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。\[ \begin{equation}\begin{split}【1/1】求函数{ln(\frac{(1 - 2x)}{(1 + {x}^{2})})}^{4} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = ln^{4}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( ln^{4}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})\right)}{dx}\\=&\frac{4ln^{3}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})(-2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x - \frac{2}{(x^{2} + 1)} + (\frac{-(2x + 0)}{(x^{2} + 1)^{2}}))}{(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})}\\=&\frac{16x^{2}ln^{3}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})}{(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})(x^{2} + 1)^{2}} - \frac{8ln^{3}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})}{(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})(x^{2} + 1)} - \frac{8xln^{3}(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})}{(\frac{-2x}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)})(x^{2} + 1)^{2}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!