人邮高数 第6章 第6-3-17题
📝 题目
17．设 $z=z(x, y)$ 由方程 $x^{2}+y^{2}+z^{2}=y f(z)$ 所确定（其中 $y f^{\prime} \neq 2 z$ ），试求 $\displaystyle \frac{\partial z}{\partial x} 、 \frac{\partial z}{\partial y}$ ．
💡 答案解析
[AI解答]
[AI解答] 已知隐函数方程 $$ x^{2}+y^{2}+z^{2}=y f(z) $$ 其中 $y f'(z) \neq 2z$。要求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。
**步骤1：对方程两边关于 $x$ 求偏导（视 $z$ 为 $x,y$ 的函数）** $$ \frac{\partial}{\partial x}(x^{2}) + \frac{\partial}{\partial x}(y^{2}) + \frac{\partial}{\partial x}(z^{2}) = \frac{\partial}{\partial x}\big(y f(z)\big) $$ 得到 $$ 2x + 0 + 2z \frac{\partial z}{\partial x} = y f'(z) \frac{\partial z}{\partial x} $$
**步骤2：整理出 $\frac{\partial z}{\partial x}$** 将含 $\frac{\partial z}{\partial x}$ 的项移到一边： $$ 2z \frac{\partial z}{\partial x} - y f'(z) \frac{\partial z}{\partial x} = -2x $$ 即 $$ \big(2z - y f'(z)\big) \frac{\partial z}{\partial x} = -2x $$ 因此 $$ \frac{\partial z}{\partial x} = \frac{-2x}{2z - y f'(z)} $$ 由条件 $y f'(z) \neq 2z$，分母不为零。
**步骤3：对方程两边关于 $y$ 求偏导** $$ \frac{\partial}{\partial y}(x^{2}) + \frac{\partial}{\partial y}(y^{2}) + \frac{\partial}{\partial y}(z^{2}) = \frac{\partial}{\partial y}\big(y f(z)\big) $$ 得到 $$ 0 + 2y + 2z \frac{\partial z}{\partial y} = f(z) + y f'(z) \frac{\partial z}{\partial y} $$
**步骤4：整理出 $\frac{\partial z}{\partial y}$** 将含 $\frac{\partial z}{\partial y}$ 的项移到一边： $$ 2z \frac{\partial z}{\partial y} - y f'(z) \frac{\partial z}{\partial y} = f(z) - 2y $$ 即 $$ \big(2z - y f'(z)\big) \frac{\partial z}{\partial y} = f(z) - 2y $$ 因此 $$ \frac{\partial z}{\partial y} = \frac{f(z) - 2y}{2z - y f'(z)} $$
**最终结果** $$ \boxed{\frac{\partial z}{\partial x} = \frac{-2x}{2z - y f'(z)},\quad \frac{\partial z}{\partial y} = \frac{f(z) - 2y}{2z - y f'(z)}} $$
难度：★★☆☆☆