人邮高数 第6章 第6-3-20题
📝 题目
20．设 $z=x y+u, u=\varphi(x, y)$ ，求 $\displaystyle \frac{\partial z}{\partial x} 、 \frac{\partial^{2} z}{\partial x^{2}} 、 \frac{\partial^{2} z}{\partial x \partial y}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知： $$ z = x y + u, \quad u = \varphi(x, y) $$ 其中 $\varphi$ 具有二阶连续偏导数。
首先求一阶偏导 $\displaystyle \frac{\partial z}{\partial x}$： $$ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x y) + \frac{\partial u}{\partial x} = y + \frac{\partial \varphi}{\partial x} $$ 记 $\displaystyle \varphi_x = \frac{\partial \varphi}{\partial x}$，则： $$ \frac{\partial z}{\partial x} = y + \varphi_x $$
再求二阶偏导 $\displaystyle \frac{\partial^2 z}{\partial x^2}$： $$ \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left( y + \varphi_x \right) = 0 + \frac{\partial \varphi_x}{\partial x} = \varphi_{xx} $$ 其中 $\displaystyle \varphi_{xx} = \frac{\partial^2 \varphi}{\partial x^2}$。
最后求混合偏导 $\displaystyle \frac{\partial^2 z}{\partial x \partial y}$： $$ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left( \frac{\partial z}{\partial x} \right) = \frac{\partial}{\partial y}\left( y + \varphi_x \right) = 1 + \frac{\partial \varphi_x}{\partial y} = 1 + \varphi_{xy} $$ 其中 $\displaystyle \varphi_{xy} = \frac{\partial^2 \varphi}{\partial y \partial x}$，由连续性假设有 $\varphi_{xy} = \varphi_{yx}$。
因此结果为： $$ \boxed{\frac{\partial z}{\partial x} = y + \varphi_x,\quad \frac{\partial^{2} z}{\partial x^{2}} = \varphi_{xx},\quad \frac{\partial^{2} z}{\partial x \partial y} = 1 + \varphi_{xy}} $$
难度：★☆☆☆☆