本次共计算 1 个题目:每一题对 x 求 2 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数ln(\frac{(1 - 2x)}{(1 + 3x)}) 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = ln(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( ln(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})\right)}{dx}\\=&\frac{(-2(\frac{-(3 + 0)}{(3x + 1)^{2}})x - \frac{2}{(3x + 1)} + (\frac{-(3 + 0)}{(3x + 1)^{2}}))}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})}\\=&\frac{6x}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{3}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{2}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})(3x + 1)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{6x}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{3}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{2}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})(3x + 1)}\right)}{dx}\\=&\frac{6(\frac{-2(3 + 0)}{(3x + 1)^{3}})x}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} + \frac{6(\frac{-(-2(\frac{-(3 + 0)}{(3x + 1)^{2}})x - \frac{2}{(3x + 1)} + (\frac{-(3 + 0)}{(3x + 1)^{2}}))}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}})x}{(3x + 1)^{2}} + \frac{6}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{3(\frac{-2(3 + 0)}{(3x + 1)^{3}})}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{3(\frac{-(-2(\frac{-(3 + 0)}{(3x + 1)^{2}})x - \frac{2}{(3x + 1)} + (\frac{-(3 + 0)}{(3x + 1)^{2}}))}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}})}{(3x + 1)^{2}} - \frac{2(\frac{-(-2(\frac{-(3 + 0)}{(3x + 1)^{2}})x - \frac{2}{(3x + 1)} + (\frac{-(3 + 0)}{(3x + 1)^{2}}))}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}})}{(3x + 1)} - \frac{2(\frac{-(3 + 0)}{(3x + 1)^{2}})}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})}\\=&\frac{-36x}{(3x + 1)^{3}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{36x^{2}}{(3x + 1)^{4}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}} + \frac{12x}{(3x + 1)^{3}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}} + \frac{36x}{(3x + 1)^{4}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}} + \frac{12x}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}(3x + 1)^{3}} + \frac{12}{(3x + 1)^{2}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{9}{(3x + 1)^{4}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}} + \frac{18}{(3x + 1)^{3}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})} - \frac{6}{(3x + 1)^{3}(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}} - \frac{6}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}(3x + 1)^{3}} - \frac{4}{(\frac{-2x}{(3x + 1)} + \frac{1}{(3x + 1)})^{2}(3x + 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!