubo inikwe apha ngezantsi.

Yintoni jikelele ngakumbi: ukwenyuselwa Kwempikiswano okanye umehluko yayo

Apha kuyimfuneko ukwenza ezinye kucaciswa. Ukumelwa ixabiso f '(x) mahluko Δh kunokwenzeka xa kuqwalaselwa x ye, classes. Kodwa msebenzi inokuba entsokothileyo, apho x ingaba umsebenzi we t ingxabano. Ke imbonakaliso yebinzana mahluko of f '(x) Δh, njengokuba umthetho, akunakwenzeka; ngaphandle kwakwimeko ukuxhomekeka yomgama x = xa + b.

Njengoko wokubala f '(x) DX = Dy, ngoko kwimeko ezimeleyo argument x (ngoko DX = Δh) kwimeko ukuxhomekeka parametric lwe x t, kuba umehluko.

Umzekelo, ibinzana 2 x Δh kukuba y = x ² umehluko yayo xa x lo mbandela. Ngoku x = t ² bacinga t ingxabano. Emva koko y = x ² = t ^4.

Oku kulandelwa (t + Δt) ² = t ² + 2tΔt + Δt ^2. Ngoko Δh = 2tΔt + Δt ^2. Ngenxa yoko: 2xΔh = 2t ² (2tΔt + Δt ^2).

Eli binzana ayikho umlinganiselo ukuba Δt, ngoko ke ngoku 2xΔh akayi umehluko. Izakufunyanwa ukusuka equation y = x ² = t ^4. Kuyinto dy ngokulinganayo = 4t ³ Δt.

Ukuba ngaba sithatha 2xdx ibinzana, kodwa umahluko y = x ^{2 ubuncinane} kuzo naziphi t ingxabano. Eneneni, xa x = t ² ukufumana DX = 2tΔt.

Ngoko 2xdx = 2t ² 2tΔt = 4t ³ .DELTA.t, t. E. nomahluko elithi obhalwe ngeyantlukwano ezimbini ezahlukeneyo idibana.

Ukufaka ngo ukwehluka

Ukuba f '(x) ≠ 0, ngoko Δu kunye dy elingana (xa Δh → 0); ukuba f '(x) = 0 (intsingiselo dy = 0), kungenjalo ilingana.

Umzekelo, xa y = x ^2, ngoko Δu = (x + Δh) ² ─ x ² = 2xΔh + Δh ² kunye dy = 2xΔh. Ukuba x = 3, ke ngoko siye Δu = 6Δh + Δh ² kunye dy = 6Δh ukuba alingana ngenxa Δh ² → 0, xa x = 0 ixabiso Δu = Δh ² kunye dy = 0 azikho ilingana.

Le nyaniso, kunye nesakhiwo olulula umahluko (m. E. Linearity ngokuphathelele Δh), isoloko isetyenziswa ukubala ilingane, kwingcinga yokuba Δu ≈ dy ngenxa amancinci Δh. Fumana umsebenzi umahluko idla kulula ukubala ixabiso ngqo le okukodwa.

Ngokomzekelo, siye sibe ebuntsimbi ityhubhu ngohlangothi x = 10.00 cm. Xa ngokushushubeza ngohlangothi elide kwi Δh = 0.001 cm. Ukwanda njani umthamo cube V? Thina V = x ^2, ukuze ivuma = 3x ² = Δh 3 ∙ ∙ ⁰ 10 2/01 = 3 (cm ^3). Ukwanda ΔV elingana umehluko ivuma, kangangokuba ΔV = 3 cm ^3. yokubala Full kunganika ΔV = 10,01 ─ ³ 10 ³ = 3.003001. Kodwa ke ngenxa yokuba zonke manharhu ngaphandle ongathembekanga yokuqala; Ngoko ke, kubalulekile noko siya kwi-3 cm ^3.

Ngokucacileyo, le ndlela iluncedo kuphela xa kunokwenzeka ukuqikelela ixabiso singaneli kunabela sisiphoso.

umsebenzi umehluko: imizekelo

Makhe zama ukufumana umahluko lomsebenzi y = x ^3, ngokufumana leyo ithe. Makhe ukunika ingxoxo okukodwa Δu kunye nokuchaza.

