Nomehluko - Iyini? Indlela yokuthola umehluko umsebenzi?

Kanye nemikhiqizo imisebenzi yabo nomehluko - ke ezinye imiqondo eyisisekelo -calculus umehluko, engxenyeni eyinhloko sokuhlaziya zezibalo. Njengoba atehlukaniseki, bobabili ukuthi emakhulwini amaningana eminyaka kusetshenziswa kakhulu ekuxazululeni cishe zonke izinkinga ezaphakama Yiqiniso yomsebenzi zesayensi kanye lobuchwepheshe.

Ukuvela nomqondo umehluko

Ngokokuqala ngqá wakwenza kwacaca ukuthi umehluko onjalo, omunye wabasunguli (kanye Isaakom Nyutonom) umehluko calculus sezibalo esidumile German Gotfrid Vilgelm Leybnits. Ngaphambi ukuthi zezibalo ngekhulu le-17. esetshenziswa umqondo okungacaci kakhulu futhi ezingacacile abanye "okuphelele" kuncane yanoma yimuphi umsebenzi eyaziwa, emele lwenani elincane kakhulu njalo kodwa hhayi ilingane no-zero, ngezansi okuyinto uyayazisa umsebenzi akukwazi nje. Ngakho kwaba isinyathelo esisodwa kuphela isingeniso lwemibono increments encanyana ye kwemisebenzi kanye ziyakhula yabo ehlukene imisebenzi ezingase zivezwe ngokuya nemikhiqizo lakamuva. Futhi lesi sinyathelo sathathwa cishe kanyekanye okungenhla ososayensi ezimbili ezinkulu.

Ngokusekelwe isidingo ukubhekana esiphuthumayo esisebenzayo Mechanics izinkinga zalo isayensi ngokushesha ekuthuthukiseni umkhakha nobuchwepheshe, Newton kanye Leibniz wadala zezindlela ezivame lokuthola imisebenzi ngesilinganiso ushintsho (ikakhulukazi ngokuphathelene ijubane mechanical emzimbeni trajectory eyaziwa), okwaholela ukwethulwa imiqondo enjalo, njengoba umsebenzi esuselwe kanye umehluko, futhi wathola inkinga algorithm ephambene izixazululo njengoba aziwa ngayinye se (variable) ngesivinini siwela ukuthola indlela okuye kwabangela ukuba nomqondo ebalulekile Ala.

Ngo imisebenzi Leibniz futhi Newton umqondo wokuqala kwakubonakala sengathi nomehluko - kuyinto wakhona anyuswe-agumenti eziyisisekelo Δh ziyakhula Δu imisebenzi ukuthi singasetshenziswa ngempumelelo ukubala ukubaluleka lakamuva. Ngamanye amazwi, baye bathola ukuthi umsebenzi anyuswe kungaba nganoma yingasiphi (kusizinda yayo definition), uzwakalise ngokuya esuselwe yayo njengendlela Δu = y '(x) Δh + αΔh lapho Δh α - okusele ivame zero lapho Δh → 0, isheshe ngaphandle Δh langempela.

Ngokusho abasunguli izingabunjalo, nomehluko - yilokho kanye eside kuqala ziyakhula kwanoma iyiphi imisebenzi. Ngisho ngaphandle kokuba ecacile umqondo umkhawulo ukulandelanisa ziqondwa intuitively ukuthi ukubaluleka umehluko we esuselwe ivame ukusebenza lapho Δh → 0 - Δu / Δh → y '(x).

Ngokungafani Newton, owayeneminyaka ngokuyinhloko se-physics kanye apharathasi zezibalo kubhekwa njengento ithuluzi asizayo cwaningo nezinkinga eziningi ezingokomzimba, Leibniz banaka kakhudlwana toolkit, kufaka hlangana irherho kwezimpawu ezibukwayo futhi eqondakalayo amagugu zezibalo. Nguye ezihlongozwayo notation esezingeni nomehluko umsebenzi dy = y '(x) dx, dx, kanye esuselwe umsebenzi baphikisana ngokuthi ubuhlobo babo y' (x) = dy / dx.

