本次共计算 1 个题目：每一题对 d 求 4 阶导数。
    注意，变量是区分大小写的。
$$\begin{equation}\begin{split}【1/1】求函数\frac{(dln({d}^{2} + 1))}{(dd)} 关于 d 的 4 阶导数：\\\end{split}\end{equation}$$ $$\begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{ln(d^{2} + 1)}{d}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{ln(d^{2} + 1)}{d}\right)}{dd}\\=&\frac{-ln(d^{2} + 1)}{d^{2}} + \frac{(2d + 0)}{d(d^{2} + 1)}\\=&\frac{-ln(d^{2} + 1)}{d^{2}} + \frac{2}{(d^{2} + 1)}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{-ln(d^{2} + 1)}{d^{2}} + \frac{2}{(d^{2} + 1)}\right)}{dd}\\=&\frac{--2ln(d^{2} + 1)}{d^{3}} - \frac{(2d + 0)}{d^{2}(d^{2} + 1)} + 2(\frac{-(2d + 0)}{(d^{2} + 1)^{2}})\\=&\frac{2ln(d^{2} + 1)}{d^{3}} - \frac{2}{(d^{2} + 1)d} - \frac{4d}{(d^{2} + 1)^{2}}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( \frac{2ln(d^{2} + 1)}{d^{3}} - \frac{2}{(d^{2} + 1)d} - \frac{4d}{(d^{2} + 1)^{2}}\right)}{dd}\\=&\frac{2*-3ln(d^{2} + 1)}{d^{4}} + \frac{2(2d + 0)}{d^{3}(d^{2} + 1)} - \frac{2(\frac{-(2d + 0)}{(d^{2} + 1)^{2}})}{d} - \frac{2*-1}{(d^{2} + 1)d^{2}} - 4(\frac{-2(2d + 0)}{(d^{2} + 1)^{3}})d - \frac{4}{(d^{2} + 1)^{2}}\\=&\frac{-6ln(d^{2} + 1)}{d^{4}} + \frac{6}{(d^{2} + 1)d^{2}} + \frac{16d^{2}}{(d^{2} + 1)^{3}}\\\\ &\color{blue}{函数的第 4 阶导数：} \\&\frac{d\left( \frac{-6ln(d^{2} + 1)}{d^{4}} + \frac{6}{(d^{2} + 1)d^{2}} + \frac{16d^{2}}{(d^{2} + 1)^{3}}\right)}{dd}\\=&\frac{-6*-4ln(d^{2} + 1)}{d^{5}} - \frac{6(2d + 0)}{d^{4}(d^{2} + 1)} + \frac{6(\frac{-(2d + 0)}{(d^{2} + 1)^{2}})}{d^{2}} + \frac{6*-2}{(d^{2} + 1)d^{3}} + 16(\frac{-3(2d + 0)}{(d^{2} + 1)^{4}})d^{2} + \frac{16*2d}{(d^{2} + 1)^{3}}\\=&\frac{24ln(d^{2} + 1)}{d^{5}} - \frac{24}{(d^{2} + 1)d^{3}} - \frac{12}{(d^{2} + 1)^{2}d} - \frac{96d^{3}}{(d^{2} + 1)^{4}} + \frac{32d}{(d^{2} + 1)^{3}}\\ \end{split}\end{equation}$$

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