13.6-bo'lim: Zanjir qoidasi
Biz oxirgi ikki bo'limda bitta o'zgaruvchining funktsiyalari uchun standart zanjir qoidasidan foydalandik. Endi zanjir qoidasini yanada murakkab holatlarga uzaytirish vaqti keldi. Buni amalga oshirishdan oldin, keling, bitta o'zgaruvchining funktsiyalari uchun zanjir qoidasi belgilarini ko'rib chiqaylik.
Ko'pchilikka tanish bo'lgan belgi quyidagicha.
\[F\left( x \right) = f\left( {g\left( x \right)} \right)\hspace{0.5in}F'\left( x \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right)\]Muqobil yozuv mavjud, ammo, ehtimol, Calculus I-da ko'p ishlatilmasa ham, bu nuqtada qulayroqdir, chunki u biz ushbu bo'limda ishlatmoqchi bo'lgan belgiga mos keladi. Mana.
\[{\mbox{If}}\hspace{0.25in}y = f\left( x \right)\hspace{0.25in}{\mbox{and }}\hspace{0.25in}x = g\left( t \right)\hspace{0.25in}{\mbox{then }}\hspace{0.25in}\,\,\,\,\frac{{dy}}{{dt}} = \frac{{dy}}{{dx}}\frac{{dx}}{{dt}}\]E'tibor bering, \(\frac{{dy}}{{dt}}\) hosilasi bu erda haqiqatan ham ma'noga ega, chunki agar biz \(x\) ga ulanadigan bo'lsak, \(y\) haqiqatan ham \(t\) funktsiyasi bo'lar edi. Zanjir qoidasining bu shaklini eslab qolish usullaridan biri shuni ta'kidlash kerakki, agar biz o'ng tomondagi ikkita hosilani kasrlar deb hisoblasak, \(dx\) ikkala tomondan bir xil hosilani olish uchun bekor qiladi.
Xo'sh, biz buni yo'lga qo'yganimizdan so'ng, keling, ushbu kursda duch kelishimiz kerak bo'lgan yanada murakkab zanjir qoidalariga o'taylik.
Ko'p o'zgaruvchilar hisobidagi ko'plab mavzularda bo'lgani kabi, biz ko'rib chiqayotgan o'zgaruvchilar soniga qarab, aslida juda ko'p turli xil formulalar mavjud. Shunday qilib, keling, ushbu muhokamani ikkita o'zgaruvchining funktsiyasidan boshlaylik, \(z = f\left( {x,y} \right)\). Shu nuqtadan boshlab biz ko'rib chiqishimiz mumkin bo'lgan juda ko'p turli xil imkoniyatlar mavjud. Butun fikrni umumlashtirishdan oldin biz ikkita alohida holatni ko'rib chiqamiz.
1-holat: \(z = f\left( {x,y} \right)\), \(x = g\left( t \right)\), \(y = h\left( t \right)\) va \(\displaystyle \frac{{dz}}{{dt}}\) ni hisoblang.
Bu holat biz yuqorida ko'rib chiqqan Calculus I ning standart zanjir qoidasiga o'xshaydi. Bu holda biz oddiy hosilani hisoblaymiz, chunki \(z\) haqiqatan ham \(t\) funktsiyasi bo'ladi, agar biz \(x\) va \(y\) o'rniga o'rinli bo'lsak.
Bu holat uchun zanjir qoidasi:
\[\frac{{dz}}{{dt}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}}\]Shunday qilib, biz bu erda qilayotgan narsamiz \(f\) ni undagi har bir o'zgaruvchiga nisbatan farqlash va keyin ularning har birini \(t\) ga nisbatan ushbu o'zgaruvchining hosilasiga ko'paytirishdir. Yakuniy qadam bularning barchasini qo'shishdir.
Keling, bir nechta misollarni ko'rib chiqaylik.