Inchazelo yesimanje

Uyini umehluko ngokuya nesifundo sezibalo zanamuhla? Kuhlobene eduze kakhulu ukuba nomqondo anyuswe onhlobonhlobo. Uma variable y ithatha Inani lokuqala y y = _1, ke y = y _2, y umehluko ₂ ─ y ₁ ubizwa ngokuthi wokwengeza Inani y. Isengezo ungaba muhle. engemihle futhi zero. Igama elithi "anyuswe" lumiswa Δ, Δu ukuqopha (funda 'delta y') lisho ukubaluleka y wokwengeza. ngakho Δu = y ₂ ─ y _1.

Uma kungenzeka ukubaluleka Δu umsebenzi ngokungenasizathu y = f (x) lokumelwa njengoba Δu = A Δh + α, lapho A akukho ukuncika Δh, t. E. A = const for x unikezwa, kanye α eside lapho Δh → 0 ivame ukuba kuba ngisho ngokushesha kunokuba Δh uqobo, khona-ke ( "master") sokuqala elithi ezilinganiselwe Δh, futhi ngenxa y = f (x) umehluko, okhonjiswe dy noma DF (x) (funda "y de", "de EFF kusukela X"). Ngakho-ke nomehluko - a "main" komugqa mayelana izingxenye ziyakhula imisebenzi Δh.

incazelo mechanical

Let s = f (t) - ibanga emgceni locondzile ezihambayo iphuzu impahla kusukela isikhundla sokuqala (t - isikhathi sokuhamba). Okwengeza Δs - kuyindlela iphuzu ngesikhathi isikhawu Δt, kanye ds umehluko = f '(t) Δt - lendlela, okuyilapho owawuzoba ngoba esifanayo Δt, uma lalilokhu f isivinini' (t), okuye kwafinyelelwa kuyo isikhathi t . Lapho i kuncane Δt ds endleleni engokomfanekiso ihlukile Δs langempela infinitesimally kokuba lesezingeni lelisetulu maqondana Δt. Uma ijubane ngesikhathi t ayilingani kuqanda, cishe ukubaluleka ds kunikeza ezincane noma ukuchema kuzo iphuzu.

incazelo weJiyomethri

Vumela L umugqa iyona igrafu y = f (x). Khona-ke Δ x = MQ, Δu = QM '(bheka. UMdwebo ngezansi). Tangent MN iphula Δu uthathe izingxenye ezimbili, umbz futhi NM '. Okokuqala futhi Δh kuyinto ezilinganiselwe umbz = MQ ∙ TG (i-engeli QMN) = Δh f '(x), t. E umbz kuyinto umehluko dy.

Ingxenye yesibili umehluko Δu NM'daet ─ dy, lapho Δh → 0 NM obuphelele 'incipha ngisho ngokushesha kunokuba anyuswe lwempikiswano, ie it has oda nobuncane ephakeme kunaleyo Δh. Kulokhu, uma f '(x) ≠ 0 (tangent non-parallel OX) izingxenye QM'i umbz okulingana; ngamanye amagama NM 'incipha ngokushesha (oda nobuncane of ephakeme yayo) ngaphandle Imininingwane anyuswe Δu = QM'. Lokhu kubonakala kahle ku-Figure (esondela ingxenye M'k M NM'sostavlyaet zonke ezincane ngamaphesenti QM 'ingxenye).

Ngakho, imidwebo Ukwahlukaniswa umsebenzi ngokungenasizathu uyalingana anyuswe we ahlelembise ka tangent.

Esuselwe futhi umehluko

Isici eThemini yokukhuluma anyuswe umsebenzi ilingana ukubaluleka f yayo esuselwe '(x). Ngakho, ubuhlobo ezilandelayo - dy = f '(x) Δh noma DF (x) = f' (x) Δh.

Kuyaziwa ukuthi anyuswe lwempikiswano ezimele ilingana umehluko yayo Δh = dx. Ngenxa yalokho, singasho ukubhala: f '(x) dx = dy.

