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。\[ \begin{equation}\begin{split}【1/1】求函数{(1 + 2x)}^{(\frac{3}{sin(x)})} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = (2x + 1)^{(\frac{3}{sin(x)})}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( (2x + 1)^{(\frac{3}{sin(x)})}\right)}{dx}\\=&((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))\\=&\frac{-3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)sin(x)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)sin(x)}\right)}{dx}\\=&\frac{-3((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))ln(2x + 1)cos(x)}{sin^{2}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}(2 + 0)cos(x)}{(2x + 1)sin^{2}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)*-2cos(x)cos(x)}{sin^{3}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)*-sin(x)}{sin^{2}(x)} + \frac{6(\frac{-(2 + 0)}{(2x + 1)^{2}})(2x + 1)^{(\frac{3}{sin(x)})}}{sin(x)} + \frac{6((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))}{(2x + 1)sin(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}*-cos(x)}{(2x + 1)sin^{2}(x)}\\=&\frac{9(2x + 1)^{(\frac{3}{sin(x)})}ln^{2}(2x + 1)cos^{2}(x)}{sin^{4}(x)} - \frac{36(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{(2x + 1)sin^{3}(x)} - \frac{12(2x + 1)^{(\frac{3}{sin(x)})}cos(x)}{(2x + 1)sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos^{2}(x)}{sin^{3}(x)} + \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)}{sin(x)} + \frac{36(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)^{2}sin^{2}(x)} - \frac{12(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)^{2}sin(x)}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!