- \(z = x{{\bf{e}}^{xy}}\), \(x = {t^2}\), \(y = {t^{ - 1}}\)
- \(z = {x^2}{y^3} + y\cos x\), \(x = \ln \left( {{t^2}} \right)\), \(y = \sin \left( {4t} \right)\)
a \(z = x{{\bf{e}}^{xy}}\), \(x = {t^2}\), \(y = {t^{ - 1}}\) Yechim
Bu erda formuladan foydalanishdan boshqa ko'p narsa qilishning iloji yo'q.
\[\begin{align*}\frac{{dz}}{{dt}} & = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}}\\ & = \left( {{{\bf{e}}^{xy}} + yx{{\bf{e}}^{xy}}} \right)\left( {2t} \right) + {x^2}{{\bf{e}}^{xy}}\left( { - {t^{ - 2}}} \right)\\ & = 2t\left( {{{\bf{e}}^{xy}} + yx{{\bf{e}}^{xy}}} \right) - {t^{ - 2}}{x^2}{{\bf{e}}^{xy}}\end{align*}\]Shunday qilib, texnik jihatdan biz hosilani hisobladik. Ammo, ehtimol, biz \(x\) va \(y\) o'rnini bosishimiz kerak, chunki bizda allaqachon \(t\) hosilasi mavjud. Buni qilish,
\[\frac{{dz}}{{dt}} = 2t\left( {{{\bf{e}}^t} + t{{\bf{e}}^t}} \right) - {t^{ - 2}}{t^4}{{\bf{e}}^t} = 2t{{\bf{e}}^t} + {t^2}{{\bf{e}}^t}\]Esda tutingki, bu holda asl funktsiyada \(x\) va \(y\) ni almashtirish va biz odatdagidek hosilani hisoblash osonroq bo'lishi mumkin edi. Taqqoslash uchun buni qilaylik.
\[z = {t^2}{{\bf{e}}^t}\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\frac{{dz}}{{dt}} = 2t{{\bf{e}}^t} + {t^2}{{\bf{e}}^t}\]Kamroq ish uchun bir xil natija. Ammo shuni yodda tutingki, ko'pincha birinchi navbatda almashtirishni amalga oshirish ko'proq ish bo'ladi.
b \(z = {x^2}{y^3} + y\cos x\), \(x = \ln \left( {{t^2}} \right)\), \(y = \sin \left( {4t} \right)\) Yechim
Xo'sh, bu holda birinchi navbatda almashtirishni amalga oshirish ko'proq ish bo'ladi, shuning uchun biz avval zanjir qoidasidan foydalanamiz, keyin esa almashtiramiz.
\[\begin{align*}\frac{{dz}}{{dt}} & = \left( {2x{y^3} - y\sin x} \right)\left( {\frac{2}{t}} \right) + \left( {3{x^2}{y^2} + \cos x} \right)\left( {4\cos \left( {4t} \right)} \right)\\ & = \frac{{4{{\sin }^3}\left( {4t} \right)\ln {t^2} - 2\sin \left( {4t} \right)\sin \left( {\ln {t^2}} \right)}}{t} + 4\cos \left( {4t} \right)\left( {3{{\sin }^2}\left( {4t} \right){{\left[ {\ln {t^2}} \right]}^2} + \cos \left( {\ln {t^2}} \right)} \right)\end{align*}\]E'tibor bering, ba'zida yakuniy javobning jiddiy tartibsizligi tufayli biz birinchi qadamni biroz soddalashtiramiz va javobni \(x\), \(y\) va \(t\) shartlarida qoldiramiz. Bu vaziyatga, sinfga va o'qituvchiga bog'liq, ammo o'qituvchingiz bilan gaplashmasdan o'rnini bosmaslikdan ehtiyot bo'ling.
Endi, keyingi holatga o'tishdan oldin tezda ko'rib chiqishimiz kerak bo'lgan alohida holat bor. Aytaylik, bizda quyidagi vaziyat bor,
\[z = f\left( {x,y} \right)\hspace{0.5in}y = g\left( x \right)\]Bu holda \(\frac{{dz}}{{dx}}\) uchun zanjir qoidasi quyidagicha bo'ladi:
\[\frac{{dz}}{{dx}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dx}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}} = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dx}}\]Birinchi muddatda biz haqiqatdan foydalanamiz,
\[\frac{{dx}}{{dx}} = \frac{d}{{dx}}\left( x \right) = 1\]Keling, bir misolni tez ko'rib chiqaylik.
