题目:设 \( e^z - xyz = 0 \), 求 \(\frac{\partial^2 z}{\partial x \partial y}\)
时间: 2024-04-07 22:26:57 浏览: 233
作业4(8.4-8.6).docx
这是一个求偏导数的问题。给定方程 \( e^z - xyz = 0 \),我们需要求解 \(\frac{\partial^2 z}{\partial x \partial y}\)。
为了求解这个偏导数,我们可以先对方程两边分别对 x 和 y 进行求导。首先对 x 求偏导数,得到:
\(\frac{\partial}{\partial x}(e^z - xyz) = \frac{\partial}{\partial x}(0)\)
应用链式法则,我们可以得到:
\(e^z \frac{\partial z}{\partial x} - yz - xy \frac{\partial z}{\partial x} = 0\)
整理后,我们可以得到:
\((e^z - x) \frac{\partial z}{\partial x} = yz\)
接下来,对上式两边再次对 y 求偏导数,得到:
\(\frac{\partial}{\partial y}((e^z - x) \frac{\partial z}{\partial x}) = \frac{\partial}{\partial y}(yz)\)
应用链式法则,我们可以得到:
\((e^z - x) \frac{\partial^2 z}{\partial x \partial y} + \frac{\partial z}{\partial x} \frac{\partial}{\partial y}(e^z - x) = z\)
整理后,我们可以得到:
\((e^z - x) \frac{\partial^2 z}{\partial x \partial y} + \frac{\partial z}{\partial x} e^z = z\)
最后,我们可以解出 \(\frac{\partial^2 z}{\partial x \partial y}\):
\(\frac{\partial^2 z}{\partial x \partial y} = \frac{z - \frac{\partial z}{\partial x} e^z}{e^z - x}\)
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