本次共计算 2 个题目：每一题对 a 求 4 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/2】求函数\frac{(-b + sqrt({b}^{2} - 4ac))}{(2a)} 关于 a 的 4 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ca)}{a}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} + \frac{\frac{1}{2}sqrt(b^{2} - 4ca)}{a}\right)}{da}\\=&\frac{\frac{-1}{2}b*-1}{a^{2}} + \frac{\frac{1}{2}*-sqrt(b^{2} - 4ca)}{a^{2}} + \frac{\frac{1}{2}(0 - 4c)*\frac{1}{2}}{a(b^{2} - 4ca)^{\frac{1}{2}}}\\=&\frac{b}{2a^{2}} - \frac{sqrt(b^{2} - 4ca)}{2a^{2}} - \frac{c}{(b^{2} - 4ca)^{\frac{1}{2}}a}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{b}{2a^{2}} - \frac{sqrt(b^{2} - 4ca)}{2a^{2}} - \frac{c}{(b^{2} - 4ca)^{\frac{1}{2}}a}\right)}{da}\\=&\frac{b*-2}{2a^{3}} - \frac{-2sqrt(b^{2} - 4ca)}{2a^{3}} - \frac{(0 - 4c)*\frac{1}{2}}{2a^{2}(b^{2} - 4ca)^{\frac{1}{2}}} - \frac{(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a} - \frac{c*-1}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}}\\=&\frac{-b}{a^{3}} + \frac{sqrt(b^{2} - 4ca)}{a^{3}} + \frac{2c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}} - \frac{2c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( \frac{-b}{a^{3}} + \frac{sqrt(b^{2} - 4ca)}{a^{3}} + \frac{2c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}} - \frac{2c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a}\right)}{da}\\=&\frac{-b*-3}{a^{4}} + \frac{-3sqrt(b^{2} - 4ca)}{a^{4}} + \frac{(0 - 4c)*\frac{1}{2}}{a^{3}(b^{2} - 4ca)^{\frac{1}{2}}} + \frac{2(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a^{2}} + \frac{2c*-2}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} - \frac{2(\frac{\frac{-3}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{5}{2}}})c^{2}}{a} - \frac{2c^{2}*-1}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}}\\=&\frac{3b}{a^{4}} - \frac{3sqrt(b^{2} - 4ca)}{a^{4}} - \frac{6c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} + \frac{6c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}} - \frac{12c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a}\\\\ &\color{blue}{函数的第 4 阶导数：} \\&\frac{d\left( \frac{3b}{a^{4}} - \frac{3sqrt(b^{2} - 4ca)}{a^{4}} - \frac{6c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} + \frac{6c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}} - \frac{12c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a}\right)}{da}\\=&\frac{3b*-4}{a^{5}} - \frac{3*-4sqrt(b^{2} - 4ca)}{a^{5}} - \frac{3(0 - 4c)*\frac{1}{2}}{a^{4}(b^{2} - 4ca)^{\frac{1}{2}}} - \frac{6(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a^{3}} - \frac{6c*-3}{(b^{2} - 4ca)^{\frac{1}{2}}a^{4}} + \frac{6(\frac{\frac{-3}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{5}{2}}})c^{2}}{a^{2}} + \frac{6c^{2}*-2}{(b^{2} - 4ca)^{\frac{3}{2}}a^{3}} - \frac{12(\frac{\frac{-5}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{7}{2}}})c^{3}}{a} - \frac{12c^{3}*-1}{(b^{2} - 4ca)^{\frac{5}{2}}a^{2}}\\=&\frac{-12b}{a^{5}} + \frac{12sqrt(b^{2} - 4ca)}{a^{5}} + \frac{24c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{4}} - \frac{24c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{3}} + \frac{48c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a^{2}} - \frac{120c^{4}}{(b^{2} - 4ca)^{\frac{7}{2}}a}\\ \end{split}\end{equation} \]

