📝 题目
13．利用一阶全微分形式的不变性求函数 $u=\frac{x}{x^{2}+y^{2}+z^{2}}$ 的偏导数．
💡 答案与解析
[AI解答]
[AI解答] 我们要求函数 $$ u = \frac{x}{x^{2}+y^{2}+z^{2}} $$ 的偏导数，利用一阶全微分形式的不变性，即先求全微分 $\mathrm{d}u$，再从中直接读出各偏导系数。
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**第一步：将函数写成便于微分的形式** 令 $$ u = x \cdot (x^{2}+y^{2}+z^{2})^{-1} $$ 设 $$ R^{2} = x^{2}+y^{2}+z^{2} $$ 则 $$ u = x R^{-2} $$
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**第二步：对两边取全微分** 利用乘积法则： $$ \mathrm{d}u = \mathrm{d}x \cdot R^{-2} + x \cdot \mathrm{d}(R^{-2}) $$ 而 $$ \mathrm{d}(R^{-2}) = -2 R^{-3} \, \mathrm{d}R $$ 又 $$ \mathrm{d}R = \frac{\partial R}{\partial x}\mathrm{d}x + \frac{\partial R}{\partial y}\mathrm{d}y + \frac{\partial R}{\partial z}\mathrm{d}z $$ 由于 $R = (x^{2}+y^{2}+z^{2})^{1/2}$，有 $$ \frac{\partial R}{\partial x} = \frac{x}{R},\quad \frac{\partial R}{\partial y} = \frac{y}{R},\quad \frac{\partial R}{\partial z} = \frac{z}{R} $$ 所以 $$ \mathrm{d}R = \frac{x}{R}\mathrm{d}x + \frac{y}{R}\mathrm{d}y + \frac{z}{R}\mathrm{d}z $$
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**第三步：代入并整理** 先计算 $$ x \cdot \mathrm{d}(R^{-2}) = x \cdot (-2 R^{-3}) \left( \frac{x}{R}\mathrm{d}x + \frac{y}{R}\mathrm{d}y + \frac{z}{R}\mathrm{d}z \right) $$ 即 $$ = -\frac{2x}{R^{4}} \left( x\mathrm{d}x + y\mathrm{d}y + z\mathrm{d}z \right) $$ 而第一项为 $$ R^{-2} \mathrm{d}x $$ 因此 $$ \mathrm{d}u = \frac{1}{R^{2}}\mathrm{d}x - \frac{2x^{2}}{R^{4}}\mathrm{d}x - \frac{2xy}{R^{4}}\mathrm{d}y - \frac{2xz}{R^{4}}\mathrm{d}z $$ 合并 $\mathrm{d}x$ 的系数： $$ \frac{1}{R^{2}} - \frac{2x^{2}}{R^{4}} = \frac{R^{2} - 2x^{2}}{R^{4}} $$ 而 $R^{2} = x^{2}+y^{2}+z^{2}$，所以 $$ R^{2} - 2x^{2} = y^{2}+z^{2} - x^{2} $$ 于是 $$ \mathrm{d}u = \frac{y^{2}+z^{2}-x^{2}}{(x^{2}+y^{2}+z^{2})^{2}}\mathrm{d}x - \frac{2xy}{(x^{2}+y^{2}+z^{2})^{2}}\mathrm{d}y - \frac{2xz}{(x^{2}+y^{2}+z^{2})^{2}}\mathrm{d}z $$
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**第四步：由全微分形式直接读出偏导数** 由 $$ \mathrm{d}u = \frac{\partial u}{\partial x}\mathrm{d}x + \frac{\partial u}{\partial y}\mathrm{d}y + \frac{\partial u}{\partial z}\mathrm{d}z $$ 对比系数得： $$ \displaystyle\frac{\partial u}{\partial x} = \frac{y^{2}+z^{2}-x^{2}}{(x^{2}+y^{2}+z^{2})^{2}} $$ $$ \displaystyle\frac{\partial u}{\partial y} = -\frac{2xy}{(x^{2}+y^{2}+z^{2})^{2}} $$ $$ \displaystyle\frac{\partial u}{\partial z} = -\frac{2xz}{(x^{2}+y^{2}+z^{2})^{2}} $$
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**最终答案** $$ \boxed{\displaystyle\frac{\partial u}{\partial x} = \frac{y^{2}+z^{2}-x^{2}}{(x^{2}+y^{2}+z^{2})^{2}},\quad \frac{\partial u}{\partial y} = -\frac{2xy}{(x^{2}+y^{2}+z^{2})^{2}},\quad \frac{\partial u}{\partial z} = -\frac{2xz}{(x^{2}+y^{2}+z^{2})^{2}}} $$
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难度：★★☆☆☆ （计算过程稍长，但方法固定，属于中等偏易题型）