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。\[ \begin{equation}\begin{split}【1/1】求函数(\frac{(2x - 1)}{(1 + x)})ln(\frac{(2x - 1)}{(1 + x)}) - \frac{(2x - 1)}{(1 + x)} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{2xln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} - \frac{ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} - \frac{2x}{(x + 1)} + \frac{1}{(x + 1)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{2xln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} - \frac{ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} - \frac{2x}{(x + 1)} + \frac{1}{(x + 1)}\right)}{dx}\\=&2(\frac{-(1 + 0)}{(x + 1)^{2}})xln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)}) + \frac{2ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} + \frac{2x(2(\frac{-(1 + 0)}{(x + 1)^{2}})x + \frac{2}{(x + 1)} - (\frac{-(1 + 0)}{(x + 1)^{2}}))}{(x + 1)(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})} - (\frac{-(1 + 0)}{(x + 1)^{2}})ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)}) - \frac{(2(\frac{-(1 + 0)}{(x + 1)^{2}})x + \frac{2}{(x + 1)} - (\frac{-(1 + 0)}{(x + 1)^{2}}))}{(x + 1)(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})} - 2(\frac{-(1 + 0)}{(x + 1)^{2}})x - \frac{2}{(x + 1)} + (\frac{-(1 + 0)}{(x + 1)^{2}})\\=&\frac{-2xln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)^{2}} + \frac{2ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)} - \frac{4x^{2}}{(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})(x + 1)^{3}} + \frac{4x}{(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})(x + 1)^{2}} + \frac{4x}{(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})(x + 1)^{3}} + \frac{ln(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})}{(x + 1)^{2}} - \frac{1}{(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})(x + 1)^{3}} - \frac{2}{(\frac{2x}{(x + 1)} - \frac{1}{(x + 1)})(x + 1)^{2}} + \frac{2x}{(x + 1)^{2}} - \frac{1}{(x + 1)^{2}} - \frac{2}{(x + 1)}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!