本次共计算 1 个题目：每一题对 x 求 3 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数sin(x){(arctan(x))}^{2} 关于 x 的 3 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = sin(x)arctan^{2}(x)\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( sin(x)arctan^{2}(x)\right)}{dx}\\=&cos(x)arctan^{2}(x) + sin(x)(\frac{2arctan(x)(1)}{(1 + (x)^{2})})\\=&cos(x)arctan^{2}(x) + \frac{2sin(x)arctan(x)}{(x^{2} + 1)}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( cos(x)arctan^{2}(x) + \frac{2sin(x)arctan(x)}{(x^{2} + 1)}\right)}{dx}\\=&-sin(x)arctan^{2}(x) + cos(x)(\frac{2arctan(x)(1)}{(1 + (x)^{2})}) + 2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})sin(x)arctan(x) + \frac{2cos(x)arctan(x)}{(x^{2} + 1)} + \frac{2sin(x)(\frac{(1)}{(1 + (x)^{2})})}{(x^{2} + 1)}\\=&-sin(x)arctan^{2}(x) + \frac{4cos(x)arctan(x)}{(x^{2} + 1)} - \frac{4xsin(x)arctan(x)}{(x^{2} + 1)^{2}} + \frac{2sin(x)}{(x^{2} + 1)^{2}}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( -sin(x)arctan^{2}(x) + \frac{4cos(x)arctan(x)}{(x^{2} + 1)} - \frac{4xsin(x)arctan(x)}{(x^{2} + 1)^{2}} + \frac{2sin(x)}{(x^{2} + 1)^{2}}\right)}{dx}\\=&-cos(x)arctan^{2}(x) - sin(x)(\frac{2arctan(x)(1)}{(1 + (x)^{2})}) + 4(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})cos(x)arctan(x) + \frac{4*-sin(x)arctan(x)}{(x^{2} + 1)} + \frac{4cos(x)(\frac{(1)}{(1 + (x)^{2})})}{(x^{2} + 1)} - 4(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})xsin(x)arctan(x) - \frac{4sin(x)arctan(x)}{(x^{2} + 1)^{2}} - \frac{4xcos(x)arctan(x)}{(x^{2} + 1)^{2}} - \frac{4xsin(x)(\frac{(1)}{(1 + (x)^{2})})}{(x^{2} + 1)^{2}} + 2(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})sin(x) + \frac{2cos(x)}{(x^{2} + 1)^{2}}\\=&-cos(x)arctan^{2}(x) - \frac{6sin(x)arctan(x)}{(x^{2} + 1)} - \frac{12xcos(x)arctan(x)}{(x^{2} + 1)^{2}} + \frac{6cos(x)}{(x^{2} + 1)^{2}} + \frac{16x^{2}sin(x)arctan(x)}{(x^{2} + 1)^{3}} - \frac{4sin(x)arctan(x)}{(x^{2} + 1)^{2}} - \frac{12xsin(x)}{(x^{2} + 1)^{3}}\\ \end{split}\end{equation} \]

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