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。\[ \begin{equation}\begin{split}【1/1】求函数\frac{1}{({x}^{2}{(x - 1)}^{2})} 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{1}{(x - 1)^{2}x^{2}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{1}{(x - 1)^{2}x^{2}}\right)}{dx}\\=&\frac{(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x^{2}} + \frac{-2}{(x - 1)^{2}x^{3}}\\=&\frac{-2}{(x - 1)^{3}x^{2}} - \frac{2}{(x - 1)^{2}x^{3}}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-2}{(x - 1)^{3}x^{2}} - \frac{2}{(x - 1)^{2}x^{3}}\right)}{dx}\\=&\frac{-2(\frac{-3(1 + 0)}{(x - 1)^{4}})}{x^{2}} - \frac{2*-2}{(x - 1)^{3}x^{3}} - \frac{2(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x^{3}} - \frac{2*-3}{(x - 1)^{2}x^{4}}\\=&\frac{6}{(x - 1)^{4}x^{2}} + \frac{8}{(x - 1)^{3}x^{3}} + \frac{6}{(x - 1)^{2}x^{4}}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{6}{(x - 1)^{4}x^{2}} + \frac{8}{(x - 1)^{3}x^{3}} + \frac{6}{(x - 1)^{2}x^{4}}\right)}{dx}\\=&\frac{6(\frac{-4(1 + 0)}{(x - 1)^{5}})}{x^{2}} + \frac{6*-2}{(x - 1)^{4}x^{3}} + \frac{8(\frac{-3(1 + 0)}{(x - 1)^{4}})}{x^{3}} + \frac{8*-3}{(x - 1)^{3}x^{4}} + \frac{6(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x^{4}} + \frac{6*-4}{(x - 1)^{2}x^{5}}\\=&\frac{-24}{(x - 1)^{5}x^{2}} - \frac{36}{(x - 1)^{4}x^{3}} - \frac{36}{(x - 1)^{3}x^{4}} - \frac{24}{(x - 1)^{2}x^{5}}\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{-24}{(x - 1)^{5}x^{2}} - \frac{36}{(x - 1)^{4}x^{3}} - \frac{36}{(x - 1)^{3}x^{4}} - \frac{24}{(x - 1)^{2}x^{5}}\right)}{dx}\\=&\frac{-24(\frac{-5(1 + 0)}{(x - 1)^{6}})}{x^{2}} - \frac{24*-2}{(x - 1)^{5}x^{3}} - \frac{36(\frac{-4(1 + 0)}{(x - 1)^{5}})}{x^{3}} - \frac{36*-3}{(x - 1)^{4}x^{4}} - \frac{36(\frac{-3(1 + 0)}{(x - 1)^{4}})}{x^{4}} - \frac{36*-4}{(x - 1)^{3}x^{5}} - \frac{24(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x^{5}} - \frac{24*-5}{(x - 1)^{2}x^{6}}\\=&\frac{120}{(x - 1)^{6}x^{2}} + \frac{192}{(x - 1)^{5}x^{3}} + \frac{216}{(x - 1)^{4}x^{4}} + \frac{192}{(x - 1)^{3}x^{5}} + \frac{120}{(x - 1)^{2}x^{6}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!