本次共计算 1 个题目:每一题对 P 求 2 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数(p - f)(\frac{1}{2} + \frac{(s(p - q) + a(hm - kn))}{(2(ts - ab))}) + (hm - g)(\frac{1}{2} + \frac{(t(hm - kn) + b(p - q))}{(2(ts - ab))}) 关于 P 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{1}{2}p - \frac{psq}{(2st - 2ab)} + \frac{p^{2}s}{(2st - 2ab)} + \frac{pahm}{(2st - 2ab)} - \frac{pakn}{(2st - 2ab)} - \frac{1}{2}f - \frac{pfs}{(2st - 2ab)} + \frac{fsq}{(2st - 2ab)} - \frac{fahm}{(2st - 2ab)} + \frac{fakn}{(2st - 2ab)} + \frac{1}{2}hm - \frac{hmtg}{(2st - 2ab)} - \frac{hmknt}{(2st - 2ab)} + \frac{phmb}{(2st - 2ab)} - \frac{qhmb}{(2st - 2ab)} - \frac{1}{2}g + \frac{h^{2}m^{2}t}{(2st - 2ab)} + \frac{kntg}{(2st - 2ab)} - \frac{pbg}{(2st - 2ab)} + \frac{qbg}{(2st - 2ab)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{1}{2}p - \frac{psq}{(2st - 2ab)} + \frac{p^{2}s}{(2st - 2ab)} + \frac{pahm}{(2st - 2ab)} - \frac{pakn}{(2st - 2ab)} - \frac{1}{2}f - \frac{pfs}{(2st - 2ab)} + \frac{fsq}{(2st - 2ab)} - \frac{fahm}{(2st - 2ab)} + \frac{fakn}{(2st - 2ab)} + \frac{1}{2}hm - \frac{hmtg}{(2st - 2ab)} - \frac{hmknt}{(2st - 2ab)} + \frac{phmb}{(2st - 2ab)} - \frac{qhmb}{(2st - 2ab)} - \frac{1}{2}g + \frac{h^{2}m^{2}t}{(2st - 2ab)} + \frac{kntg}{(2st - 2ab)} - \frac{pbg}{(2st - 2ab)} + \frac{qbg}{(2st - 2ab)}\right)}{dP}\\=&0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})psq + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})p^{2}s + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pahm + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pakn + 0 + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pfs + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fsq + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fahm + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fakn + 0 + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})hmtg + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})hmknt + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})phmb + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})qhmb + 0 + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})h^{2}m^{2}t + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})kntg + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pbg + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})qbg + 0\\=& - 0\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( - 0\right)}{dP}\\=& - 0\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!