同济高数 第9章 第9-2-4题
📝 题目
4．设 $\displaystyle z=\mathrm{e}^{-\left(\frac{1}{x}+\frac{1}{y}\right)}$ ，求证 $\displaystyle x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=2 z$ ．
💡 答案解析
[AI解答]
[AI解答] 已知 $$ z = e^{-\left(\frac{1}{x}+\frac{1}{y}\right)}. $$ 先求偏导数 $\frac{\partial z}{\partial x}$。将 $y$ 视为常数，利用链式法则： $$ \frac{\partial z}{\partial x} = e^{-\left(\frac{1}{x}+\frac{1}{y}\right)} \cdot \frac{\partial}{\partial x}\left(-\frac{1}{x} - \frac{1}{y}\right) = z \cdot \left( \frac{1}{x^{2}} \right), $$ 因为 $$ \frac{\partial}{\partial x}\left(-\frac{1}{x}\right) = \frac{1}{x^{2}}. $$ 所以 $$ \frac{\partial z}{\partial x} = \frac{z}{x^{2}}. $$
同理，对 $y$ 求偏导： $$ \frac{\partial z}{\partial y} = e^{-\left(\frac{1}{x}+\frac{1}{y}\right)} \cdot \frac{\partial}{\partial y}\left(-\frac{1}{x} - \frac{1}{y}\right) = z \cdot \left( \frac{1}{y^{2}} \right), $$ 即 $$ \frac{\partial z}{\partial y} = \frac{z}{y^{2}}. $$
现在计算 $$ x^{2} \frac{\partial z}{\partial x} + y^{2} \frac{\partial z}{\partial y} = x^{2} \cdot \frac{z}{x^{2}} + y^{2} \cdot \frac{z}{y^{2}} = z + z = 2z. $$
因此等式成立。
难度：★☆☆☆☆