Yechim
Biz faqat formulaga qo'shamiz.
\[\begin{align*}\frac{{dz}}{{dx}} & = \left( {\ln \left( {xy} \right) + x\frac{y}{{xy}}} \right) + \left( {x\frac{x}{{xy}} + 3{y^2}} \right)\left( { - 2x\sin \left( {{x^2} + 1} \right)} \right)\\ & = \ln \left( {x\cos \left( {{x^2} + 1} \right)} \right) + 1 - 2x\sin \left( {{x^2} + 1} \right)\left( {\frac{x}{{\cos \left( {{x^2} + 1} \right)}} + 3{{\cos }^2}\left( {{x^2} + 1} \right)} \right)\\ & = \ln \left( {x\cos \left( {{x^2} + 1} \right)} \right) + 1 - 2{x^2}\tan \left( {{x^2} + 1} \right) - 6x\sin \left( {{x^2} + 1} \right){\cos ^2}\left( {{x^2} + 1} \right)\end{align*}\]Endi ikkinchi holatni ko'rib chiqaylik.
2-holat: \(z = f\left( {x,y} \right)\), \(x = g\left( {s,t} \right)\), \(y = h\left( {s,t} \right)\) va \(\displaystyle \frac{{\partial z}}{{\partial s}}\) va \(\displaystyle \frac{{\partial z}}{{\partial t}}\) ni hisoblang.
Bu holda, agar biz \(x\) va \(y\) o'rnini bosadigan bo'lsak, biz \(z\) \(s\) va \(t\) funksiyasi ekanligini bilib olamiz va shuning uchun biz bu erda qisman hosilalarni hisoblashimiz va ulardan ikkitasi bo'lishi mantiqiy bo'ladi.
Mana bu ikkala holat uchun zanjir qoidasi.
\[\frac{{\partial z}}{{\partial s}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial s}}\hspace{0.5in}\hspace{0.25in}\frac{{\partial z}}{{\partial t}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial t}}\]Shuning uchun, bu biz ko'rib chiqqan birinchi holatga juda o'xshashligi ajablanarli emas. Mana shunday zanjir qoidasiga tezkor misol.
Yechim
Mana \(\displaystyle \frac{{\partial z}}{{\partial s}}\) uchun zanjir qoidasi.
\[\begin{align*}\frac{{\partial z}}{{\partial s}} & = \left( {2{{\bf{e}}^{2r}}\sin \left( {3\theta } \right)} \right)\left( t \right) + \left( {3{{\bf{e}}^{2r}}\cos \left( {3\theta } \right)} \right)\frac{s}{{\sqrt {{s^2} + {t^2}} }}\\ & = t\left( {2{{\bf{e}}^{2\left( {st - {t^2}} \right)}}\sin \left( {3\sqrt {{s^2} + {t^2}} } \right)} \right) + \frac{{3s{{\bf{e}}^{2\left( {st - {t^2}} \right)}}\cos \left( {3\sqrt {{s^2} + {t^2}} } \right)}}{{\sqrt {{s^2} + {t^2}} }}\end{align*}\]Endi \(\displaystyle \frac{{\partial z}}{{\partial t}}\) uchun zanjir qoidasi.
\[\begin{align*}\frac{{\partial z}}{{\partial t}} & = \left( {2{{\bf{e}}^{2r}}\sin \left( {3\theta } \right)} \right)\left( {s - 2t} \right) + \left( {3{{\bf{e}}^{2r}}\cos \left( {3\theta } \right)} \right)\frac{t}{{\sqrt {{s^2} + {t^2}} }}\\ & = \left( {s - 2t} \right)\left( {2{{\bf{e}}^{2\left( {st - {t^2}} \right)}}\sin \left( {3\sqrt {{s^2} + {t^2}} } \right)} \right) + \frac{{3t{{\bf{e}}^{2\left( {st - {t^2}} \right)}}\cos \left( {3\sqrt {{s^2} + {t^2}} } \right)}}{{\sqrt {{s^2} + {t^2}} }}\end{align*}\]Xo'sh, endi biz zanjir qoidasi uchun bir nechta holatlarni ko'rganimizdan so'ng, zanjir qoidasining umumiy versiyasini ko'rib chiqamiz.
