题目
设=(u)^v，=(u)^v，=(u)^v，则=(u)^vA.=(u)^vB.=(u)^vC.=(u)^vD.=(u)^v
设
，
，
，则
A.
B.
C.
D.
题目解答
答案
已知，，，即
，根据导数公式得
，则
，故选项是B。
解析
步骤 1：计算偏导数$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$
首先，根据链式法则，我们有：
$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$
$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$
其中，$z = u^v$，$u = \ln \sqrt{x^2 + y^2}$，$v = \arctan \frac{y}{x}$。
步骤 2：计算$\frac{\partial z}{\partial u}$和$\frac{\partial z}{\partial v}$
$$\frac{\partial z}{\partial u} = v u^{v-1}$$
$$\frac{\partial z}{\partial v} = u^v \ln u$$
步骤 3：计算$\frac{\partial u}{\partial x}$，$\frac{\partial u}{\partial y}$，$\frac{\partial v}{\partial x}$和$\frac{\partial v}{\partial y}$
$$\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}$$
$$\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2}$$
$$\frac{\partial v}{\partial x} = -\frac{y}{x^2 + y^2}$$
$$\frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2}$$
步骤 4：计算$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$在点$(e^2, 0)$的值
$$\frac{\partial z}{\partial x} = v u^{v-1} \frac{x}{x^2 + y^2} + u^v \ln u \left(-\frac{y}{x^2 + y^2}\right)$$
$$\frac{\partial z}{\partial y} = v u^{v-1} \frac{y}{x^2 + y^2} + u^v \ln u \left(\frac{x}{x^2 + y^2}\right)$$
在点$(e^2, 0)$，$u = \ln \sqrt{e^4} = 2$，$v = \arctan 0 = 0$，因此：
$$\frac{\partial z}{\partial x} = 0 \cdot 2^{-1} \cdot \frac{e^2}{e^4} + 2^0 \ln 2 \cdot 0 = 0$$
$$\frac{\partial z}{\partial y} = 0 \cdot 2^{-1} \cdot 0 + 2^0 \ln 2 \cdot \frac{e^2}{e^4} = \frac{\ln 2}{e^2}$$
步骤 5：计算$dz$
$$dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$$
$$dz = 0 \cdot dx + \frac{\ln 2}{e^2} dy$$
首先，根据链式法则，我们有：
$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$
$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$
其中，$z = u^v$，$u = \ln \sqrt{x^2 + y^2}$，$v = \arctan \frac{y}{x}$。
步骤 2：计算$\frac{\partial z}{\partial u}$和$\frac{\partial z}{\partial v}$
$$\frac{\partial z}{\partial u} = v u^{v-1}$$
$$\frac{\partial z}{\partial v} = u^v \ln u$$
步骤 3：计算$\frac{\partial u}{\partial x}$，$\frac{\partial u}{\partial y}$，$\frac{\partial v}{\partial x}$和$\frac{\partial v}{\partial y}$
$$\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}$$
$$\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2}$$
$$\frac{\partial v}{\partial x} = -\frac{y}{x^2 + y^2}$$
$$\frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2}$$
步骤 4：计算$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$在点$(e^2, 0)$的值
$$\frac{\partial z}{\partial x} = v u^{v-1} \frac{x}{x^2 + y^2} + u^v \ln u \left(-\frac{y}{x^2 + y^2}\right)$$
$$\frac{\partial z}{\partial y} = v u^{v-1} \frac{y}{x^2 + y^2} + u^v \ln u \left(\frac{x}{x^2 + y^2}\right)$$
在点$(e^2, 0)$，$u = \ln \sqrt{e^4} = 2$，$v = \arctan 0 = 0$，因此：
$$\frac{\partial z}{\partial x} = 0 \cdot 2^{-1} \cdot \frac{e^2}{e^4} + 2^0 \ln 2 \cdot 0 = 0$$
$$\frac{\partial z}{\partial y} = 0 \cdot 2^{-1} \cdot 0 + 2^0 \ln 2 \cdot \frac{e^2}{e^4} = \frac{\ln 2}{e^2}$$
步骤 5：计算$dz$
$$dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$$
$$dz = 0 \cdot dx + \frac{\ln 2}{e^2} dy$$