\[ \begin{equation}\begin{split}【2/2】求函数\frac{(-b - sqrt({b}^{2} - 4ac))}{(2a)} 关于 a 的 4 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ca)}{a}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{\frac{-1}{2}b}{a} - \frac{\frac{1}{2}sqrt(b^{2} - 4ca)}{a}\right)}{da}\\=&\frac{\frac{-1}{2}b*-1}{a^{2}} - \frac{\frac{1}{2}*-sqrt(b^{2} - 4ca)}{a^{2}} - \frac{\frac{1}{2}(0 - 4c)*\frac{1}{2}}{a(b^{2} - 4ca)^{\frac{1}{2}}}\\=&\frac{b}{2a^{2}} + \frac{sqrt(b^{2} - 4ca)}{2a^{2}} + \frac{c}{(b^{2} - 4ca)^{\frac{1}{2}}a}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{b}{2a^{2}} + \frac{sqrt(b^{2} - 4ca)}{2a^{2}} + \frac{c}{(b^{2} - 4ca)^{\frac{1}{2}}a}\right)}{da}\\=&\frac{b*-2}{2a^{3}} + \frac{-2sqrt(b^{2} - 4ca)}{2a^{3}} + \frac{(0 - 4c)*\frac{1}{2}}{2a^{2}(b^{2} - 4ca)^{\frac{1}{2}}} + \frac{(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a} + \frac{c*-1}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}}\\=&\frac{-b}{a^{3}} - \frac{sqrt(b^{2} - 4ca)}{a^{3}} - \frac{2c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}} + \frac{2c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( \frac{-b}{a^{3}} - \frac{sqrt(b^{2} - 4ca)}{a^{3}} - \frac{2c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{2}} + \frac{2c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a}\right)}{da}\\=&\frac{-b*-3}{a^{4}} - \frac{-3sqrt(b^{2} - 4ca)}{a^{4}} - \frac{(0 - 4c)*\frac{1}{2}}{a^{3}(b^{2} - 4ca)^{\frac{1}{2}}} - \frac{2(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a^{2}} - \frac{2c*-2}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} + \frac{2(\frac{\frac{-3}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{5}{2}}})c^{2}}{a} + \frac{2c^{2}*-1}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}}\\=&\frac{3b}{a^{4}} + \frac{3sqrt(b^{2} - 4ca)}{a^{4}} + \frac{6c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} - \frac{6c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}} + \frac{12c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a}\\\\ &\color{blue}{函数的第 4 阶导数：} \\&\frac{d\left( \frac{3b}{a^{4}} + \frac{3sqrt(b^{2} - 4ca)}{a^{4}} + \frac{6c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{3}} - \frac{6c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{2}} + \frac{12c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a}\right)}{da}\\=&\frac{3b*-4}{a^{5}} + \frac{3*-4sqrt(b^{2} - 4ca)}{a^{5}} + \frac{3(0 - 4c)*\frac{1}{2}}{a^{4}(b^{2} - 4ca)^{\frac{1}{2}}} + \frac{6(\frac{\frac{-1}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{3}{2}}})c}{a^{3}} + \frac{6c*-3}{(b^{2} - 4ca)^{\frac{1}{2}}a^{4}} - \frac{6(\frac{\frac{-3}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{5}{2}}})c^{2}}{a^{2}} - \frac{6c^{2}*-2}{(b^{2} - 4ca)^{\frac{3}{2}}a^{3}} + \frac{12(\frac{\frac{-5}{2}(0 - 4c)}{(b^{2} - 4ca)^{\frac{7}{2}}})c^{3}}{a} + \frac{12c^{3}*-1}{(b^{2} - 4ca)^{\frac{5}{2}}a^{2}}\\=&\frac{-12b}{a^{5}} - \frac{12sqrt(b^{2} - 4ca)}{a^{5}} - \frac{24c}{(b^{2} - 4ca)^{\frac{1}{2}}a^{4}} + \frac{24c^{2}}{(b^{2} - 4ca)^{\frac{3}{2}}a^{3}} - \frac{48c^{3}}{(b^{2} - 4ca)^{\frac{5}{2}}a^{2}} + \frac{120c^{4}}{(b^{2} - 4ca)^{\frac{7}{2}}a}\\ \end{split}\end{equation} \]

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