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。\[ \begin{equation}\begin{split}【1/1】求函数ln(1 + {tan(x)}^{2}) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = ln(tan^{2}(x) + 1)\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( ln(tan^{2}(x) + 1)\right)}{dx}\\=&\frac{(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)}\\=&\frac{2tan(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{2tan(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\right)}{dx}\\=&2(\frac{-(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{2}})tan(x)sec^{2}(x) + \frac{2sec^{2}(x)(1)sec^{2}(x)}{(tan^{2}(x) + 1)} + \frac{2tan(x)*2sec^{2}(x)tan(x)}{(tan^{2}(x) + 1)}\\=&\frac{-4tan^{2}(x)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} + \frac{2sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{4tan^{2}(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{-4tan^{2}(x)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} + \frac{2sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{4tan^{2}(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\right)}{dx}\\=&-4(\frac{-2(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{3}})tan^{2}(x)sec^{4}(x) - \frac{4*2tan(x)sec^{2}(x)(1)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{4tan^{2}(x)*4sec^{4}(x)tan(x)}{(tan^{2}(x) + 1)^{2}} + 2(\frac{-(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{2}})sec^{4}(x) + \frac{2*4sec^{4}(x)tan(x)}{(tan^{2}(x) + 1)} + 4(\frac{-(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{2}})tan^{2}(x)sec^{2}(x) + \frac{4*2tan(x)sec^{2}(x)(1)sec^{2}(x)}{(tan^{2}(x) + 1)} + \frac{4tan^{2}(x)*2sec^{2}(x)tan(x)}{(tan^{2}(x) + 1)}\\=&\frac{16tan^{3}(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{3}} - \frac{12tan(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{24tan^{3}(x)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} + \frac{16tan(x)sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{8tan^{3}(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{16tan^{3}(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{3}} - \frac{12tan(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{24tan^{3}(x)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} + \frac{16tan(x)sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{8tan^{3}(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\right)}{dx}\\=&16(\frac{-3(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{4}})tan^{3}(x)sec^{6}(x) + \frac{16*3tan^{2}(x)sec^{2}(x)(1)sec^{6}(x)}{(tan^{2}(x) + 1)^{3}} + \frac{16tan^{3}(x)*6sec^{6}(x)tan(x)}{(tan^{2}(x) + 1)^{3}} - 12(\frac{-2(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{3}})tan(x)sec^{6}(x) - \frac{12sec^{2}(x)(1)sec^{6}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{12tan(x)*6sec^{6}(x)tan(x)}{(tan^{2}(x) + 1)^{2}} - 24(\frac{-2(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{3}})tan^{3}(x)sec^{4}(x) - \frac{24*3tan^{2}(x)sec^{2}(x)(1)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{24tan^{3}(x)*4sec^{4}(x)tan(x)}{(tan^{2}(x) + 1)^{2}} + 16(\frac{-(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{2}})tan(x)sec^{4}(x) + \frac{16sec^{2}(x)(1)sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{16tan(x)*4sec^{4}(x)tan(x)}{(tan^{2}(x) + 1)} + 8(\frac{-(2tan(x)sec^{2}(x)(1) + 0)}{(tan^{2}(x) + 1)^{2}})tan^{3}(x)sec^{2}(x) + \frac{8*3tan^{2}(x)sec^{2}(x)(1)sec^{2}(x)}{(tan^{2}(x) + 1)} + \frac{8tan^{3}(x)*2sec^{2}(x)tan(x)}{(tan^{2}(x) + 1)}\\=&\frac{-96tan^{4}(x)sec^{8}(x)}{(tan^{2}(x) + 1)^{4}} + \frac{96tan^{2}(x)sec^{8}(x)}{(tan^{2}(x) + 1)^{3}} + \frac{192tan^{4}(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{3}} - \frac{12sec^{8}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{176tan^{2}(x)sec^{6}(x)}{(tan^{2}(x) + 1)^{2}} - \frac{112tan^{4}(x)sec^{4}(x)}{(tan^{2}(x) + 1)^{2}} + \frac{16sec^{6}(x)}{(tan^{2}(x) + 1)} + \frac{88tan^{2}(x)sec^{4}(x)}{(tan^{2}(x) + 1)} + \frac{16tan^{4}(x)sec^{2}(x)}{(tan^{2}(x) + 1)}\\ \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!