人邮高数 第6章 第6-3-7题
📝 题目
7．设 $w=f(x+x y+x y z)$ ，求 $\displaystyle \frac{\partial w}{\partial x} 、 \frac{\partial w}{\partial y} 、 \frac{\partial w}{\partial z}$ ．
💡 答案解析
[AI解答]
[AI解答]
设 $u = x + xy + xyz$，则 $w = f(u)$，其中 $f$ 为可微函数。
首先求 $\displaystyle \frac{\partial w}{\partial x}$： 由链式法则， $$ \frac{\partial w}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x}. $$ 计算 $\displaystyle \frac{\partial u}{\partial x}$： $$ u = x + xy + xyz, $$ 对 $x$ 求偏导时，$y$ 和 $z$ 视为常数： $$ \frac{\partial u}{\partial x} = 1 + y + yz. $$ 因此， $$ \frac{\partial w}{\partial x} = f'(u) \cdot (1 + y + yz). $$
其次求 $\displaystyle \frac{\partial w}{\partial y}$： $$ \frac{\partial w}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y}. $$ 计算 $\displaystyle \frac{\partial u}{\partial y}$： $$ \frac{\partial u}{\partial y} = 0 + x + xz = x(1+z). $$ 因此， $$ \frac{\partial w}{\partial y} = f'(u) \cdot x(1+z). $$
最后求 $\displaystyle \frac{\partial w}{\partial z}$： $$ \frac{\partial w}{\partial z} = f'(u) \cdot \frac{\partial u}{\partial z}. $$ 计算 $\displaystyle \frac{\partial u}{\partial z}$： $$ \frac{\partial u}{\partial z} = 0 + 0 + xy = xy. $$ 因此， $$ \frac{\partial w}{\partial z} = f'(u) \cdot xy. $$
综上，结果为： $$ \frac{\partial w}{\partial x} = f'(x+xy+xyz)(1+y+yz), $$ $$ \frac{\partial w}{\partial y} = f'(x+xy+xyz) \cdot x(1+z), $$ $$ \frac{\partial w}{\partial z} = f'(x+xy+xyz) \cdot xy. $$
难度：★☆☆☆☆