题目
9.已知 =dfrac (x-y)({e)^x} ,下列选项成立的是 ()-|||-A. dfrac (partial z)(partial x)-dfrac (partial z)(partial y)=0 B. dfrac (partial z)(partial x)+dfrac (partial z)(partial y)=0 C. dfrac (partial z)(partial x)-dfrac (partial z)(partial z)=-z D. dfrac (partial z)(partial x)+dfrac (partial z)(partial y)=-z
题目解答
答案
D. $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}=-z$
解析
步骤 1：计算 $\dfrac {\partial z}{\partial x}$
根据 $z=\dfrac {x-y}{{e}^{x}}$，对 $x$ 求偏导数，得到 $\dfrac {\partial z}{\partial x}=\dfrac {1}{{e}^{x}}-\dfrac {x-y}{{e}^{x}}=\dfrac {1-x+y}{{e}^{x}}$。
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
根据 $z=\dfrac {x-y}{{e}^{x}}$，对 $y$ 求偏导数，得到 $\dfrac {\partial z}{\partial y}=-\dfrac {1}{{e}^{x}}$。
步骤 3：计算 $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
将步骤 1 和步骤 2 的结果相加，得到 $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}=\dfrac {1-x+y}{{e}^{x}}-\dfrac {1}{{e}^{x}}=\dfrac {1-x+y-1}{{e}^{x}}=\dfrac {-x+y}{{e}^{x}}=-z$。
根据 $z=\dfrac {x-y}{{e}^{x}}$，对 $x$ 求偏导数，得到 $\dfrac {\partial z}{\partial x}=\dfrac {1}{{e}^{x}}-\dfrac {x-y}{{e}^{x}}=\dfrac {1-x+y}{{e}^{x}}$。
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
根据 $z=\dfrac {x-y}{{e}^{x}}$，对 $y$ 求偏导数，得到 $\dfrac {\partial z}{\partial y}=-\dfrac {1}{{e}^{x}}$。
步骤 3：计算 $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}$
将步骤 1 和步骤 2 的结果相加，得到 $\dfrac {\partial z}{\partial x}+\dfrac {\partial z}{\partial y}=\dfrac {1-x+y}{{e}^{x}}-\dfrac {1}{{e}^{x}}=\dfrac {1-x+y-1}{{e}^{x}}=\dfrac {-x+y}{{e}^{x}}=-z$。