Zanjir qoidasi
Faraz qilaylik, \(z\) \(n\) o‘zgaruvchilari \({x_1},{x_2}, \ldots ,{x_n}\) funksiyasi va bu o‘zgaruvchilarning har biri o‘z navbatida \(m\) o‘zgaruvchilari \({t_1},{t_2}, \ldots ,{t_m}\) funksiyasi. Keyin \({t_i}\), \(i = 1,2, \ldots ,m\) har qanday o'zgaruvchisi uchun bizda quyidagilar mavjud:
\[\frac{{\partial z}}{{\partial {t_i}}} = \frac{{\partial z}}{{\partial {x_1}}}\frac{{\partial {x_1}}}{{\partial {t_i}}} + \frac{{\partial z}}{{\partial {x_2}}}\frac{{\partial {x_2}}}{{\partial {t_i}}} + \cdots + \frac{{\partial z}}{{\partial {x_n}}}\frac{{\partial {x_n}}}{{\partial {t_i}}}\]Voy. Buni eslash juda ko'p. Aslida biz bo'limda muhokama qilgan yoki keyingi misollarda ko'rib chiqiladigan barcha zanjir qoidalarini yaratishning osonroq yo'li mavjud. Biz qurishimiz mumkin daraxt diagrammasi bu bizga har qanday vaziyat uchun zanjir qoidasini beradi. Bularning qanday ishlashini ko'rish uchun keling, orqaga qaytaylik va \(z = f\left( {x,y} \right)\), \(x = g\left( {s,t} \right)\), \(y = h\left( {s,t} \right)\) ni hisobga olgan holda \(\frac{{\partial z}}{{\partial s}}\) zanjiri qoidasini ko'rib chiqaylik. Biz bu nima ekanligini allaqachon bilamiz, lekin agar biz allaqachon bilsak, bu daraxt diagrammasini ko'rsatishga yordam berishi mumkin javob. Malumot uchun bu holat uchun zanjir qoidasi,
\[\frac{{\partial z}}{{\partial s}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial s}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial s}}\]Mana bu holat uchun daraxt diagrammasi.
Biz yuqoridan funktsiyaning o'zi va shu nuqtadan filialni boshlaymiz. Birinchi bo'limlar to'plami funksiyadagi o'zgaruvchilar uchun. Ushbu so'nggi nuqtalarning har biridan biz \(x\) va \(y\) funksiyasi bo'lgan o'zgaruvchilarni beradigan boshqa filiallar to'plamini chiqaramiz. Biz har bir harfni chiziq bilan bog'laymiz va har bir satr ko'rsatilgandek qisman hosilani ifodalaydi. E'tibor bering, qisman hosila sonidagi harf daraxtning yuqori "tuguni" va qisman hosilaning maxrajidagi harf daraxtning pastki "tugun"idir.
Zanjir qoidasini olish uchun undan foydalanish uchun biz pastdan boshlaymiz va o‘zgaruvchi bilan tugaydigan har bir novdaga (bu holda \(s\)) nisbatan hosila olishni istaymiz, biz daraxt bo‘ylab yuqoriga ko‘tarilib, yuqoriga tegmagunimizcha, bu shoxlar to‘plami bo‘ylab ko‘rgan hosilalarni ko‘paytiramiz. Buni \(s\) bilan tugaydigan har bir filial uchun bajarganimizdan so'ng, biz ushbu vaziyat uchun zanjir qoidasini olish uchun natijalarni qo'shamiz.
E'tibor bering, biz har doim hosilalarni daraxtga qo'ymaymiz. Ba'zi daraxtlar biroz kattaroq / tartibsiz bo'lib qoladi va shuning uchun biz hosilalarni qo'ymaymiz. Har bir filialda qanday lotin bo'lishi kerakligini eslang va siz ularni yozmasdan ham yaxshi bo'lasiz.
