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。\[ \begin{equation}\begin{split}【1/1】求函数ln(sin(x)){\frac{1}{(π - 2x)}}^{2} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{ln(sin(x))}{(π - 2x)^{2}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{ln(sin(x))}{(π - 2x)^{2}}\right)}{dx}\\=&(\frac{-2(0 - 2)}{(π - 2x)^{3}})ln(sin(x)) + \frac{cos(x)}{(π - 2x)^{2}(sin(x))}\\=&\frac{4ln(sin(x))}{(π - 2x)^{3}} + \frac{cos(x)}{(π - 2x)^{2}sin(x)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{4ln(sin(x))}{(π - 2x)^{3}} + \frac{cos(x)}{(π - 2x)^{2}sin(x)}\right)}{dx}\\=&4(\frac{-3(0 - 2)}{(π - 2x)^{4}})ln(sin(x)) + \frac{4cos(x)}{(π - 2x)^{3}(sin(x))} + \frac{(\frac{-2(0 - 2)}{(π - 2x)^{3}})cos(x)}{sin(x)} + \frac{-cos(x)cos(x)}{(π - 2x)^{2}sin^{2}(x)} + \frac{-sin(x)}{(π - 2x)^{2}sin(x)}\\=&\frac{24ln(sin(x))}{(π - 2x)^{4}} + \frac{8cos(x)}{(π - 2x)^{3}sin(x)} - \frac{cos^{2}(x)}{(π - 2x)^{2}sin^{2}(x)} - \frac{1}{(π - 2x)^{2}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!