题目
1.求下列各函数的一阶偏导数:-|||-(1) =(x)^2ln ((x)^2+(y)^2);
题目解答
答案
解析
步骤 1：计算 $\dfrac {\partial z}{\partial x}$
首先，我们对函数 $z={x}^{2}\ln ({x}^{2}+{y}^{2})$ 关于 $x$ 求偏导数。根据乘积法则，我们有：
$$\dfrac {\partial z}{\partial x} = \dfrac {\partial}{\partial x}({x}^{2})\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {\partial}{\partial x}(\ln ({x}^{2}+{y}^{2}))$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}\dfrac {\partial}{\partial x}({x}^{2}+{y}^{2})$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}(2x)$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + \dfrac {2{x}^{3}}{{x}^{2}+{y}^{2}}$$
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
接下来，我们对函数 $z={x}^{2}\ln ({x}^{2}+{y}^{2})$ 关于 $y$ 求偏导数。根据乘积法则，我们有：
$$\dfrac {\partial z}{\partial y} = {x}^{2}\dfrac {\partial}{\partial y}(\ln ({x}^{2}+{y}^{2}))$$
$$= {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}\dfrac {\partial}{\partial y}({x}^{2}+{y}^{2})$$
$$= {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}(2y)$$
$$= \dfrac {2{x}^{2}y}{{x}^{2}+{y}^{2}}$$
首先，我们对函数 $z={x}^{2}\ln ({x}^{2}+{y}^{2})$ 关于 $x$ 求偏导数。根据乘积法则，我们有：
$$\dfrac {\partial z}{\partial x} = \dfrac {\partial}{\partial x}({x}^{2})\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {\partial}{\partial x}(\ln ({x}^{2}+{y}^{2}))$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}\dfrac {\partial}{\partial x}({x}^{2}+{y}^{2})$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}(2x)$$
$$= 2x\ln ({x}^{2}+{y}^{2}) + \dfrac {2{x}^{3}}{{x}^{2}+{y}^{2}}$$
步骤 2：计算 $\dfrac {\partial z}{\partial y}$
接下来，我们对函数 $z={x}^{2}\ln ({x}^{2}+{y}^{2})$ 关于 $y$ 求偏导数。根据乘积法则，我们有：
$$\dfrac {\partial z}{\partial y} = {x}^{2}\dfrac {\partial}{\partial y}(\ln ({x}^{2}+{y}^{2}))$$
$$= {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}\dfrac {\partial}{\partial y}({x}^{2}+{y}^{2})$$
$$= {x}^{2}\dfrac {1}{{x}^{2}+{y}^{2}}(2y)$$
$$= \dfrac {2{x}^{2}y}{{x}^{2}+{y}^{2}}$$