Keling, bir nechta zanjir qoidalarini yozaylik.
- \(\displaystyle \frac{{dw}}{{dt}}\) \(w = f\left( {x,y,z} \right)\), \(x = {g_1}\left( t \right)\), \(y = {g_2}\left( t \right)\) va \(z = {g_3}\left( t \right)\) uchun
- \(\displaystyle \frac{{\partial w}}{{\partial r}}\), \(w = f\left( {x,y,z} \right)\), \(x = {g_1}\left( {s,t,r} \right)\), \(y = {g_2}\left( {s,t,r} \right)\) va \(z = {g_3}\left( {s,t,r} \right)\) uchun
a \(\displaystyle \frac{{dw}}{{dt}}\) \(w = f\left( {x,y,z} \right)\), \(x = {g_1}\left( t \right)\), \(y = {g_2}\left( t \right)\) va \(z = {g_3}\left( t \right)\) uchun Yechim
Shunday qilib, bizga birinchi navbatda daraxt diagrammasi kerak bo'ladi, shuning uchun uni olamiz.
Bundan ko'rinib turibdiki, bu holat uchun zanjir qoidasi shunday bo'lishi kerak:
\[\frac{{dw}}{{dt}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial f}}{{\partial z}}\frac{{dz}}{{dt}}\]Bu, albatta, biz yuqorida ko'rgan ikkita o'zgaruvchan holatning tabiiy kengaytmasidir.
b \(\displaystyle \frac{{\partial w}}{{\partial r}}\), \(w = f\left( {x,y,z} \right)\), \(x = {g_1}\left( {s,t,r} \right)\), \(y = {g_2}\left( {s,t,r} \right)\) va \(z = {g_3}\left( {s,t,r} \right)\) uchun Yechim
Mana bu holat uchun daraxt diagrammasi.
Bundan hosila shunday ko'rinadi,
\[\frac{{\partial w}}{{\partial r}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial r}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial r}} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial r}}\]Shunday qilib, agar biz daraxt diagrammasini yozishimiz mumkin bo'lsa va ularni yozish odatda yomon bo'lmasa, biz duch kelishimiz mumkin bo'lgan har qanday o'rnatish uchun zanjir qoidasini bajarishimiz mumkin.
Endi biz ushbu murakkab vaziyatlarning birinchi hosilalarini qanday olish kerakligini ko'rdik, ammo yuqori tartibli hosilalar haqida nima deyish mumkin? Bularni qanday qilamiz? Bu bilan qanday kurashishni misol bilan ko'rish eng osondir.
Yechim
Ikkinchi hosilani topish haqida o'ylashdan oldin bizga birinchi hosila kerak bo'ladi, shuning uchun uni olamiz. Bu holat biz yuqorida ko'rib chiqqan ikkinchi holatga to'g'ri keladi, shuning uchun bizga yangi daraxt diagrammasi kerak emas. Mana birinchi hosila.
