本次共计算 1 个题目:每一题对 x 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{(({a}^{2} + {b}^{2})(d - g) + ({c}^{2} + {d}^{2})(g - b) + ({f}^{2} + {g}^{2})(d - b))}{(2a(d - g) - 2b(c - f) + 2cg - 2fd)} - x 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{a^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{a^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{b^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{b^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{gc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{d^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bd^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{df^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bf^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{dg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - x\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{a^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{a^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{b^{2}d}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{b^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{gc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bc^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{d^{2}g}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bd^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{df^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bf^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} + \frac{dg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - \frac{bg^{2}}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)} - x\right)}{dx}\\=&(\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})a^{2}d + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})a^{2}g + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})b^{2}d + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})b^{2}g + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})gc^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bc^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})d^{2}g + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bd^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})df^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bf^{2} + 0 + (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})dg^{2} + 0 - (\frac{-(0 + 0 + 0 + 0 + 0 + 0)}{(2ad - 2ag - 2bc + 2bf + 2gc - 2df)^{2}})bg^{2} + 0 - 1\\=& - 1\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!