多元函数微积分 / 偏导数 / 常规计算与结论整理
设二元函数$z = \left(x - y\right) \sin \left(x y\right)$,求$x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y}$.
正确答案
$\frac{\partial z}{\partial x} = \sin \left(x y\right) + \left(x - y\right) \cos \left(x y\right) \cdot y = \sin \left(x y\right) + y \left(x - y\right) \cos \left(x y\right)$,
$\frac{\partial z}{\partial y} = - \sin \left(x y\right) + \left(x - y\right) \cos \left(x y\right) \cdot x = - \sin \left(x y\right) + x \left(x - y\right) \cos \left(x y\right)$,
则$x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} =$$x \sin \left(x y\right) + x y \left(x - y\right) \cos \left(x y\right)$$- \left[- y \sin \left(x y\right) + x y \left(x - y\right) \cos \left(x y\right)\right] = \left(x + y\right) \sin \left(x y\right)$.
$\frac{\partial z}{\partial y} = - \sin \left(x y\right) + \left(x - y\right) \cos \left(x y\right) \cdot x = - \sin \left(x y\right) + x \left(x - y\right) \cos \left(x y\right)$,
则$x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} =$$x \sin \left(x y\right) + x y \left(x - y\right) \cos \left(x y\right)$$- \left[- y \sin \left(x y\right) + x y \left(x - y\right) \cos \left(x y\right)\right] = \left(x + y\right) \sin \left(x y\right)$.
题目解析
设 $z = (x - y)\sin(xy)$。先求一阶偏导数:
$$
\frac{\partial z}{\partial x} = \sin(xy) + (x - y)\cos(xy) \cdot y = \sin(xy) + y(x - y)\cos(xy),
$$
$$
\frac{\partial z}{\partial y} = -\sin(xy) + (x - y)\cos(xy) \cdot x = -\sin(xy) + x(x - y)\cos(xy).
$$
于是
$$
x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = x\sin(xy) + xy(x - y)\cos(xy) - \left[ -y\sin(xy) + xy(x - y)\cos(xy) \right].
$$
其中 $xy(x - y)\cos(xy)$ 项相消,余下 $x\sin(xy) + y\sin(xy) = (x + y)\sin(xy)$。故结果为 $(x + y)\sin(xy)$。
$$
\frac{\partial z}{\partial x} = \sin(xy) + (x - y)\cos(xy) \cdot y = \sin(xy) + y(x - y)\cos(xy),
$$
$$
\frac{\partial z}{\partial y} = -\sin(xy) + (x - y)\cos(xy) \cdot x = -\sin(xy) + x(x - y)\cos(xy).
$$
于是
$$
x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = x\sin(xy) + xy(x - y)\cos(xy) - \left[ -y\sin(xy) + xy(x - y)\cos(xy) \right].
$$
其中 $xy(x - y)\cos(xy)$ 项相消,余下 $x\sin(xy) + y\sin(xy) = (x + y)\sin(xy)$。故结果为 $(x + y)\sin(xy)$。