题目
已知 (x,y)=(x)^2arctan dfrac (y)(x)-(y)^2arctan dfrac (x)(y) ,求 dfrac ({a)^2f}(axpartial y)
题目解答
答案
解析
步骤 1：计算 $\dfrac {\partial f}{\partial x}$
首先，我们对函数 $f(x,y)={x}^{2}\arctan \dfrac {y}{x}-{y}^{2}\arctan \dfrac {x}{y}$ 关于 $x$ 求偏导数。根据链式法则和乘积法则，我们得到：
$$\dfrac {\partial f}{\partial x}=2x\arctan \dfrac {y}{x}+\dfrac {{x}^{2}}{1+{(\dfrac {y}{x})}^{2}}\cdot (-\dfrac {y}{{x}^{2}})-\dfrac {{y}^{2}}{1+{(\dfrac {x}{y})}^{2}}\cdot \dfrac {1}{y}$$
$$=2x\arctan \dfrac {y}{x}-\dfrac {{x}^{2}y}{{x}^{2}+{y}^{2}}-\dfrac {{y}^{3}}{{x}^{2}+{y}^{2}}$$
步骤 2：计算 $\dfrac {{\partial }^{2}f}{\partial x\partial y}$
接下来，我们对 $\dfrac {\partial f}{\partial x}$ 关于 $y$ 求偏导数。根据链式法则和乘积法则，我们得到：
$$\dfrac {{\partial }^{2}f}{\partial x\partial y}=\dfrac {2x}{1+{(\dfrac {y}{x})}^{2}}\cdot \dfrac {1}{x}-\dfrac {{x}^{2}({x}^{2}+{y}^{2})-{x}^{2}y\cdot 2y}{{({x}^{2}+{y}^{2})}^{2}}-\dfrac {3{y}^{2}({x}^{2}+{y}^{2})-{y}^{3}\cdot 2y}{{({x}^{2}+{y}^{2})}^{2}}$$
$$=\dfrac {{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$$
首先，我们对函数 $f(x,y)={x}^{2}\arctan \dfrac {y}{x}-{y}^{2}\arctan \dfrac {x}{y}$ 关于 $x$ 求偏导数。根据链式法则和乘积法则，我们得到：
$$\dfrac {\partial f}{\partial x}=2x\arctan \dfrac {y}{x}+\dfrac {{x}^{2}}{1+{(\dfrac {y}{x})}^{2}}\cdot (-\dfrac {y}{{x}^{2}})-\dfrac {{y}^{2}}{1+{(\dfrac {x}{y})}^{2}}\cdot \dfrac {1}{y}$$
$$=2x\arctan \dfrac {y}{x}-\dfrac {{x}^{2}y}{{x}^{2}+{y}^{2}}-\dfrac {{y}^{3}}{{x}^{2}+{y}^{2}}$$
步骤 2：计算 $\dfrac {{\partial }^{2}f}{\partial x\partial y}$
接下来，我们对 $\dfrac {\partial f}{\partial x}$ 关于 $y$ 求偏导数。根据链式法则和乘积法则，我们得到：
$$\dfrac {{\partial }^{2}f}{\partial x\partial y}=\dfrac {2x}{1+{(\dfrac {y}{x})}^{2}}\cdot \dfrac {1}{x}-\dfrac {{x}^{2}({x}^{2}+{y}^{2})-{x}^{2}y\cdot 2y}{{({x}^{2}+{y}^{2})}^{2}}-\dfrac {3{y}^{2}({x}^{2}+{y}^{2})-{y}^{3}\cdot 2y}{{({x}^{2}+{y}^{2})}^{2}}$$
$$=\dfrac {{x}^{2}-{y}^{2}}{{x}^{2}+{y}^{2}}$$