There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
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\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{ln(x)}{(x - 1)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{ln(x)}{(x - 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{ln(x)}{(x - 1)}\right)}{dx}\\=&(\frac{-(1 + 0)}{(x - 1)^{2}})ln(x) + \frac{1}{(x - 1)(x)}\\=&\frac{-ln(x)}{(x - 1)^{2}} + \frac{1}{(x - 1)x}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-ln(x)}{(x - 1)^{2}} + \frac{1}{(x - 1)x}\right)}{dx}\\=&-(\frac{-2(1 + 0)}{(x - 1)^{3}})ln(x) - \frac{1}{(x - 1)^{2}(x)} + \frac{(\frac{-(1 + 0)}{(x - 1)^{2}})}{x} + \frac{-1}{(x - 1)x^{2}}\\=&\frac{2ln(x)}{(x - 1)^{3}} - \frac{2}{(x - 1)^{2}x} - \frac{1}{(x - 1)x^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{2ln(x)}{(x - 1)^{3}} - \frac{2}{(x - 1)^{2}x} - \frac{1}{(x - 1)x^{2}}\right)}{dx}\\=&2(\frac{-3(1 + 0)}{(x - 1)^{4}})ln(x) + \frac{2}{(x - 1)^{3}(x)} - \frac{2(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x} - \frac{2*-1}{(x - 1)^{2}x^{2}} - \frac{(\frac{-(1 + 0)}{(x - 1)^{2}})}{x^{2}} - \frac{-2}{(x - 1)x^{3}}\\=&\frac{-6ln(x)}{(x - 1)^{4}} + \frac{6}{(x - 1)^{3}x} + \frac{3}{(x - 1)^{2}x^{2}} + \frac{2}{(x - 1)x^{3}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-6ln(x)}{(x - 1)^{4}} + \frac{6}{(x - 1)^{3}x} + \frac{3}{(x - 1)^{2}x^{2}} + \frac{2}{(x - 1)x^{3}}\right)}{dx}\\=&-6(\frac{-4(1 + 0)}{(x - 1)^{5}})ln(x) - \frac{6}{(x - 1)^{4}(x)} + \frac{6(\frac{-3(1 + 0)}{(x - 1)^{4}})}{x} + \frac{6*-1}{(x - 1)^{3}x^{2}} + \frac{3(\frac{-2(1 + 0)}{(x - 1)^{3}})}{x^{2}} + \frac{3*-2}{(x - 1)^{2}x^{3}} + \frac{2(\frac{-(1 + 0)}{(x - 1)^{2}})}{x^{3}} + \frac{2*-3}{(x - 1)x^{4}}\\=&\frac{24ln(x)}{(x - 1)^{5}} - \frac{24}{(x - 1)^{4}x} - \frac{12}{(x - 1)^{3}x^{2}} - \frac{8}{(x - 1)^{2}x^{3}} - \frac{6}{(x - 1)x^{4}}\\ \end{split}\end{equation} \]

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