Question - Partial Derivatives of the Function
mathpartial derivativesmixed partial derivativesfirst-order partial derivativesecond-order partial derivative
Solution:
题目要求我们计算函数 $$ z = x\ln(y) $$ 的混合偏导数 $$ \frac{\partial^3 z}{\partial x^2\partial y} $$ 和 $$ \frac{\partial^3 z}{\partial x\partial y^2} $$。首先，我们需要找到 $$ z $$ 关于 $$ x $$ 和 $$ y $$ 的一阶偏导数：$$\frac{\partial z}{\partial x} = \ln(y)$$$$\frac{\partial z}{\partial y} = \frac{x}{y}$$然后，我们对第一个偏导数 $$ \frac{\partial z}{\partial x} $$ 分别对 $$ x $$ 和 $$ y $$ 求导：$$\frac{\partial^2 z}{\partial x^2} = 0$$$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} (\ln(y)) = \frac{1}{y}$$因为 $$ \frac{\partial^2 z}{\partial x^2} = 0 $$， 所以 $$ \frac{\partial^3 z}{\partial x^2\partial y} = 0 $$。再对 $$ \frac{\partial z}{\partial y} = \frac{x}{y} $$ 关于 $$ y $$ 求导：$$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{x}{y}\right) = -\frac{x}{y^2}$$然后对这个结果关于 $$ x $$ 求导：$$\frac{\partial^3 z}{\partial x\partial y^2} = \frac{\partial}{\partial x} \left(-\frac{x}{y^2}\right) = -\frac{1}{y^2}$$所以，$$\frac{\partial^3 z}{\partial x^2\partial y} : \frac{\partial^3 z}{\partial x\partial y^2} = 0 : -\frac{1}{y^2}$$既然 $$ \frac{\partial^3 z}{\partial x^2\partial y} = 0 $$，这个比值就是未定义的，因为我们不能除以零。所以最后的结果是未定义的。
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