Xo'sh, biz ushbu lotinni aniqlash uchun eslatmalardagi "formulalar" dan (harflarga kichik o'zgartirishlar bilan) foydalanishimiz mumkin. Mana bu muammo uchun ish.

\[\begin{align*}\frac{{\partial z}}{{\partial u}} & = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial u}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\\ & = \left[ { - 2{x^{ - 3}}{y^6} - 4} \right]\left[ {2uv} \right] + \left[ {6{x^{ - 2}}{y^5}} \right]\left[ { - 3} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{2uv\left( { - 2{u^{ - 6}}{v^{ - 3}}{{\left( {v - 3u} \right)}^6} - 4} \right) - 18{u^{ - 4}}{v^{ - 2}}{{\left( {v - 3u} \right)}^5}}}\end{align*}\] \[\begin{align*}\frac{{\partial z}}{{\partial v}} & = \frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial z}}{{\partial y}}\frac{{\partial y}}{{\partial v}}\\ & = \left[ { - 2{x^{ - 3}}{y^6} - 4} \right]\left[ {{u^2}} \right] + \left[ {6{x^{ - 2}}{y^5}} \right]\left[ 1 \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{{u^2}\left( { - 2{u^{ - 6}}{v^{ - 3}}{{\left( {v - 3u} \right)}^6} - 4} \right) + 6{u^{ - 4}}{v^{ - 2}}{{\left( {v - 3u} \right)}^5}}}\end{align*}\]

Ikkinchi bosqichda biz qavslar qo'shdik, bu "formula" da qaysi atama qaysi lotindan kelganligini aniq ko'rsatish uchun.

Shuningdek, biz \(u\) va \(v\) da tenglamani olish uchun uchinchi bosqichda \(x\) va \(y\) ni uladik. Ulardan ba'zilari uchun, yakuniy formulaning chalkashligi sababli, javobni biroz "yaxshiroq" saqlash uchun \(x\) va \(y\)-ni yolg'iz qoldirib, \(u\) va \(v\) nuqtai nazaridan ta'rifini tan olish osonroq bo'lishi mumkin edi. Ehtimol, siz o'qituvchingizdan bu borada uning afzalliklarini so'rashingiz kerak.