📝 题目
5. 2.1 求下列函数的偏导数:
(1) $u = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}$ ; (2) $u = \tan \frac{{x}^{2}}{y}$ ;
(3) $u = \sin \left( {x\cos y}\right)$ ; (4) $u = {\mathrm{e}}^{x/y}$ ;
(5) $u = \ln \sqrt{{x}^{2} + {y}^{2}}$ ; (6) $u = \arctan \frac{x + y}{1 - {xy}}$ ;
(7) $u = {\left( \frac{x}{y}\right) }^{x}$ ; (8) $u = \arccos \frac{z}{\sqrt{{x}^{2} + {y}^{2}}}$ .
💡 答案与解析
5. 2.1 (1) ${u}_{x}^{\prime } = \frac{{y}^{2}}{{\left( {x}^{2} + {y}^{2}\right) }^{\frac{3}{2}}},{u}_{y}^{\prime } = - \frac{xy}{{\left( {x}^{2} + {y}^{2}\right) }^{\frac{3}{2}}}$ ;
(2) ${u}_{x}^{\prime } = 2\frac{x}{y}{\sec }^{2}\frac{{x}^{2}}{y},{u}_{y}^{\prime } = - \frac{{x}^{2}}{{y}^{2}}{\sec }^{2}\frac{{x}^{2}}{y}$ ;
(3) ${u}_{x}^{\prime } = \cos y\cos \left( {x\cos y}\right) ,{u}_{y}^{\prime } = - x\sin y\cos \left( {x\cos y}\right)$ ;
(4) ${u}_{x}^{\prime } = \frac{1}{y}{\mathrm{e}}^{\frac{x}{y}} - \frac{x}{{y}^{2}}{\mathrm{e}}^{\frac{x}{y}}$ ;
(5) ${u}_{x}^{\prime } = \frac{x}{{x}^{2} + {y}^{2}},{u}_{y}^{\prime } = \frac{y}{{x}^{2} + {y}^{2}}$ ;
(6) $u = \arctan x + \arctan y,{u}_{x}^{\prime } = \frac{1}{{x}^{2} + 1},{u}_{y}^{\prime } = \frac{1}{{y}^{2} + 1}$ ;
(7) ${u}_{x}^{\prime } = \frac{z}{x}{\left( \frac{x}{y}\right) }^{z},{u}_{y}^{\prime } = - \frac{z}{y}{\left( \frac{x}{y}\right) }^{z},{u}_{z}^{\prime } = {\left( \frac{x}{y}\right) }^{z}\ln \frac{x}{y}$ ;
(8) ${u}_{x}^{\prime } = \frac{xz}{\left( {{x}^{2} + {y}^{2}}\right) \sqrt{{x}^{2} + {y}^{2} - {z}^{2}}},{u}_{y}^{\prime } = \frac{yz}{\left( {{x}^{2} + {y}^{2}}\right) \sqrt{{x}^{2} + {y}^{2} - {z}^{2}}}$ ,
$$ {u}_{z}^{\prime } = - \frac{1}{\sqrt{{x}^{2} + {y}^{2} - {z}^{2}}}. $$