📝 题目
5. 3.1 求由下列方程所定义的函数 $y$ 的一阶、二阶导数:
(1) $\ln \sqrt{{x}^{2} + {y}^{2}} = \arctan \frac{y}{x}$ ; (2) ${xy} + {2}^{y} = 0$ .
💡 答案与解析
5. 3.1 (1) ${y}^{\prime } = \frac{x + y}{x - y},{y}^{\prime \prime } = \frac{2{x}^{2} + {y}^{2}}{{\left( x - y\right) }^{3}}$ ;
(2) ${y}^{\prime } = - \frac{y}{{2}^{y}\ln 2 + x}$ ,
$$ {y}^{\prime \prime } = \frac{1}{{\left( {2}^{y}\ln 2 + x\right) }^{3}} \cdot \left( {{2xy} + y{2}^{y + 1}\ln 2 - {y}^{2}{2}^{y}{\left( \ln 2\right) }^{2}}\right) . $$