题目
若函数z=(x-y)/(x+y),则dz=( ) A. (2(xdy-ydx))/((x+y)^2) B. (2(ydy-xdx))/((x+y)^2) C. (2(ydx-xdy))/((x+y)^2) D. (2(xdy-ydx))/((x+y)^2)
若函数$z=\frac{x-y}{x+y}$,则dz=( )
A. $\frac{2(xdy-ydx)}{(x+y)^{2}}$
B. $\frac{2(ydy-xdx)}{(x+y)^{2}}$
C. $\frac{2(ydx-xdy)}{(x+y)^{2}}$
D. $\frac{2(xdy-ydx)}{(x+y)^{2}}$
A. $\frac{2(xdy-ydx)}{(x+y)^{2}}$
B. $\frac{2(ydy-xdx)}{(x+y)^{2}}$
C. $\frac{2(ydx-xdy)}{(x+y)^{2}}$
D. $\frac{2(xdy-ydx)}{(x+y)^{2}}$
题目解答
答案
将函数 $ z = \frac{x - y}{x + y} $ 重写为:
\[
z = 1 - \frac{2y}{x + y}
\]
求偏导数:
\[
\frac{\partial z}{\partial x} = \frac{2y}{(x + y)^2}, \quad \frac{\partial z}{\partial y} = -\frac{2x}{(x + y)^2}
\]
全微分:
\[
dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy = \frac{2y dx - 2x dy}{(x + y)^2} = \frac{2(y dx - x dy)}{(x + y)^2}
\]
答案:$\boxed{C}$
解析
本题考查多元函数全微分的计算。解题思路是先对函数$z=\frac{x - y}{x + y}$进行适当变形,然后分别求出$z$对$x$和$y$的偏导数,最后根据全微分公式$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$计算$dz$。
- 对函数进行变形:
将函数$z = \frac{x - y}{x + y}$变形为$z = 1 - \frac{2y}{x + y}$。 - 求$z$对$x$的偏导数$\frac{\partial z}{\partial x}$:
对$z = 1 - \frac{2y}{x + y}$关于$x$求偏导数,因为$1$关于$x$的偏导数为$0$,$-\frac{2y}{x + y}$关于$x$求偏导数时,把$y$看作常数,根据除法求导公式$(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}$,这里$u = -2y$($u^\prime = 0$),$v = x + y$($v^\prime = 1$),可得:
$\frac{\partial z}{\partial x}=\frac{0\times(x + y)-(-2y)\times1}{(x + y)^2}=\frac{2y}{(x + y)^2}$ - 求$z$对$y$的偏导数$\frac{\partial z}{\partial y}$:
对$z = 1 - \frac{2y}{x + y}$关于$y$求偏导数,因为$1$关于$y$的偏导数为$0$,$-\frac{2y}{x + y}$关于$y$求偏导数时,把$x$看作常数,根据除法求导公式$(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}$,这里$u = -2y$($u^\prime = -2$),$v = x + y$($v^\prime = 1$),可得:
$\frac{\partial z}{\partial y}=\frac{-2\times(x + y)-(-2y)\times1}{(x + y)^2}=\frac{-2x - 2y + 2y}{(x + y)^2}=-\frac{2x}{(x + y)^2}$ - 计算全微分$dz$:
根据全微分公式$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$,将$\frac{\partial z}{\partial x}=\frac{2y}{(x + y)^2}$和$\frac{\partial z}{\partial y}=-\frac{2x}{(x + y)^2}$代入可得:
$dz=\frac{2y}{(x + y)^2}dx-\frac{2x}{(x + y)^2}dy=\frac{2y dx - 2x dy}{(x + y)^2}=\frac{2(y dx - x dy)}{(x + y)^2}$