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：$\boxed{C}$