Xo'sh, biz ushbu lotinni aniqlash uchun eslatmalardan "formula" dan foydalanishimiz mumkin. Mana bu muammo uchun ish.

\[\begin{align*}\frac{{dz}}{{dt}} & = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\\ & = \left[ { - 2xy\sin \left( {y{x^2}} \right)} \right]\left[ {4{t^3} - 2} \right] + \left[ { - {x^2}\sin \left( {y{x^2}} \right)} \right]\left[ { - 6{t^5}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 2\left( {{t^4} - 2t} \right)\left( {1 - {t^6}} \right)\left( {4{t^3} - 2} \right)\sin \left( {\left( {1 - {t^6}} \right){{\left( {{t^4} - 2t} \right)}^2}} \right) + 6{t^5}{{\left( {{t^4} - 2t} \right)}^2}\sin \left( {\left( {1 - {t^6}} \right){{\left( {{t^4} - 2t} \right)}^2}} \right)}}\end{align*}\]

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

Shuningdek, biz \(t\) da tenglamani olish uchun uchinchi bosqichda \(x\) va \(y\) ni uladik. Ulardan ba'zilari uchun yakuniy formulaning chalkashligi tufayli \(x\) va \(y\)-ni yolg'iz qoldirib, javobni biroz "yaxshiroq" saqlash uchun \(t\) nuqtai nazaridan ularning ta'rifini tan olish osonroq bo'lishi mumkin edi. Ehtimol, siz o'qituvchingizdan bu borada uning afzalliklarini so'rashingiz kerak.