本次共计算 1 个题目:每一题对 x 求 4 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数xth(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( xth(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}})))))\right)}{dx}\\=&th(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))) + \frac{x(1 - th^{2}(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))))*-csc^{2}(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))(\frac{-(sec^{2}(e^{e^{cos(sqrt(t))}})(\frac{e^{e^{cos(sqrt(t))}}e^{cos(sqrt(t))}*-sin(sqrt(t))*0*\frac{1}{2}}{(t)^{\frac{1}{2}}}))}{((1 - (tan(e^{e^{cos(sqrt(t))}}))^{2})^{\frac{1}{2}})})}{ln{10}(arccos(tan(e^{e^{cos(sqrt(t))}})))}\\=&th(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}})))))\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( th(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}})))))\right)}{dx}\\=&\frac{(1 - th^{2}(cot(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))))*-csc^{2}(lg(arccos(tan(e^{e^{cos(sqrt(t))}}))))(\frac{-(sec^{2}(e^{e^{cos(sqrt(t))}})(\frac{e^{e^{cos(sqrt(t))}}e^{cos(sqrt(t))}*-sin(sqrt(t))*0*\frac{1}{2}}{(t)^{\frac{1}{2}}}))}{((1 - (tan(e^{e^{cos(sqrt(t))}}))^{2})^{\frac{1}{2}})})}{ln{10}(arccos(tan(e^{e^{cos(sqrt(t))}})))}\\=& - 0\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( - 0\right)}{dx}\\=& - 0\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( - 0\right)}{dx}\\=& - 0\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!