📝 题目
8．求下列函数的一阶偏导数（其中 $f$ 具有一阶连续偏导数）： （1）$u=f\left(x^{2}-y^{2}, \mathrm{e}^{x y}\right)$ ； （2）$u=f\left(\frac{x}{y}, \frac{y}{z}\right)$ ； （3）$u=f(x, x y, x y z)$ ．
💡 答案与解析
[AI解答]
[AI解答] 以下为各小题的详细求解步骤。
---
**（1）** $ u = f(x^2 - y^2, e^{xy}) $
设中间变量： $ v = x^2 - y^2 $，$ w = e^{xy} $，则 $ u = f(v, w) $。
由链式法则：
$$ \frac{\partial u}{\partial x} = \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial x} = f_1 \cdot (2x) + f_2 \cdot (y e^{xy}) $$
$$ \frac{\partial u}{\partial y} = \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial y} = f_1 \cdot (-2y) + f_2 \cdot (x e^{xy}) $$
其中 $ f_1 = \frac{\partial f}{\partial v} $，$ f_2 = \frac{\partial f}{\partial w} $。
---
**（2）** $ u = f\left( \frac{x}{y}, \frac{y}{z} \right) $
设 $ p = \frac{x}{y} $，$ q = \frac{y}{z} $，则 $ u = f(p, q) $。
$$ \frac{\partial u}{\partial x} = f_1 \cdot \frac{\partial p}{\partial x} + f_2 \cdot \frac{\partial q}{\partial x} = f_1 \cdot \frac{1}{y} + f_2 \cdot 0 = \frac{1}{y} f_1 $$
$$ \frac{\partial u}{\partial y} = f_1 \cdot \frac{\partial p}{\partial y} + f_2 \cdot \frac{\partial q}{\partial y} = f_1 \cdot \left(-\frac{x}{y^2}\right) + f_2 \cdot \frac{1}{z} = -\frac{x}{y^2} f_1 + \frac{1}{z} f_2 $$
$$ \frac{\partial u}{\partial z} = f_1 \cdot \frac{\partial p}{\partial z} + f_2 \cdot \frac{\partial q}{\partial z} = f_1 \cdot 0 + f_2 \cdot \left(-\frac{y}{z^2}\right) = -\frac{y}{z^2} f_2 $$
---
**（3）** $ u = f(x, xy, xyz) $
设 $ r = x $，$ s = xy $，$ t = xyz $，则 $ u = f(r, s, t) $。
$$ \frac{\partial u}{\partial x} = f_1 \cdot \frac{\partial r}{\partial x} + f_2 \cdot \frac{\partial s}{\partial x} + f_3 \cdot \frac{\partial t}{\partial x} = f_1 \cdot 1 + f_2 \cdot y + f_3 \cdot yz $$
$$ \frac{\partial u}{\partial y} = f_1 \cdot 0 + f_2 \cdot \frac{\partial s}{\partial y} + f_3 \cdot \frac{\partial t}{\partial y} = f_2 \cdot x + f_3 \cdot xz $$
$$ \frac{\partial u}{\partial z} = f_1 \cdot 0 + f_2 \cdot 0 + f_3 \cdot \frac{\partial t}{\partial z} = f_3 \cdot xy $$
其中 $ f_1, f_2, f_3 $ 分别表示对第1、2、3个中间变量的偏导数。
---
**难度评级**：★★☆☆☆ （属于基础复合函数求偏导，只需掌握链式法则即可，但需注意变量顺序与符号书写。）