13.6-bo'lim : Zanjir qoidasi
11. Quyidagi vaziyat uchun \({f_{u\,u}}\) ni aniqlang.
\[f = f\left( {x,y} \right)\hspace{0.5in}x = {u^2} + 3v,\,\,\,\,\,\,\,y = uv\]Yechim
Bunday muammolar har doim birinchi qarashda sirli ko'rinadi. Bu, ehtimol, biz funktsiyaning o'zi nima ekanligini bilmasligimizdir. Bu haqiqatan ham muammo emas. Bu shunchaki javoblar biroz chalkash bo'lishi mumkinligini anglatadi, chunki biz kamdan-kam hollarda soddalashtirish yo'lida ko'p ish qila olamiz.
Shunday qilib, bu erda birinchi qadam birinchi hosilani olishdir va bu quyidagi zanjir qoidasi formulasini talab qiladi.
\[{f_u} = \frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\]Mana birinchi hosila,
\[\frac{{\partial f}}{{\partial u}} = \frac{{\partial f}}{{\partial x}}\left[ {2u} \right] + \frac{{\partial f}}{{\partial y}}\left[ v \right] = 2u\frac{{\partial f}}{{\partial x}} + v\frac{{\partial f}}{{\partial y}}\]Bizning javobimizdagi "noma'lum" lotinlar haqida hayajonlanmang. Bunday muammolarda ular doimo mavjud bo'ladi.
2-qadam
Endi, eslatmalarda qilganimizdek, keling, yuqoridagi javobni quyidagicha qayta yozamiz:
\[\frac{\partial }{{\partial u}}\left( f \right) = 2u\frac{\partial }{{\partial x}}\left( f \right) + v\frac{\partial }{{\partial y}}\left( f \right)\]Ushbu qayta yozish bilan biz endi \(x\) va \(y\) ning har qanday funksiyasini \(u\) ga nisbatan \(x = {u^2} + 3v\) va \(y = uv\) ni farqlash uchun “formula”ga egamiz. Boshqacha qilib aytganda, bizda bunday funktsiya mavjud bo'lganda, biz qilishimiz kerak bo'lgan yagona narsa qavs ichidagi \(f\) ni bizning funktsiyamiz qanday bo'lishidan qat'i nazar, o'zgartirishdir. Bu bizga oxir-oqibat kerak bo'ladi.
3-qadam
Endi ikkinchi hosilani olamiz. Biz bilamizki, biz ikkinchi hosilani quyidagicha topamiz,
\[{f_{u\,u}} = \frac{{{\partial ^2}f}}{{\partial {u^2}}} = \frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial u}}} \right) = \frac{\partial }{{\partial u}}\left( {2u\frac{{\partial f}}{{\partial x}} + v\frac{{\partial f}}{{\partial y}}} \right)\]4-qadam
Endi eslang, \(\frac{{\partial f}}{{\partial x}}\) va \(\frac{{\partial f}}{{\partial y}}\) \(x\) va \(y\) funktsiyalari bo'lib, ular o'z navbatida muammo bayonotida aniqlangan \(u\) va \(v\) shartlarida aniqlanadi. Bu shuni anglatadiki, mahsulot qoidasini birinchi muddatda bajarishimiz kerak, chunki u \(u\) ni o'z ichiga olgan ikkita funktsiyaning mahsulotidir. Bu holda biz ikkinchi atamani hosil qilishimiz shart emas, chunki bu atamadagi birinchi funktsiya faqat \(v\) ni o'z ichiga oladi.
Mana bu ish,
\[{f_{u\,u}} = 2\frac{{\partial f}}{{\partial x}} + 2u\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + v\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial y}}} \right)\]Funktsiya faqat \(x\) va \(y\) shartlarida aniqlanganligi sababli biz ikkinchi haddadagi \(u\) va \(x\) hosilalarini “aralash tartibli” ikkinchi hosilaga “birlashtira olmaymiz”. Xuddi shu sababga ko'ra biz \(u\) va \(y\) hosilalarini uchinchi muddatda "birlashtira olmaymiz".
Ushbu holatlarning har birida bizdan \(x\) va \(y\) funksiyalarini \(u\) ga nisbatan farqlash talab qilinadi, bunda \(x\) va \(y\) \(u\) va \(v\) nuqtai nazaridan aniqlanadi.
5-qadam
Endi 2-bosqichdagi "formulani" eslang,
\[\frac{\partial }{{\partial u}}\left( f \right) = 2u\frac{\partial }{{\partial x}}\left( f \right) + v\frac{\partial }{{\partial y}}\left( f \right)\]Eslatib o'tamiz, bu bizga \(x\) va \(y\) ning har qanday funktsiyasini \(u\) ga nisbatan qanday ajratish mumkinligini aytadi, chunki \(x\) va \(y\) \(u\) va \(v\) nuqtai nazaridan aniqlangan bo'lsa, ular ushbu muammoda mavjud.
Yaxshiyamki, biz uchun \(\frac{{\partial f}}{{\partial x}}\) ham, \(\frac{{\partial f}}{{\partial y}}\) ham \(x\) va \(y\) funktsiyalari bo'lib, ular o'z navbatida \(u\) va \(v\) nuqtai nazaridan aniqlangan, chunki ular bizga kerak. Bu shuni anglatadiki, biz ushbu "formulani" 4-bosqich natijasidagi hosilalarning har biri uchun quyidagicha ishlatishimiz mumkin,
\[\begin{align*}\frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial x}}} \right) & = 2u\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + v\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial x}}} \right) = 2u\frac{{{\partial ^2}f}}{{\partial {x^2}}} + v\frac{{{\partial ^2}f}}{{\partial y\partial x}}\\ \frac{\partial }{{\partial u}}\left( {\frac{{\partial f}}{{\partial y}}} \right) & = 2u\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial y}}} \right) + v\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial y}}} \right) = 2u\frac{{{\partial ^2}f}}{{\partial x\partial y}} + v\frac{{{\partial ^2}f}}{{\partial {y^2}}}\end{align*}\]6-qadam
OK, hozir qilishimiz kerak bo'lgan yagona narsa 5-bosqichdagi natijalarni 4-bosqich natijasiga qo'yishdir va biz bajaramiz.
\[\begin{align*}{f_{u\,u}} & = 2\frac{{\partial f}}{{\partial x}} + 2u\left[ {2u\frac{{{\partial ^2}f}}{{\partial {x^2}}} + v\frac{{{\partial ^2}f}}{{\partial y\partial x}}} \right] + v\left[ {2u\frac{{{\partial ^2}f}}{{\partial x\partial y}} + v\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right]\\ & = 2\frac{{\partial f}}{{\partial x}} + 4{u^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 2uv\frac{{{\partial ^2}f}}{{\partial y\partial x}} + 2uv\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {v^2}\frac{{{\partial ^2}f}}{{\partial {y^2}}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2\frac{{\partial f}}{{\partial x}} + 4{u^2}\frac{{{\partial ^2}f}}{{\partial {x^2}}} + 4uv\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {v^2}\frac{{{\partial ^2}f}}{{\partial {y^2}}}}}\end{align*}\]E'tibor bering, biz ushbu muammo uchun ikkita aralash tartibli xususiy hosila teng deb taxmin qildik va shuning uchun bu shartlarni birlashtirdik. Agar siz buni qabul qila olmasangiz yoki buni qabul qilishni xohlamasangiz, ikkinchi qator javob bo'ladi.