同济高数 第9章 第9-4-12题
📝 题目
12．设 $z=f\left(x^{2}+y^{2}\right)$ ，其中 $f$ 具有二阶导数，求 $\displaystyle \frac{\partial^{2} z}{\partial x^{2}}, \frac{\partial^{2} z}{\partial x \partial y}, \frac{\partial^{2} z}{\partial y^{2}}$ ．
💡 答案解析
[AI解答]
[AI解答]
已知 $ z = f(x^2 + y^2) $，其中 $ f $ 具有二阶导数。令 $$ u = x^2 + y^2, $$ 则 $$ z = f(u). $$
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**第一步：求一阶偏导数**
对 $ x $ 求偏导： $$ \frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} = f'(u) \cdot 2x. $$
对 $ y $ 求偏导： $$ \frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} = f'(u) \cdot 2y. $$
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**第二步：求二阶偏导数**
1. 求 $\frac{\partial^2 z}{\partial x^2}$：
由 $\frac{\partial z}{\partial x} = 2x f'(u)$，再对 $x$ 求偏导： $$ \frac{\partial^2 z}{\partial x^2} = 2 f'(u) + 2x \cdot f''(u) \cdot \frac{\partial u}{\partial x}. $$ 而 $\frac{\partial u}{\partial x} = 2x$，所以 $$ \frac{\partial^2 z}{\partial x^2} = 2 f'(u) + 2x \cdot f''(u) \cdot 2x = 2 f'(u) + 4x^2 f''(u). $$
2. 求 $\frac{\partial^2 z}{\partial y^2}$：
同理，由 $\frac{\partial z}{\partial y} = 2y f'(u)$，再对 $y$ 求偏导： $$ \frac{\partial^2 z}{\partial y^2} = 2 f'(u) + 2y \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2 f'(u) + 2y \cdot f''(u) \cdot 2y = 2 f'(u) + 4y^2 f''(u). $$
3. 求 $\frac{\partial^2 z}{\partial x \partial y}$：
由 $\frac{\partial z}{\partial x} = 2x f'(u)$，再对 $y$ 求偏导： $$ \frac{\partial^2 z}{\partial x \partial y} = 2x \cdot f''(u) \cdot \frac{\partial u}{\partial y} = 2x \cdot f''(u) \cdot 2y = 4xy f''(u). $$
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**最终结果：** $$ \boxed{\frac{\partial^{2} z}{\partial x^{2}} = 2 f'(x^2+y^2) + 4x^2 f''(x^2+y^2)} $$ $$ \boxed{\frac{\partial^{2} z}{\partial y^{2}} = 2 f'(x^2+y^2) + 4y^2 f''(x^2+y^2)} $$ $$ \boxed{\frac{\partial^{2} z}{\partial x \partial y} = 4xy f''(x^2+y^2)} $$