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