Ukuthola (ngezinye izikhathi kuthiwa 'isinqumo ") nomehluko wenziwa imithetho efanayo for the nemikhiqizo. Uhlu wabo ngezansi.

Kuyini jikelele kakhulu: anyuswe lwempikiswano noma umehluko yayo

Lapha kubalulekile ukwenza amanye kucaciswa. Ukumelwa ukubaluleka f '(x) umehluko Δh kungenzeka uma ucabangela x njengephuzu lokuphikisa. Kodwa umsebenzi kungaba eziyinkimbinkimbi, lapho x kungaba umsebenzi t agumenti. Khona-ke umfanekiso nkulumo umehluko f '(x) Δh, njengoba umthetho, akunakwenzeka; ngaphandle uma sokuncika komugqa x = at + b.

Ngokuqondene f ifomula '(x) dx = dy ke esimweni ezimele agumenti x (ke dx = Δh) esimweni ukwencika parametric x t, kuba umehluko.

Ngokwesibonelo, inkulumo ethi 2 x Δh iwukuba y = x ² Umehluko yayo lapho x impikiswano. Thina manje x = t ² futhi bacabanga t agumenti. Khona-ke y = x ² = t ^4.

Lokhu kulandelwa (t + Δt) ² = t ² + 2tΔt + Δt ^2. Yingakho Δh = 2tΔt + Δt ^2. Yingakho: 2xΔh = 2t ² (2tΔt + Δt ^2).

Le nkulumo akuyona wakhona Δt, ngakho manje 2xΔh akuyona umehluko. Kungatholakala kubhalwe kusukela equation y = x ² = t ^4. Kuyinto dy alinganayo = 4t ³ Δt.

Uma sithatha 2xdx isisho, kuba y umehluko = x ² nganoma yisiphi t agumenti. Ngempela, lapho x = t ² ukuthola dx = 2tΔt.

Ngakho 2xdx = 2t ² 2tΔt = 4t ³ .DELTA.t, t. E. nomehluko inkulumo eqoshiwe eziguquguqukayo ezimbili ezihlukene luqondane.

Njengoba ezothatha isikhundla ziyakhula nomehluko

Uma f '(x) ≠ 0 ke Δu futhi dy okulingana (lapho Δh → 0); uma f '(x) = 0 (okusho futhi dy = 0), awayona okulingana.

Ngokwesibonelo, uma y = x ^2, bese Δu = (x + Δh) ² ─ x ² = 2xΔh + Δh ² futhi dy = 2xΔh. Uma x = 3, khona-ke siba Δu = 6Δh + Δh ² futhi dy = 6Δh ukuthi kukhona okulingana ngenxa Δh ² → 0, lapho x = 0 Inani Δu = Δh ² futhi dy = 0 akuzona okulingana.

Ngokuqinisekile, leli qiniso kanye nesakhiwo okulula umehluko (m. E. Linearity maqondana Δh), ngokuvamile sisetshenziswa ukubala eseduzane, ngoba kucatshangwa ukuthi Δu ≈ dy ngoba Δh encane. Thola umsebenzi umehluko ngokuvamile kulula kuka ukubala inani esiqondile wokwengeza.

Ngokwesibonelo, sinalo metallic cube ne onqenqemeni x = 10.00 cm. On amalahle onqenqemeni yinde ku Δh = 0,001 cm. Lwakhula kanjani umthamo cube V? Sine V = x ^2, ukuze DV = 3x ² = Δh 3 ∙ ∙ ⁰ 10 2/01 = 3 (cm ^3). Ukwandiswa ΔV okulingana umehluko DV, ukuze ΔV = 3 cm ^3. ukubala Okugcwele ebengakhipha ³ ΔV = 10,01 ─ Mashi ¹⁰ = 3.003001. Kodwa umphumela wonke amadijithi ngaphandle alithembekile lokuqala; Ngakho-ke, kusadingeka wokuqoqa kuya ku-3 cm ^3.

Ngokusobala, le ndlela kubalulekile kuphela uma kungenzeka ukulinganisa ukubaluleka wanginika yisiphambeko.

umsebenzi Ukwahlukaniswa: izibonelo

Ake sizame ukuthola umehluko y umsebenzi = x ^3, lokuthola esuselwe. Asiphumele agumenti anyuswe Δu kanye nokuchaza.

