方企勤 第五章 多元函数微分学 第5.3题
📝 题目
5.3.9 求函数 $z = f\left( {x + y,z + y}\right)$ 的二阶偏导数.
💡 答案解析
5.3.9 ${z}_{xx}^{\prime \prime } = \frac{{\left( 1 - {f}_{2}^{\prime }\right) }^{2}{f}_{11}^{\prime \prime } + 2{f}_{1}^{\prime }\left( {1 - {f}_{2}^{\prime }}\right) {f}_{12}^{\prime \prime } + {f}_{1}^{\prime }2{f}_{22}^{\prime \prime }}{{\left( 1 - {f}_{2}^{\prime }\right) }^{3}}$ ,
$$ {z}_{xy}^{\prime \prime } = \frac{{\left( 1 - {f}_{2}^{\prime }\right) }^{2}{f}_{11}^{\prime \prime } + \left( {1 - {f}_{2}^{\prime }}\right) \left( {1 + 2{f}_{1}^{\prime }}\right) {f}_{12}^{\prime \prime } + {f}_{1}^{\prime }\left( {1 + {f}_{1}^{\prime }}\right) {f}_{22}^{\prime \prime }}{{\left( 1 - {f}_{2}^{\prime }\right) }^{3}}, $$
$$ {z}_{yy}^{\prime \prime } = \frac{{\left( 1 - {f}_{2}^{\prime }\right) }^{2}{f}_{11}^{\prime \prime } + 2\left( {1 - {f}_{2}^{\prime }}\right) \left( {1 + {f}_{1}^{\prime }}\right) {f}_{12}^{\prime \prime } + {\left( 1 + {f}_{1}^{\prime }\right) }^{2}{f}_{22}^{\prime \prime }}{{\left( 1 - {f}_{2}^{\prime }\right) }^{3}}. $$