Δu = (Δh + x) ³ ─ x ³ = 3x ² + Δh (Δh 3xΔh ² + ^3).

Apha, 'ngumlingani A = 3x ² akuxhomekekanga Δh, kangangokuba kwikota yokuqala umlinganiselo Δh, elinye ilungu 3xΔh Δh ² + ³ xa Δh → 0 kunciphisa ngokukhawuleza ngaphezu isongezo sika ingxabano. Ngenxa yoko, ilungu 3x ² Δh ke umahluko ye y = x ^3:

dy = 3x ² Δh = 3x ² DX okanye d (x ³⁾ = 3x ² DX.

Enigcoba d (x ³⁾ / DX = 3x ^2.

Dy Ngoku ufumane umsebenzi y = 1 / x yi ithe. Emva koko d (1 / x) / DX = ─1 / x ^2. Ngoko dy = ─ Δh / x ^2.

Ukwehluka imisebenzi esisiseko aljibra inikwe apha ngasezantsi.

izibalo elimisiweyo Ukusebenzisa umehluko

Ukuze nokuvavanya umsebenzi f (x), kwaye uvumelane yayo f '(x) kwi x = a kudla ngokuba nzima, kodwa ukuba benze okufanayo kwi endhawini ye x = a ayikho lula. Emva koko kuza kubanceda ibinzana eyiyona

f (a + Δh) ≈ f '(a) Δh + f (a).

Oku kunika inani elisondeleyo kwitheko yokunyuka asakhasayo umehluko layo Δh f '(a) Δh.

Ngoko ke, le fomyula unika ibinzana omalunga ukuba umsebenzi ekupheleni ingongoma inxalenye ubude Δh njengesixa ixabiso layo kwi loqalo lwencopho lwenxalenye (x = a) kunye umahluko kwi loqalo lwencopho efanayo. Ukuchaneka indlela yokuqingqa amaxabiso umsebenzi ngezantsi ibonisa umzobo.

Nangona kunjalo eyaziwa kunye ibinzana kanye ukuba ixabiso lomsebenzi x = a + Δh anikwe ifomula nokunyuswa aphelayo (okanye, kungenjalo, ifomula आयडेंटीफिकेशन iSazisi kaThixo)

f (a Δh +) ≈ f '(ξ) Δh + f (a),

apho indawo x = a + ξ na lekhefu evela x = a ku x = a + Δh, nangona indawo yayo ngqo ayaziwa. Le ndlela yokubala ngqo ivumela ukuhlola ngesiphoso formula uqikelelo. Ukuba sibeka kwi आयडेंटीफिकेशन iSazisi formula ξ = Δh / 2, nangona liyayeka ukuba zichanile, kodwa, njengokuba umthetho, indlela engcono kakhulu ibinzana lokuqala ngokwemiqathango umahluko.

ifomula Evaluation imposiso ngokusebenzisa umehluko

Wokulinganisa izixhobo , umgaqo, engeyizizo, uze uzise idatha yokukala ehambelana yemposiso. Bona luphawulwa ngokunciphisa isiphoso elililo, okanye, ngamafutshane, impazamo umda - HIV, kunene ngokucacileyo impazamo ixabiso elililo (okanye eyona ilingana kuyo). Ukucutha impazamo isihlobo ibizwa QUOTIENT afunyenwe ngokwahlula oko ixabiso elililo ixabiso elilinganiselweyo.

Makhe ngqo ifomula y = f (x) umsebenzi ezisetyenziswa vychislyaeniya y, kodwa ke ixabiso x na isiphumo womlinganiselo, ngoko kuzisa impazamo y. Emva koko, ukuba ufumane i umda impazamo ingundoqo │Δu│funktsii y, usebenzisa ifomula

│Δu│≈│dy│ = │ f '(x) ││Δh│,

apho │Δh│yavlyaetsya Imposiso ebekiwe ingxabano. │Δu│ ubungakanani kufuneka uzakufakwa phezulu, njengoko ubalo alunalo ngokwayo ngokokubekwa ukwenyuselwa kwi ukubala umahluko.