设z=sin(uv)，u=x+y，v=x-y，则(partial z)/(partial y)=【】 A. 2xcos(x^2-y^2)B. -2xcos(x^2-y^2)C. -2ycos(x^2-y^2)D. 2ycos(x^2-y^2)

为了求解 $ \frac{\partial z}{\partial y} $，我们需要使用链式法则。给定 $ z = \sin(uv) $， $ u = x + y $， $ v = x - y $，我们首先需要找到 $ z $ 关于 $ y $ 的偏导数。 链式法则告诉我们： \[ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \] 首先，我们计算 $ \frac{\partial z}{\partial u} $ 和 $ \frac{\partial z}{\partial v} $： \[ \frac{\partial z}{\partial u} = \frac{\partial}{\partial u} \sin(uv) = v \cos(uv) \] \[ \frac{\partial z}{\partial v} = \frac{\partial}{\partial v} \sin(uv) = u \cos(uv) \] 接下来，我们计算 $ \frac{\partial u}{\partial y} $ 和 $ \frac{\partial v}{\partial y} $： \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x + y) = 1 \] \[ \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} (x - y) = -1 \] 现在，将这些偏导数代入链式法则公式中： \[ \frac{\partial z}{\partial y} = (v \cos(uv)) \cdot 1 + (u \cos(uv)) \cdot (-1) \] \[ \frac{\partial z}{\partial y} = v \cos(uv) - u \cos(uv) \] \[ \frac{\partial z}{\partial y} = (v - u) \cos(uv) \] 将 $ u = x + y $ 和 $ v = x - y $ 代入： \[ \frac{\partial z}{\partial y} = [(x - y) - (x + y)] \cos((x + y)(x - y)) \] \[ \frac{\partial z}{\partial y} = (x - y - x - y) \cos(x^2 - y^2) \] \[ \frac{\partial z}{\partial y} = (-2y) \cos(x^2 - y^2) \] \[ \frac{\partial z}{\partial y} = -2y \cos(x^2 - y^2) \] 因此，正确答案是： \[ \boxed{C} \]