Asosiy tarkibga o'tish

13.6-bo'lim : Zanjir qoidasi

5. Quyidagi ma'lumotlarni hisobga olgan holda, \({z_t}\) va \({z_p}\) ni aniqlash uchun zanjir qoidasidan foydalaning.

\[z = 4y\sin \left( {2x} \right)\,\hspace{0.5in}x = 3u - p,\,\,\,\,y = {p^2}u,\,\,\,\,\,\,u = {t^2} + 1\]
Yechim

Xo'sh, bizda buning uchun eslatmalardan formula yo'q, shuning uchun biz birinchi navbatda xulosa qilishimiz kerak. Buning uchun bizga quyidagi daraxt diagrammasi kerak bo'ladi.

A tree diagram for the chain rule.  At the top is \(z\), which branches down to \(x\) by \(\frac{{\partial z}}{{\partial x}}\) and to \(y\) by \(\frac{{\partial z}}{{\partial y}}\).  Each of \(x\) and \(y\) then branches down to \(u\) and to \(p\), and under each \(u\) there is a further branch down to \(t\) by \(\frac{{du}}{{dt}}\).
2-qadam

Mana bizdan topishimiz so'ralayotgan ikkita hosila uchun formulalar.

\[{z_t} = \frac{{\partial z}}{{\partial t}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial u}}\frac{{du}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\frac{{du}}{{dt}}\hspace{0.5in}{z_p} = \frac{{\partial z}}{{\partial p}} = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial p}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial p}}\]
3-qadam

Mana bu muammo uchun ish.

\[\begin{align*}{z_t} & = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial u}}\frac{{du}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\frac{{du}}{{dt}}\\ & = \left[ {8y\cos \left( {2x} \right)} \right]\left[ 3 \right]\left[ {2t} \right] + \left[ {4\sin \left( {2x} \right)} \right]\left[ {{p^2}} \right]\left[ {2t} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{48ty\cos \left( {2x} \right) + 8t{p^2}\sin \left( {2x} \right)}}\end{align*}\] \[\begin{align*}{z_p} & = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial p}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial p}}\\ & = \left[ {8y\cos \left( {2x} \right)} \right]\left[ { - 1} \right] + \left[ {4\sin \left( {2x} \right)} \right]\left[ {2pu} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 8y\cos \left( {2x} \right) + 8pu\sin \left( {2x} \right)}}\end{align*}\]

Har bir lotinning ikkinchi bosqichida biz “formula”da qaysi atama qaysi hosiladan kelganligini aniq ko'rsatish uchun qavslar qo'shdik.

Bundan tashqari, biz barcha orqa almashtirishni tugatganimizdan so'ng, har bir hosiladan kelib chiqadigan tartibsizlik tufayli bu lotinlarda hech qanday "orqa almashtirish" qilmadik.