本次共计算 1 个题目：每一题对 x 求 4 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数\frac{(x + ln(x + 1))}{x} + 1 关于 x 的 4 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{ln(x + 1)}{x} + 2\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{ln(x + 1)}{x} + 2\right)}{dx}\\=&\frac{-ln(x + 1)}{x^{2}} + \frac{(1 + 0)}{x(x + 1)} + 0\\=& - \frac{ln(x + 1)}{x^{2}} + \frac{1}{(x + 1)x}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( - \frac{ln(x + 1)}{x^{2}} + \frac{1}{(x + 1)x}\right)}{dx}\\=& - \frac{-2ln(x + 1)}{x^{3}} - \frac{(1 + 0)}{x^{2}(x + 1)} + \frac{(\frac{-(1 + 0)}{(x + 1)^{2}})}{x} + \frac{-1}{(x + 1)x^{2}}\\=&\frac{2ln(x + 1)}{x^{3}} - \frac{2}{(x + 1)x^{2}} - \frac{1}{(x + 1)^{2}x}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( \frac{2ln(x + 1)}{x^{3}} - \frac{2}{(x + 1)x^{2}} - \frac{1}{(x + 1)^{2}x}\right)}{dx}\\=&\frac{2*-3ln(x + 1)}{x^{4}} + \frac{2(1 + 0)}{x^{3}(x + 1)} - \frac{2(\frac{-(1 + 0)}{(x + 1)^{2}})}{x^{2}} - \frac{2*-2}{(x + 1)x^{3}} - \frac{(\frac{-2(1 + 0)}{(x + 1)^{3}})}{x} - \frac{-1}{(x + 1)^{2}x^{2}}\\=& - \frac{6ln(x + 1)}{x^{4}} + \frac{6}{(x + 1)x^{3}} + \frac{3}{(x + 1)^{2}x^{2}} + \frac{2}{(x + 1)^{3}x}\\\\ &\color{blue}{函数的第 4 阶导数：} \\&\frac{d\left( - \frac{6ln(x + 1)}{x^{4}} + \frac{6}{(x + 1)x^{3}} + \frac{3}{(x + 1)^{2}x^{2}} + \frac{2}{(x + 1)^{3}x}\right)}{dx}\\=& - \frac{6*-4ln(x + 1)}{x^{5}} - \frac{6(1 + 0)}{x^{4}(x + 1)} + \frac{6(\frac{-(1 + 0)}{(x + 1)^{2}})}{x^{3}} + \frac{6*-3}{(x + 1)x^{4}} + \frac{3(\frac{-2(1 + 0)}{(x + 1)^{3}})}{x^{2}} + \frac{3*-2}{(x + 1)^{2}x^{3}} + \frac{2(\frac{-3(1 + 0)}{(x + 1)^{4}})}{x} + \frac{2*-1}{(x + 1)^{3}x^{2}}\\=&\frac{24ln(x + 1)}{x^{5}} - \frac{24}{(x + 1)x^{4}} - \frac{12}{(x + 1)^{2}x^{3}} - \frac{8}{(x + 1)^{3}x^{2}} - \frac{6}{(x + 1)^{4}x}\\ \end{split}\end{equation} \]

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