设 =(e)^x+ycos (y-x) ,求 dfrac (partial z)(partial x) ,(ax)/(a)y.

步骤 1：计算 $\dfrac {\partial z}{\partial x}$
根据多元函数的偏导数定义，我们对 $z$ 关于 $x$ 求偏导数。这里 $z$ 是 $x$ 和 $y$ 的函数，$z={e}^{x+y}\cos (y-x)$。我们使用乘积法则和链式法则来计算偏导数。
$$
\dfrac {\partial z}{\partial x} = \dfrac {\partial }{\partial x}({e}^{x+y})\cos (y-x) + {e}^{x+y}\dfrac {\partial }{\partial x}(\cos (y-x))
$$
$$
= {e}^{x+y}\cos (y-x) + {e}^{x+y}(-\sin (y-x))(-1)
$$
$$
= {e}^{x+y}\cos (y-x) + {e}^{x+y}\sin (y-x)
$$
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
同样地，我们对 $z$ 关于 $y$ 求偏导数。使用乘积法则和链式法则来计算偏导数。
$$
\dfrac {\partial z}{\partial y} = \dfrac {\partial }{\partial y}({e}^{x+y})\cos (y-x) + {e}^{x+y}\dfrac {\partial }{\partial y}(\cos (y-x))
$$
$$
= {e}^{x+y}\cos (y-x) + {e}^{x+y}(-\sin (y-x))(1)
$$
$$
= {e}^{x+y}\cos (y-x) - {e}^{x+y}\sin (y-x)
$$