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