本次共计算 1 个题目:每一题对 x 求 2 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{(x + arccot(x))}{ln(1 + {x}^{2})} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{x}{ln(x^{2} + 1)} + \frac{arccot(x)}{ln(x^{2} + 1)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{x}{ln(x^{2} + 1)} + \frac{arccot(x)}{ln(x^{2} + 1)}\right)}{dx}\\=&\frac{1}{ln(x^{2} + 1)} + \frac{x*-(2x + 0)}{ln^{2}(x^{2} + 1)(x^{2} + 1)} + \frac{-(2x + 0)arccot(x)}{ln^{2}(x^{2} + 1)(x^{2} + 1)} + \frac{(\frac{(1)}{(1 + (x)^{2})})}{ln(x^{2} + 1)}\\=&\frac{1}{ln(x^{2} + 1)} - \frac{2xarccot(x)}{(x^{2} + 1)ln^{2}(x^{2} + 1)} - \frac{2x^{2}}{(x^{2} + 1)ln^{2}(x^{2} + 1)} + \frac{1}{(x^{2} + 1)ln(x^{2} + 1)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{1}{ln(x^{2} + 1)} - \frac{2xarccot(x)}{(x^{2} + 1)ln^{2}(x^{2} + 1)} - \frac{2x^{2}}{(x^{2} + 1)ln^{2}(x^{2} + 1)} + \frac{1}{(x^{2} + 1)ln(x^{2} + 1)}\right)}{dx}\\=&\frac{-(2x + 0)}{ln^{2}(x^{2} + 1)(x^{2} + 1)} - \frac{2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})xarccot(x)}{ln^{2}(x^{2} + 1)} - \frac{2arccot(x)}{(x^{2} + 1)ln^{2}(x^{2} + 1)} - \frac{2x*-2(2x + 0)arccot(x)}{(x^{2} + 1)ln^{3}(x^{2} + 1)(x^{2} + 1)} - \frac{2x(\frac{(1)}{(1 + (x)^{2})})}{(x^{2} + 1)ln^{2}(x^{2} + 1)} - \frac{2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x^{2}}{ln^{2}(x^{2} + 1)} - \frac{2*2x}{(x^{2} + 1)ln^{2}(x^{2} + 1)} - \frac{2x^{2}*-2(2x + 0)}{(x^{2} + 1)ln^{3}(x^{2} + 1)(x^{2} + 1)} + \frac{(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})}{ln(x^{2} + 1)} + \frac{-(2x + 0)}{(x^{2} + 1)ln^{2}(x^{2} + 1)(x^{2} + 1)}\\=& - \frac{2x}{(x^{2} + 1)^{2}ln(x^{2} + 1)} + \frac{4x^{2}arccot(x)}{(x^{2} + 1)^{2}ln^{2}(x^{2} + 1)} - \frac{2arccot(x)}{(x^{2} + 1)ln^{2}(x^{2} + 1)} + \frac{8x^{2}arccot(x)}{(x^{2} + 1)^{2}ln^{3}(x^{2} + 1)} - \frac{4x}{(x^{2} + 1)^{2}ln^{2}(x^{2} + 1)} + \frac{4x^{3}}{(x^{2} + 1)^{2}ln^{2}(x^{2} + 1)} - \frac{6x}{(x^{2} + 1)ln^{2}(x^{2} + 1)} + \frac{8x^{3}}{(x^{2} + 1)^{2}ln^{3}(x^{2} + 1)}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!