题目
1.设 ^2+(y)^2+(z)^2-z=0, 求 dfrac ({a)^2z}(a{y)^2}
题目解答
答案
解析
步骤 1：求偏导数 $\dfrac{\partial z}{\partial y}$
给定方程 ${x}^{2}+{y}^{2}+{z}^{2}-z=0$，对 $y$ 求偏导数，得到：
$2y + 2z\dfrac{\partial z}{\partial y} - \dfrac{\partial z}{\partial y} = 0$
整理得到：
$\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$
步骤 2：求二阶偏导数 $\dfrac{{\partial}^{2}z}{\partial {y}^{2}}$
对 $\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$ 再次对 $y$ 求偏导数，得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z) - 2y(-2)\dfrac{\partial z}{\partial y}}{(1-2z)^{2}}$
将 $\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$ 代入上式，得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z) + 4y\dfrac{2y}{1-2z}}{(1-2z)^{2}}$
化简得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z)^{2} + 8y^{2}}{(1-2z)^{3}}$
给定方程 ${x}^{2}+{y}^{2}+{z}^{2}-z=0$，对 $y$ 求偏导数，得到：
$2y + 2z\dfrac{\partial z}{\partial y} - \dfrac{\partial z}{\partial y} = 0$
整理得到：
$\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$
步骤 2：求二阶偏导数 $\dfrac{{\partial}^{2}z}{\partial {y}^{2}}$
对 $\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$ 再次对 $y$ 求偏导数，得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z) - 2y(-2)\dfrac{\partial z}{\partial y}}{(1-2z)^{2}}$
将 $\dfrac{\partial z}{\partial y} = \dfrac{2y}{1-2z}$ 代入上式，得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z) + 4y\dfrac{2y}{1-2z}}{(1-2z)^{2}}$
化简得到：
$\dfrac{{\partial}^{2}z}{\partial {y}^{2}} = \dfrac{2(1-2z)^{2} + 8y^{2}}{(1-2z)^{3}}$