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

Lapha, Coefficient A = 3x ² akuxhomekile Δh, ukuze eside sokuqala Δh ezilinganiselwe, omunye ilungu 3xΔh Δh ² + ³ lapho Δh → 0 incipha ngokushesha kunokuba anyuswe lwempikiswano. Ngenxa yalokho, ilungu 3x ² Δh iyona umehluko y = x ^3:

dy = 3x ² Δh = 3x ² dx noma d (x ³⁾ = 3x ² dx.

Kulokhu d (x ³⁾ / dx = 3x ^2.

Dy Manje ukuthola y umsebenzi = 1 / x yi esuselwe. Khona-ke d (1 / x) / dx = ─1 / x ^2. Ngakho-ke dy = ─ Δh / x ^2.

Nomehluko imisebenzi algebraic eziyisisekelo banikwa ngezansi.

izibalo lesilinganisiwe usebenzisa umehluko

Ukuze ukuhlola f umsebenzi (x), kanye esuselwe yayo f '(x) e ngokuvamile x = a kunzima, kodwa ukwenza okufanayo eduze x = a akulula. Khona-ke basisize okuvela kuyo le nkulumo eseduze

f (a + Δh) ≈ f '(a) Δh + f (a).

Lokhu kunikeza i value silinganisiwe lo mcimbi ozoba se ziyakhula ezincane ngokusebenzisa umehluko yayo Δh f '(a) Δh.

Ngakho-ke, lokhu ifomula enikeza inkulumo ethile eseduze ukuze umsebenzi ekupheleni iphuzu ingxenye ubude Δh njengoba isamba ukubaluleka kwalo ngesikhathi esizoqala kuso ukubala ingxenye (x = a) kanye umehluko e okufanayo yekucala. Ukunemba indlela yokunquma abakuzuzile umsebenzi ngezansi sibonisa umdwebo.

Nokho eyaziwa nenkulumo eqondile ukubaluleka umsebenzi x = a + Δh inikezwe ngefomula ziyakhula ezilinganiselwe (noma, kungenjalo, ifomula Lagrange sika)

f (a Δh +) ≈ f '(ξ) Δh + f (a),

lapho iphuzu x = a + ξ uku isikhawu kusuka x = a x = a + Δh, nakuba isikhundla salo ngqo alwaziwa. Ifomula ngqo ivumela ukuhlola isiphambeko ifomula eseduze. Uma sibeka ku Lagrange ifomula ξ = Δh / 2, nakuba eyeka ukuba olunembile, kodwa linikeza, njengoba umthetho, indlela engcono kakhulu kunaleyo Inkulumo yokuqala ngokuya umehluko.

Ukuhlolwa amafomula iphutha ngokusebenzisa umehluko

Ukulinganisa amathuluzi , isimiso, olungalungile, futhi imenze idatha isilinganiso elihambisana iphutha. Basuke izimpawu nokukhawulela iphutha ngokuphelele, noma, ngamafuphi, iphutha umkhawulo - HIV, beyizidlova ngokucacile iphutha inani eliphelele (noma okungenani kulinganiswe nakho). Linganisa iphutha isihlobo ubizwa ngokuthi quotient etholwe ngokuyihlukanisa by inani eliphelele ukubaluleka kulinganiswa.

Ake ngqo ifomula y = f (x) umsebenzi esetshenziswa vychislyaeniya y, kodwa ukubaluleka x kuwumphumela nesilinganiso, ngakho kuletha iphutha y. Khona-ke, ukuthola nokukhawulela iphutha ngokuphelele │Δu│funktsii y, kusetshenziswa indlela

│Δu│≈│dy│ = │ f '(x) ││Δh│,

lapho │Δh│yavlyaetsya iphutha abekelwe emaceleni agumenti. │Δu│ ubuningi kumele imikiswe phezulu, njengoba ukubala olungalungile ngokwawo esikhundleni the isengezo kwi ekubalweni umehluko.