怀旧之偶尔做道题(7): 数学分析考研题

已知z=f(x^2-2y,g(xy))z=f(x^2-2y,g(xy)),且f二阶连续可微,g二阶可微,求\frac{\partial^2z}{\partial{x}\partial{y}}.

 

(中科大2020年数学分析考研试题)

解:

怀旧之偶尔做道题(7): 数学分析考研题_第1张图片

 

知识点:

z = f(u(x,y),v(x,y)) \\ \frac{\partial{z}}{\partial{x}}=\frac{\partial{z}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{z}}{\partial{v}}\frac{\partial{v}}{\partial{x}} \\ \frac{\partial{z}}{\partial{y}}=\frac{\partial{z}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{z}}{\partial{v}}\frac{\partial{v}}{\partial{y}}

进一步,二阶偏微分为:

你可能感兴趣的:(数学与算法趣题,数学分析,多元函数微积分,偏微分)