13.2-bo'lim : Xususiy hosilalar
8. Hammasini toping 1st quyidagi funksiyaning xususiy hosilalarini tartiblang.
\[z = \frac{{{p^2}\left( {r + 1} \right)}}{{{t^3}}} + pr\,{{\bf{e}}^{2p + 3r + 4t}}\]Yechim
Bu muammo uchun bizda uchta 1 bo'ladist hisoblash uchun xususiy hosilalarni buyurtma qiling. Mana ular,
\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial z}}{{\partial p}} & = {z_p} = \frac{{2p\left( {r + 1} \right)}}{{{t^3}}} + r\,{{\bf{e}}^{2p + 3r + 4t}} + 2pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial r}} & = {z_r} = \frac{{{p^2}}}{{{t^3}}} + p\,{{\bf{e}}^{2p + 3r + 4t}} + 3pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial t}} & = {z_t} = - \frac{{3{p^2}\left( {r + 1} \right)}}{{{t^4}}} + 4pr\,{{\bf{e}}^{2p + 3r + 4t}}\end{align*}}\]Hosila uchun ishlatiladigan belgi muhim emas, shuning uchun biz ikkala shakl bilan tanish ekanligimizga ishonch hosil qilish uchun ikkalasini ham ishlatdik.