题目

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)

设$z=\sin(uv)$，$u=x+y$，$v=x-y$，则$\frac{\partial z}{\partial y}=$【】

  • A. $2x\cos(x^2-y^2)$
  • B. $-2x\cos(x^2-y^2)$
  • C. $-2y\cos(x^2-y^2)$
  • D. $2y\cos(x^2-y^2)$

题目解答

答案

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： \[ \boxed{C} \]

解析

本题考查复合函数求偏导数的知识，解题思路是利用复合函数求偏导的链式法则来计算$\frac{\partial z}{\partial y}$。

已知$z = \sin(uv)$，$u = x + y$，$v = x - y$，根据复合函数求偏导的链式法则$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$。

  1. 计算$\frac{\partial z}{\partial u}$和$\frac{\partial z}{\partial v}$：
    • 对$z = \sin(uv)$关于$u$求偏导数，把$v$看作常数，根据求导公式$(\sin t)^\prime=\cos t$，可得$\frac{\partial z}{\partial u}=\frac{\partial}{\partial u}\sin(uv)=v\cos(uv)$。
    • 对$z = \sin(uv)$关于$v$求偏导数，把$u$看作常数，同理可得$\frac{\partial z}{\partial v}=\frac{\partial}{\partial v}\sin(uv)=u\cos(uv)$。
  2. 计算$\frac{\partial u}{\partial y}$和$\frac{\partial v}{\partial y}$：
    • 对$u = x + y$关于$y$求偏导数，把$x$看作常数，可得$\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}(x + y)=1$。
    • 对$v = x - y$关于$y$求偏导数，把$x$看作常数，可得$\frac{\partial v}{\partial y}=\frac{\partial}{\partial y}(x - y)= -1$。
  3. 将上述偏导数代入链式法则公式：
    • $\frac{\partial z}{\partial y}=(v\cos(uv))\times1+(u\cos(uv))\times(-1)=v\cos(uv)-u\cos(uv)=(v - u)\cos(uv)$。
  4. 把$u = x + y$和$v = x - y$代入上式：
    • $\frac{\partial z}{\partial y}=[(x - y)-(x + y)]\cos((x + y)(x - y))$。
    • 先化简中括号内的式子：$(x - y)-(x + y)=x - y - x - y=-2y$。
    • 再根据平方差公式$(a + b)(a - b)=a^2 - b^2$，可得$(x + y)(x - y)=x^2 - y^2$。
    • 所以$\frac{\partial z}{\partial y}=-2y\cos(x^2 - y^2)$。