本次共计算 1 个题目:每一题对 q 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数(q - f)(\frac{1}{2} + \frac{(s(q - p) + a(kn - hm))}{(2(ts - ab))}) + (kn - g)(\frac{1}{2} + \frac{(t(kn - hm) + b(q - p))}{(2(ts - ab))}) 关于 q 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{1}{2}q + \frac{sq^{2}}{(2st - 2ab)} - \frac{spq}{(2st - 2ab)} + \frac{aknq}{(2st - 2ab)} - \frac{ahmq}{(2st - 2ab)} - \frac{1}{2}f - \frac{fsq}{(2st - 2ab)} + \frac{fsp}{(2st - 2ab)} - \frac{fakn}{(2st - 2ab)} + \frac{fahm}{(2st - 2ab)} + \frac{1}{2}kn - \frac{kntg}{(2st - 2ab)} - \frac{knhmt}{(2st - 2ab)} + \frac{knbq}{(2st - 2ab)} - \frac{pknb}{(2st - 2ab)} - \frac{1}{2}g + \frac{k^{2}n^{2}t}{(2st - 2ab)} + \frac{hmtg}{(2st - 2ab)} - \frac{bgq}{(2st - 2ab)} + \frac{pbg}{(2st - 2ab)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{1}{2}q + \frac{sq^{2}}{(2st - 2ab)} - \frac{spq}{(2st - 2ab)} + \frac{aknq}{(2st - 2ab)} - \frac{ahmq}{(2st - 2ab)} - \frac{1}{2}f - \frac{fsq}{(2st - 2ab)} + \frac{fsp}{(2st - 2ab)} - \frac{fakn}{(2st - 2ab)} + \frac{fahm}{(2st - 2ab)} + \frac{1}{2}kn - \frac{kntg}{(2st - 2ab)} - \frac{knhmt}{(2st - 2ab)} + \frac{knbq}{(2st - 2ab)} - \frac{pknb}{(2st - 2ab)} - \frac{1}{2}g + \frac{k^{2}n^{2}t}{(2st - 2ab)} + \frac{hmtg}{(2st - 2ab)} - \frac{bgq}{(2st - 2ab)} + \frac{pbg}{(2st - 2ab)}\right)}{dq}\\=&\frac{1}{2} + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})sq^{2} + \frac{s*2q}{(2st - 2ab)} - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})spq - \frac{sp}{(2st - 2ab)} + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})aknq + \frac{akn}{(2st - 2ab)} - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})ahmq - \frac{ahm}{(2st - 2ab)} + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fsq - \frac{fs}{(2st - 2ab)} + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fsp + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fakn + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})fahm + 0 + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})kntg + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})knhmt + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})knbq + \frac{knb}{(2st - 2ab)} - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pknb + 0 + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})k^{2}n^{2}t + 0 + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})hmtg + 0 - (\frac{-(0 + 0)}{(2st - 2ab)^{2}})bgq - \frac{bg}{(2st - 2ab)} + (\frac{-(0 + 0)}{(2st - 2ab)^{2}})pbg + 0\\=&\frac{2sq}{(2st - 2ab)} - \frac{sp}{(2st - 2ab)} + \frac{akn}{(2st - 2ab)} - \frac{ahm}{(2st - 2ab)} - \frac{fs}{(2st - 2ab)} + \frac{knb}{(2st - 2ab)} - \frac{bg}{(2st - 2ab)} + \frac{1}{2}\\ \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!