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。\[ \begin{equation}\begin{split}【1/1】求函数(x{e}^{x}){\frac{1}{(x + ln(x))}}^{2} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{x{e}^{x}}{(x + ln(x))^{2}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{x{e}^{x}}{(x + ln(x))^{2}}\right)}{dx}\\=&(\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}})x{e}^{x} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}}\\=& - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x{e}^{x}}{(x + ln(x))^{3}} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x{e}^{x}}{(x + ln(x))^{3}} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\right)}{dx}\\=& - 2(\frac{-3(1 + \frac{1}{(x)})}{(x + ln(x))^{4}}){e}^{x} - \frac{2({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{3}} - 2(\frac{-3(1 + \frac{1}{(x)})}{(x + ln(x))^{4}})x{e}^{x} - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{3}} + (\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}}){e}^{x} + \frac{({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}} + (\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}})x{e}^{x} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}}\\=&\frac{6{e}^{x}}{(x + ln(x))^{4}x} + \frac{12{e}^{x}}{(x + ln(x))^{4}} - \frac{8{e}^{x}}{(x + ln(x))^{3}} + \frac{6x{e}^{x}}{(x + ln(x))^{4}} - \frac{4x{e}^{x}}{(x + ln(x))^{3}} - \frac{2{e}^{x}}{(x + ln(x))^{3}x} + \frac{2{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!