人邮高数 第6章 第6-3-10题
📝 题目
10．设 $z=f\left(x^{2}+y^{2}\right)$ ，其中函数 $f$ 有二阶连续偏导数，求 $\displaystyle \frac{\partial^{2} z}{\partial x^{2}} 、 \frac{\partial^{2} z}{\partial y^{2}} 、 \frac{\partial^{2} z}{\partial x \partial y}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知 $z = f(x^2 + y^2)$，令中间变量 $u = x^2 + y^2$，则 $z = f(u)$，且 $f$ 具有二阶连续偏导数。
首先求一阶偏导数： $$ \frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} = f'(u) \cdot 2x = 2x f'(u) $$ $$ \frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} = f'(u) \cdot 2y = 2y f'(u) $$
再求二阶偏导数：
1. 对 $x$ 的二阶偏导： $$ \frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( 2x f'(u) \right) = 2 f'(u) + 2x \cdot f''(u) \cdot \frac{\partial u}{\partial x} = 2 f'(u) + 2x \cdot f''(u) \cdot 2x = 2 f'(u) + 4x^2 f''(u) $$
2. 对 $y$ 的二阶偏导： $$ \frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( 2y f'(u) \right) = 2 f'(u) + 2y \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2 f'(u) + 2y \cdot f''(u) \cdot 2y = 2 f'(u) + 4y^2 f''(u) $$
3. 混合偏导： $$ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( 2x f'(u) \right) = 2x \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2x \cdot f''(u) \cdot 2y = 4xy f''(u) $$
因此结果为： $$ \frac{\partial^{2} z}{\partial x^{2}} = 2 f'(x^2+y^2) + 4x^2 f''(x^2+y^2) $$ $$ \frac{\partial^{2} z}{\partial y^{2}} = 2 f'(x^2+y^2) + 4y^2 f''(x^2+y^2) $$ $$ \frac{\partial^{2} z}{\partial x \partial y} = 4xy f''(x^2+y^2) $$
难度：★★☆☆☆