13.4-bo'lim : Yuqori tartibli xususiy hosilalar
4. Barcha 2 ni topingnd quyidagi funksiya uchun hosilalarni tartiblang.
\[f\left( {s,t} \right) = {s^2}t + \ln \left( {{t^2} - s} \right)\]Yechim
Birinchidan, bizga ikkita 1 kerak bo'lishini bilamizst xususiy hosilalarni buyurtma qiling. Mana ular,
\[{f_s} = 2st - \frac{1}{{{t^2} - s}}\hspace{0.5in}{f_t} = {s^2} + \frac{{2t}}{{{t^2} - s}}\]2-qadam
Endi ikkinchi tartibli xususiy hosilalarning har birini hisoblaylik.
\[\begin{align*}{f_{s\,s}} & = {\left( {{f_s}} \right)_s} = 2t - \frac{1}{{{{\left( {{t^2} - s} \right)}^2}}}\\ {f_{s\,t}} & = {\left( {{f_s}} \right)_t} = 2s + \frac{{2t}}{{{{\left( {{t^2} - s} \right)}^2}}}\\ {f_{t\,s}} & = {f_{s\,t}} = 2s + \frac{{2t}}{{{{\left( {{t^2} - s} \right)}^2}}}\hspace{0.5in}{\mbox{by Clairaut's Theorem}}\\ {f_{t\,t}} & = {\left( {{f_t}} \right)_t} = \frac{{ - 2{t^2} - 2s}}{{{{\left( {{t^2} - s} \right)}^2}}}\end{align*}\]