\[\begin{align*}\frac{{\partial f}}{{\partial \theta }} & = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial \theta }} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial \theta }}\\ & = - r\sin \left( \theta \right)\frac{{\partial f}}{{\partial x}} + r\cos \left( \theta \right)\frac{{\partial f}}{{\partial y}}\end{align*}\]Xo'sh, endi bilamizki, ikkinchi hosila,
\[\frac{{{\partial ^2}f}}{{\partial {\theta ^2}}} = \frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial \theta }}} \right) = \frac{\partial }{{\partial \theta }}\left( { - r\sin \left( \theta \right)\frac{{\partial f}}{{\partial x}} + r\cos \left( \theta \right)\frac{{\partial f}}{{\partial y}}} \right)\]Bu erda masala ushbu lotin bilan to'g'ri muomala qilishdir. Ikki birinchi tartib hosilalari, \(\frac{{\partial f}}{{\partial x}}\) va \(\frac{{\partial f}}{{\partial y}}\) ikkalasi ham \(x\) va \(y\) funksiyalari bo‘lib, ular o‘z navbatida \(r\) va \(\theta \) funksiyalari bo‘lib, bu atamalarning ikkalasi ham mahsulotdir. Shunday qilib, mahsulotdan foydalanish qoidasi quyidagilarni beradi:
\[\frac{{{\partial ^2}f}}{{\partial {\theta ^2}}} = - r\cos \left( \theta \right)\frac{{\partial f}}{{\partial x}} - r\sin \left( \theta \right)\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial x}}} \right) - r\sin \left( \theta \right)\frac{{\partial f}}{{\partial y}} + r\cos \left( \theta \right)\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial y}}} \right)\]Endi biz \(\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial x}}} \right)\) va \(\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial y}}} \right)\) nima bo'lishini aniqlashimiz kerak. Bu ikkalasi ham zanjir qoidasi muammolari, chunki ikkala hosila ham \(x\) va \(y\) funksiyalaridir va biz \(\theta \) ga nisbatan hosila olishni xohlaymiz.
Bularni amalga oshirishdan oldin, biz yuqorida qilgan birinchi zanjir qoidasini biroz qayta yozamiz.
\[\begin{equation}\frac{\partial }{{\partial \theta }}\left( f \right) = - r\sin \left( \theta \right)\frac{\partial }{{\partial x}}\left( f \right) + r\cos \left( \theta \right)\frac{\partial }{{\partial y}}\left( f \right) \label{eq:eq1} \end{equation}\]E'tibor bering, biz qilgan barcha narsa hosila uchun belgini biroz o'zgartirishdir. Shu tarzda yozilgan birinchi zanjir qoidasi bilan biz \(\eqref{eq:eq1}\) ni \(x = r\cos \theta \) va \(y = r\sin \theta \) bo'lsa, \(x\) va \(y\) ning \(\theta \) ga nisbatan har qanday funktsiyasini farqlash formulasi sifatida tasavvur qilishimiz mumkin.
Biroq, biz yuqorida kerak bo'lgan ikkita yangi hosilani qilishimiz kerak bo'lgan narsadir. Birinchi tartibli qisman hosilalarning ikkalasi, \(\frac{{\partial f}}{{\partial x}}\) va \(\frac{{\partial f}}{{\partial y}}\), \(x\) va \(y\) hamda \(x = r\cos \theta \) va \(y = r\sin \theta \) funksiyalari bo‘lib, biz ushbu lotinlarni hisoblash uchun \(\eqref{eq:eq1}\) dan foydalanishimiz mumkin.
Buning uchun biz shunchaki hammasini almashtiramiz f \(\eqref{eq:eq1}\) da biz farqlamoqchi bo'lgan birinchi tartibli qisman hosila bilan. O'sha paytda biz qilishimiz kerak bo'lgan narsa - ozgina notatsion ish va biz kutayotgan formulani olamiz.
Mana \(\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial x}}} \right)\) ni hisoblash uchun \(\eqref{eq:eq1}\) dan foydalanish.
\[\begin{align*}\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial x}}} \right) & = - r\sin \left( \theta \right)\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial x}}} \right) + r\cos \left( \theta \right)\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial x}}} \right)\\ & = - r\sin \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {x^2}}} + r\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial y\partial x}}\end{align*}\]Mana \(\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial y}}} \right)\) uchun hisoblash.
\[\begin{align*}\frac{\partial }{{\partial \theta }}\left( {\frac{{\partial f}}{{\partial y}}} \right) & = - r\sin \left( \theta \right)\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial y}}} \right) + r\cos \left( \theta \right)\frac{\partial }{{\partial y}}\left( {\frac{{\partial f}}{{\partial y}}} \right)\\ & = - r\sin \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial x\partial y}} + r\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {y^2}}}\end{align*}\]Yakuniy qadam ularni ikkinchi hosilaga qayta ulash va biroz soddalashtirishdir.
