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

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