本次共计算 1 个题目:每一题对 t 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{(p - c - t(p + c + f))}{(t(b - (p + c + f)))} 关于 t 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = - \frac{pt}{(bt - pt - ct - ft)} - \frac{ct}{(bt - pt - ct - ft)} + \frac{p}{(bt - pt - ct - ft)} - \frac{c}{(bt - pt - ct - ft)} - \frac{ft}{(bt - pt - ct - ft)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( - \frac{pt}{(bt - pt - ct - ft)} - \frac{ct}{(bt - pt - ct - ft)} + \frac{p}{(bt - pt - ct - ft)} - \frac{c}{(bt - pt - ct - ft)} - \frac{ft}{(bt - pt - ct - ft)}\right)}{dt}\\=& - (\frac{-(b - p - c - f)}{(bt - pt - ct - ft)^{2}})pt - \frac{p}{(bt - pt - ct - ft)} - (\frac{-(b - p - c - f)}{(bt - pt - ct - ft)^{2}})ct - \frac{c}{(bt - pt - ct - ft)} + (\frac{-(b - p - c - f)}{(bt - pt - ct - ft)^{2}})p + 0 - (\frac{-(b - p - c - f)}{(bt - pt - ct - ft)^{2}})c + 0 - (\frac{-(b - p - c - f)}{(bt - pt - ct - ft)^{2}})ft - \frac{f}{(bt - pt - ct - ft)}\\=&\frac{pbt}{(bt - pt - ct - ft)^{2}} - \frac{p^{2}t}{(bt - pt - ct - ft)^{2}} - \frac{2pct}{(bt - pt - ct - ft)^{2}} - \frac{2pft}{(bt - pt - ct - ft)^{2}} - \frac{pb}{(bt - pt - ct - ft)^{2}} + \frac{cbt}{(bt - pt - ct - ft)^{2}} - \frac{c^{2}t}{(bt - pt - ct - ft)^{2}} - \frac{2cft}{(bt - pt - ct - ft)^{2}} + \frac{cb}{(bt - pt - ct - ft)^{2}} + \frac{pf}{(bt - pt - ct - ft)^{2}} + \frac{p^{2}}{(bt - pt - ct - ft)^{2}} - \frac{cf}{(bt - pt - ct - ft)^{2}} - \frac{c^{2}}{(bt - pt - ct - ft)^{2}} + \frac{fbt}{(bt - pt - ct - ft)^{2}} - \frac{p}{(bt - pt - ct - ft)} - \frac{c}{(bt - pt - ct - ft)} - \frac{f^{2}t}{(bt - pt - ct - ft)^{2}} - \frac{f}{(bt - pt - ct - ft)}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!