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