题目
设 z = e^u sin v，而 u = xy，v = x + y，则 (partial z)/(partial x) = ( )A. z = e^xy [y sin (x + y)+ cos (x + y)]B. z = e^xy [sin (x + y)+ cos (x + y)]C. z = e^xy [x sin (x + y)+ cos (x + y)]D. z = e^xy [x sin (x + y)+ y cos (x + y)]
设 $z = e^u \sin v$，而 $u = xy$，$v = x + y$，则 $\frac{\partial z}{\partial x} = (\quad)$
A. $z = e^{xy} [y \sin (x + y)+ \cos (x + y)]$
B. $z = e^{xy} [\sin (x + y)+ \cos (x + y)]$
C. $z = e^{xy} [x \sin (x + y)+ \cos (x + y)]$
D. $z = e^{xy} [x \sin (x + y)+ y \cos (x + y)]$
题目解答
答案
A. $z = e^{xy} [y \sin (x + y)+ \cos (x + y)]$
解析
步骤 1：应用链式法则
根据链式法则，有： \[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \] 其中，$z = e^u \sin v$，$u = xy$，$v = x + y$。
步骤 2：计算各偏导数
计算各偏导数： \[ \frac{\partial z}{\partial u} = e^u \sin v, \quad \frac{\partial z}{\partial v} = e^u \cos v, \quad \frac{\partial u}{\partial x} = y, \quad \frac{\partial v}{\partial x} = 1 \]
步骤 3：代入并简化
代入得： \[ \frac{\partial z}{\partial x} = e^u \sin v \cdot y + e^u \cos v \cdot 1 = e^{xy} [y \sin(x + y) + \cos(x + y)] \] 对应选项A。
根据链式法则，有： \[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \] 其中，$z = e^u \sin v$，$u = xy$，$v = x + y$。
步骤 2：计算各偏导数
计算各偏导数： \[ \frac{\partial z}{\partial u} = e^u \sin v, \quad \frac{\partial z}{\partial v} = e^u \cos v, \quad \frac{\partial u}{\partial x} = y, \quad \frac{\partial v}{\partial x} = 1 \]
步骤 3：代入并简化
代入得： \[ \frac{\partial z}{\partial x} = e^u \sin v \cdot y + e^u \cos v \cdot 1 = e^{xy} [y \sin(x + y) + \cos(x + y)] \] 对应选项A。