本次共计算 1 个题目:每一题对 x 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{{(tan(x)sin(x))}^{sin(x)}ln(x)}{tan(5 - ln(x)sin(x))} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{(sin(x)tan(x))^{sin(x)}ln(x)}{tan(-ln(x)sin(x) + 5)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{(sin(x)tan(x))^{sin(x)}ln(x)}{tan(-ln(x)sin(x) + 5)}\right)}{dx}\\=&\frac{((sin(x)tan(x))^{sin(x)}((cos(x))ln(sin(x)tan(x)) + \frac{(sin(x))(cos(x)tan(x) + sin(x)sec^{2}(x)(1))}{(sin(x)tan(x))}))ln(x)}{tan(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}}{(x)tan(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}ln(x)*-sec^{2}(-ln(x)sin(x) + 5)(\frac{-sin(x)}{(x)} - ln(x)cos(x) + 0)}{tan^{2}(-ln(x)sin(x) + 5)}\\=&\frac{(sin(x)tan(x))^{sin(x)}ln(sin(x)tan(x))ln(x)cos(x)}{tan(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}ln^{2}(x)cos(x)sec^{2}(-ln(x)sin(x) + 5)}{tan^{2}(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}ln(x)sin(x)sec^{2}(x)}{tan(x)tan(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}}{xtan(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}ln(x)sin(x)sec^{2}(-ln(x)sin(x) + 5)}{xtan^{2}(-ln(x)sin(x) + 5)} + \frac{(sin(x)tan(x))^{sin(x)}ln(x)cos(x)}{tan(-ln(x)sin(x) + 5)}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!