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。\[ \begin{equation}\begin{split}【1/1】求函数lg(\frac{cth(x)}{th(x)}) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = lg(\frac{cth(x)}{th(x)})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( lg(\frac{cth(x)}{th(x)})\right)}{dx}\\=&\frac{(\frac{-(1 - th^{2}(x))cth(x)}{th^{2}(x)} + \frac{(1 - cth^{2}(x))}{th(x)})}{ln{10}(\frac{cth(x)}{th(x)})}\\=&\frac{-1}{ln{10}th(x)} + \frac{th(x)}{ln{10}} + \frac{1}{ln{10}cth(x)} - \frac{cth(x)}{ln{10}}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-1}{ln{10}th(x)} + \frac{th(x)}{ln{10}} + \frac{1}{ln{10}cth(x)} - \frac{cth(x)}{ln{10}}\right)}{dx}\\=&\frac{--0}{ln^{2}{10}th(x)} - \frac{-(1 - th^{2}(x))}{ln{10}th^{2}(x)} + \frac{-0th(x)}{ln^{2}{10}} + \frac{(1 - th^{2}(x))}{ln{10}} + \frac{-0}{ln^{2}{10}cth(x)} + \frac{-(1 - cth^{2}(x))}{ln{10}cth^{2}(x)} - \frac{-0cth(x)}{ln^{2}{10}} - \frac{(1 - cth^{2}(x))}{ln{10}}\\=&\frac{1}{ln{10}th^{2}(x)} - \frac{1}{ln{10}cth^{2}(x)} - \frac{th^{2}(x)}{ln{10}} + \frac{cth^{2}(x)}{ln{10}}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{1}{ln{10}th^{2}(x)} - \frac{1}{ln{10}cth^{2}(x)} - \frac{th^{2}(x)}{ln{10}} + \frac{cth^{2}(x)}{ln{10}}\right)}{dx}\\=&\frac{-0}{ln^{2}{10}th^{2}(x)} + \frac{-2(1 - th^{2}(x))}{ln{10}th^{3}(x)} - \frac{-0}{ln^{2}{10}cth^{2}(x)} - \frac{-2(1 - cth^{2}(x))}{ln{10}cth^{3}(x)} - \frac{-0th^{2}(x)}{ln^{2}{10}} - \frac{2th(x)(1 - th^{2}(x))}{ln{10}} + \frac{-0cth^{2}(x)}{ln^{2}{10}} + \frac{2cth(x)(1 - cth^{2}(x))}{ln{10}}\\=&\frac{-2}{ln{10}th^{3}(x)} + \frac{2}{ln{10}th(x)} - \frac{2}{ln{10}cth(x)} + \frac{2}{ln{10}cth^{3}(x)} - \frac{2cth^{3}(x)}{ln{10}} + \frac{2th^{3}(x)}{ln{10}} - \frac{2th(x)}{ln{10}} + \frac{2cth(x)}{ln{10}}\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{-2}{ln{10}th^{3}(x)} + \frac{2}{ln{10}th(x)} - \frac{2}{ln{10}cth(x)} + \frac{2}{ln{10}cth^{3}(x)} - \frac{2cth^{3}(x)}{ln{10}} + \frac{2th^{3}(x)}{ln{10}} - \frac{2th(x)}{ln{10}} + \frac{2cth(x)}{ln{10}}\right)}{dx}\\=&\frac{-2*-0}{ln^{2}{10}th^{3}(x)} - \frac{2*-3(1 - th^{2}(x))}{ln{10}th^{4}(x)} + \frac{2*-0}{ln^{2}{10}th(x)} + \frac{2*-(1 - th^{2}(x))}{ln{10}th^{2}(x)} - \frac{2*-0}{ln^{2}{10}cth(x)} - \frac{2*-(1 - cth^{2}(x))}{ln{10}cth^{2}(x)} + \frac{2*-0}{ln^{2}{10}cth^{3}(x)} + \frac{2*-3(1 - cth^{2}(x))}{ln{10}cth^{4}(x)} - \frac{2*-0cth^{3}(x)}{ln^{2}{10}} - \frac{2*3cth^{2}(x)(1 - cth^{2}(x))}{ln{10}} + \frac{2*-0th^{3}(x)}{ln^{2}{10}} + \frac{2*3th^{2}(x)(1 - th^{2}(x))}{ln{10}} - \frac{2*-0th(x)}{ln^{2}{10}} - \frac{2(1 - th^{2}(x))}{ln{10}} + \frac{2*-0cth(x)}{ln^{2}{10}} + \frac{2(1 - cth^{2}(x))}{ln{10}}\\=&\frac{6}{ln{10}th^{4}(x)} - \frac{8}{ln{10}th^{2}(x)} - \frac{6th^{4}(x)}{ln{10}} + \frac{8}{ln{10}cth^{2}(x)} + \frac{6cth^{4}(x)}{ln{10}} - \frac{6}{ln{10}cth^{4}(x)} - \frac{8cth^{2}(x)}{ln{10}} + \frac{8th^{2}(x)}{ln{10}}\\ \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!