\[\begin{align*}\frac{{{\partial ^2}f}}{{\partial {\theta ^2}}} & = - r\cos \left( \theta \right)\frac{{\partial f}}{{\partial x}} - r\sin \left( \theta \right)\left( { - r\sin \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {x^2}}} + r\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial y\partial x}}} \right) - \\ & \hspace{0.25in}r\sin \left( \theta \right)\frac{{\partial f}}{{\partial y}} + r\cos \left( \theta \right)\left( { - r\sin \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial x\partial y}} + r\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right)\\ & = - r\cos \left( \theta \right)\frac{{\partial f}}{{\partial x}} + {r^2}{\sin ^2}\left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {x^2}}} - {r^2}\sin \left( \theta \right)\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial y\partial x}} - \\ & \hspace{0.25in}r\sin \left( \theta \right)\frac{{\partial f}}{{\partial y}} - {r^2}\sin \left( \theta \right)\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial x\partial y}} + {r^2}{\cos ^2}\left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {y^2}}}\\ & = - r\cos \left( \theta \right)\frac{{\partial f}}{{\partial x}} - r\sin \left( \theta \right)\frac{{\partial f}}{{\partial y}} + {r^2}{\sin ^2}\left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {x^2}}} - \\ & \hspace{0.5in}2{r^2}\sin \left( \theta \right)\cos \left( \theta \right)\frac{{{\partial ^2}f}}{{\partial y\partial x}} + {r^2}{\cos ^2}\left( \theta \right)\frac{{{\partial ^2}f}}{{\partial {y^2}}}\end{align*}\]Bu uzoq va juda tartibsiz, lekin u erda.
Ushbu bo'limning yakuniy mavzusi yashirin differentsiatsiyani qayta ko'rib chiqishdir. Zanjir qoidasining ushbu shakllari bilan yashirin farqlash aslida juda oddiy jarayonga aylanadi. dan boshlaylik bilvosita farqlash Biz hisob-kitob I kursida ko'rganmiz.
Biz \(F\left( {x,y} \right) = 0\) ko'rinishidagi funktsiyadan boshlaymiz (agar u bu shaklda bo'lmasa, bu shaklga kirish uchun hamma narsani teng belgisining bir tomoniga o'tkazing), bu erda \(y = y\left( x \right)\). Calculus I kursida bizdan \(\frac{{dy}}{{dx}}\) ni hisoblashni so'rashdi va bu ko'pincha juda tartibsiz jarayon edi. Ushbu bo'limdagi zanjir qoidasidan foydalanib, biz buni amalga oshirish uchun chiroyli oddiy formulani olishimiz mumkin. Biz \(x\) ga nisbatan ikkala tomonni farqlashdan boshlaymiz. Bu chap tomonda zanjir qoidasidan foydalanishni anglatadi va o'ng tomonda, albatta, nolga farqlanadi. Mana bular buning natijalari.
\[{F_x} + {F_y}\frac{{dy}}{{dx}} = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}}\]Ko'rsatilganidek, keyingi qilishimiz kerak bo'lgan yagona narsa \(\frac{{dy}}{{dx}}\) ni hal qilishdir va bizda yashirin farqlash uchun ishlatish uchun juda yaxshi formula mavjud. Shuni ham yodda tutingki, formulani soddalashtirish uchun biz hosilalar uchun pastki yozuvdan foydalanishga qaytdik.
Keling, tezkor misolni ko'rib chiqaylik.
Yechim
Birinchi qadam, tenglik belgisining bir tomonida nolni olishdir va buni qilish juda oson.
\[x\cos \left( {3y} \right) + {x^3}{y^5} - 3x + {{\bf{e}}^{xy}} = 0\]Endi chapdagi funktsiya formulamizda \(F\left( {x,y} \right)\), shuning uchun hosila topish uchun formuladan foydalanishimiz kifoya.
\[\frac{{dy}}{{dx}} = - \frac{{\cos \left( {3y} \right) + 3{x^2}{y^5} - 3 + y{{\bf{e}}^{xy}}}}{{ - 3x\sin \left( {3y} \right) + 5{x^3}{y^4} + x{{\bf{e}}^{xy}}}}\]Mana, boramiz. Buni eski Calculus I usuli yordamida amalga oshirish uchun ko'proq vaqt kerak bo'lardi.
