设 z = e^u sin v, u = xy, v = x + y, 求 (partial z)/(partial x), (partial z)/(partial y) = ( ) A. (partial z)/(partial x) = (partial z)/(partial u) cdot (partial u)/(partial x) + (partial z)/(partial v) cdot (partial v)/(partial x) = e^v [y sin (x + y)+ cos (x + y)] (partial z)/(partial y) = (partial z)/(partial u) cdot (partial u)/(partial y) + (partial z)/(partial v) cdot (partial v)/(partial y) = e^v [x sin (x + y)+ cos (x + y)]B. (partial z)/(partial x) = (partial z)/(partial u) cdot (partial u)/(partial x) + (partial z)/(partial v) cdot (partial v)/(partial x) = e^v [sin (x + y)+ cos (x + y)] (partial z)/(partial y) = (partial z)/(partial u) cdot (partial u)/(partial y) + (partial z)/(partial v) cdot (partial v)/(partial y) = e^v [x sin (x + y)+ cos (x + y)]C. (partial z)/(partial x) = (partial z)/(partial u) cdot (partial u)/(partial x) + (partial z)/(partial v) cdot (partial v)/(partial x) = e^v [y sin (x + y)+ cos (x + y)] (partial z)/(partial y) = (partial z)/(partial u) cdot (partial u)/(partial y) + (partial z)/(partial v) cdot (partial v)/(partial y) = e^v [sin (x + y)+ cos (x + y)]D. (partial z)/(partial x) = (partial z)/(partial u) cdot (partial u)/(partial x) + (partial z)/(partial v) cdot (partial v)/(partial x) = [y sin (x + y)+ cos (x + y)] (partial z)/(partial y) = (partial z)/(partial u) cdot (partial u)/(partial y) + (partial z)/(partial v) cdot (partial v)/(partial y) = e^v [x sin (x + y)+ cos (x + y)]

为了求解 $ z = e^u \sin v $ 关于 $ x $ 和 $ y $ 的偏导数，其中 $ u = x + y $ 和 $ v = x + y $，我们将使用链式法则。链式法则指出，如果 $ z $ 是 $ u $ 和 $ v $ 的函数，而 $ u $ 和 $ v $ 又是 $ x $ 和 $ y $ 的函数，那么 $ z $ 关于 $ x $ 的偏导数由下式给出： \[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} \] 同样，$ z $ 关于 $ y $ 的偏导数由下式给出： \[ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \] 首先，我们计算 $ z $ 关于 $ u $ 和 $ v $ 的偏导数： \[ \frac{\partial z}{\partial u} = e^u \sin v \] \[ \frac{\partial z}{\partial v} = e^u \cos v \] 接下来，我们计算 $ u $ 和 $ v $ 关于 $ x $ 和 $ y $ 的偏导数： \[ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 1 \] \[ \frac{\partial v}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = 1 \] 现在，我们可以使用链式法则来求 $ z $ 关于 $ x $ 的偏导数： \[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = e^u \sin v \cdot 1 + e^u \cos v \cdot 1 = e^u (\sin v + \cos v) \] 由于 $ u = x + y $ 和 $ v = x + y $，我们将 $ u $ 和 $ v $ 代入表达式： \[ \frac{\partial z}{\partial x} = e^{x+y} [\sin(x+y) + \cos(x+y)] \] 同样，我们使用链式法则来求 $ z $ 关于 $ y $ 的偏导数： \[ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = e^u \sin v \cdot 1 + e^u \cos v \cdot 1 = e^u (\sin v + \cos v) \] 再次，由于 $ u = x + y $ 和 $ v = x + y $，我们将 $ u $ 和 $ v $ 代入表达式： \[ \frac{\partial z}{\partial y} = e^{x+y} [\sin(x+y) + \cos(x+y)] \] 因此，答案是： \[ \frac{\partial z}{\partial x} = e^{x+y} [\sin(x+y) + \cos(x+y)] \] \[ \frac{\partial z}{\partial y} = e^{x+y} [\sin(x+y) + \cos(x+y)] \] 正确选项是 $\boxed{A}$。