Tìm vi phân cấp 1 và cấp 2 của các hợp sau u=f(s,t) với s=xy, t=x/y

Giải thích các bước giải:

$\begin{array}{l}
 + )\dfrac{{\partial u}}{{\partial x}} = \dfrac{{\partial u}}{{\partial s}}.\dfrac{{\partial s}}{{\partial x}} + \dfrac{{\partial u}}{{\partial t}}.\dfrac{{\partial t}}{{\partial x}} = y.\dfrac{{\partial u}}{{\partial s}} + \dfrac{1}{y}.\dfrac{{\partial u}}{{\partial t}}\\
 + )\dfrac{{\partial u}}{{\partial y}} = \dfrac{{\partial u}}{{\partial s}}.\dfrac{{\partial s}}{{\partial y}} + \dfrac{{\partial u}}{{\partial t}}.\dfrac{{\partial t}}{{\partial y}} = x.\dfrac{{\partial u}}{{\partial s}} - \dfrac{x}{{{y^2}}}.\dfrac{{\partial u}}{{\partial t}}\\
 + )\dfrac{{{\partial ^2}u}}{{\partial {x^2}}} = \dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial u}}{{\partial x}}} \right) = \dfrac{\partial }{{\partial x}}\left( {y.\dfrac{{\partial u}}{{\partial s}} + \dfrac{1}{y}.\dfrac{{\partial u}}{{\partial t}}} \right)\\
 = y\left( {\dfrac{{{\partial ^2}u}}{{\partial {s^2}}}.y + \dfrac{{{\partial ^2}u}}{{\partial s\partial t}}.\dfrac{1}{y}} \right) + \dfrac{1}{y}\left( {\dfrac{{{\partial ^2}u}}{{\partial t\partial s}}.y + \dfrac{{{\partial ^2}u}}{{\partial {t^2}}}.\dfrac{1}{y}} \right)\\
 = {y^2}.\dfrac{{{\partial ^2}u}}{{\partial {s^2}}} + 2.\dfrac{{{\partial ^2}u}}{{\partial s\partial t}} + \dfrac{1}{{{y^2}}}.\dfrac{{{\partial ^2}u}}{{\partial {t^2}}}\\
 + )\dfrac{{{\partial ^2}u}}{{\partial {y^2}}} = \dfrac{\partial }{{\partial y}}\left( {\dfrac{{\partial u}}{{\partial y}}} \right) = \dfrac{\partial }{{\partial y}}\left( {x.\dfrac{{\partial u}}{{\partial s}} - \dfrac{x}{{{y^2}}}.\dfrac{{\partial u}}{{\partial t}}} \right)\\
 = x\left( {\dfrac{{{\partial ^2}u}}{{\partial {s^2}}}.x + \dfrac{{{\partial ^2}u}}{{\partial s\partial t}}.\left( {\dfrac{{ - x}}{{{y^2}}}} \right)} \right) + \left( {\dfrac{{ - x}}{{{y^2}}}} \right)\left( {\dfrac{{{\partial ^2}u}}{{\partial t\partial s}}.x + \dfrac{{{\partial ^2}u}}{{\partial {t^2}}}.\left( {\dfrac{{ - x}}{{{y^2}}}} \right)} \right)\\
 = {x^2}.\dfrac{{{\partial ^2}u}}{{\partial {s^2}}} - 2.\dfrac{{{x^2}}}{{{y^2}}}.\dfrac{{{\partial ^2}u}}{{\partial s\partial t}} + \dfrac{{{x^2}}}{{{y^4}}}.\dfrac{{{\partial ^2}u}}{{\partial {t^2}}}\\
 + )\dfrac{{{\partial ^2}u}}{{\partial x\partial y}} = \dfrac{{{\partial ^2}u}}{{\partial y\partial x}} = \dfrac{\partial }{{\partial x}}\left( {\dfrac{{\partial u}}{{\partial y}}} \right)\\
 = \dfrac{\partial }{{\partial x}}\left( {x.\dfrac{{\partial u}}{{\partial s}} - \dfrac{x}{{{y^2}}}.\dfrac{{\partial u}}{{\partial t}}} \right)\\
 = \dfrac{{\partial u}}{{\partial s}} + x\left( {\dfrac{{{\partial ^2}u}}{{\partial {s^2}}}.y + \dfrac{{{\partial ^2}u}}{{\partial s\partial t}}.\dfrac{1}{y}} \right) - \dfrac{1}{{{y^2}}}.\dfrac{{\partial u}}{{\partial t}} + \left( {\dfrac{{ - x}}{{{y^2}}}} \right)\left( {\dfrac{{{\partial ^2}u}}{{\partial t\partial s}}.y + \dfrac{{{\partial ^2}u}}{{\partial {t^2}}}.\dfrac{1}{y}} \right)\\
 = \dfrac{{\partial u}}{{\partial s}} - \dfrac{1}{{{y^2}}}.\dfrac{{\partial u}}{{\partial t}} + xy.\dfrac{{{\partial ^2}u}}{{\partial {s^2}}} - \dfrac{x}{{{y^3}}}.\dfrac{{{\partial ^2}u}}{{\partial {t^2}}}
\end{array}$