Biz qisman hosilalarni birinchi marta kiritganimizda ko'rganimiz kabi, qisman hosilalarni o'z ichiga olgan yashirin farqlash muammolarini hal qilish uchun ham shunga o'xshash narsalarni qilishimiz mumkin. Bunday hollarda biz \(F\left( {x,y,z} \right) = 0\) shaklidagi funksiyadan boshlaymiz va \(z = f\left( {x,y} \right)\) deb faraz qilamiz va biz \(\frac{{\partial z}}{{\partial x}}\) va/yoki \(\frac{{\partial z}}{{\partial y}}\) ni topmoqchimiz.
Keling, \(\frac{{\partial z}}{{\partial x}}\) ni topishga harakat qilaylik. Biz \(x\) ga nisbatan ikkala tomonni farqlaymiz va biz \(y\) ni doimiy deb hisoblashimizni yodda tutishimiz kerak. Bundan tashqari, chap tomonda zanjir qoidasi talab qilinadi. Mana bu lotin.
\[\frac{{\partial F}}{{\partial x}}\frac{{\partial x}}{{\partial x}} + \frac{{\partial F}}{{\partial y}}\frac{{\partial y}}{{\partial x}} + \frac{{\partial F}}{{\partial z}}\frac{{\partial z}}{{\partial x}} = 0\]Endi bizda quyidagilar mavjud:
\[\frac{{\partial x}}{{\partial x}} = 1\hspace{0.5in}{\mbox{and }}\hspace{0.5in}\frac{{\partial y}}{{\partial x}} = 0\]Birinchisi, biz \(x\) ni \(x\) ga nisbatan shunchaki farqlayapmiz va bu 1 ekanligini bilamiz. Ikkinchisi, biz \(y\) ni doimiy deb hisoblaganimiz sababli, u nolga farqlanadi.
Bularni ulash va \(\frac{{\partial z}}{{\partial x}}\) ni hal qilish,
\[\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}}\]Shunga o'xshash dalil shuni ko'rsatish uchun ishlatilishi mumkin,
\[\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}}\]Bitta o'zgaruvchan holatda bo'lgani kabi, biz formulalarni soddalashtirish uchun hosilalar uchun obuna yozuviga o'tdik. Keling, buning misolini tezda ko'rib chiqaylik.
Yechim
Bu biz orqa tarafdagi eski yashirin farqlashdan foydalangan funktsiyalardan biri edi Qisman hosilalar bo'limi. Orqaga qaytib, ikkalasi o'rtasidagi farqni ko'rishni xohlashingiz mumkin.
Avval hamma narsani bir tomondan olaylik.
\[{x^2}\sin \left( {2y - 5z} \right) - 1 - y\cos \left( {6zx} \right) = 0\]Endi chapdagi funktsiya \(F\left( {x,y,z} \right)\) va shuning uchun biz qilishimiz kerak bo'lgan narsa hosilalarni topish uchun yuqorida ishlab chiqilgan formulalardan foydalanishdir.
\[\frac{{\partial z}}{{\partial x}} = - \frac{{2x\sin \left( {2y - 5z} \right) + 6yz\sin \left( {6zx} \right)}}{{ - 5{x^2}\cos \left( {2y - 5z} \right) + 6yx\sin \left( {6zx} \right)}}\] \[\frac{{\partial z}}{{\partial y}} = - \frac{{2{x^2}\cos \left( {2y - 5z} \right) - \cos \left( {6zx} \right)}}{{ - 5{x^2}\cos \left( {2y - 5z} \right) + 6yx\sin \left( {6zx} \right)}}\]Agar siz orqaga qaytsangiz va bu javoblarni biz birinchi marta topganlarimiz bilan solishtirsangiz, ular boshqacha ko'rinishini sezasiz. Ammo, agar siz bu erda javoblarimiz oldida turgan minus belgisini hisobga olsangiz, ular aslida bir xil ekanligini ko'rasiz.