There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ ((\frac{ln(1 - 2x)}{(1 + {x}^{2})}))\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{ln(-2x + 1)}{(x^{2} + 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{ln(-2x + 1)}{(x^{2} + 1)}\right)}{dx}\\=&(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})ln(-2x + 1) + \frac{(-2 + 0)}{(x^{2} + 1)(-2x + 1)}\\=&\frac{-2xln(-2x + 1)}{(x^{2} + 1)^{2}} - \frac{2}{(-2x + 1)(x^{2} + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2xln(-2x + 1)}{(x^{2} + 1)^{2}} - \frac{2}{(-2x + 1)(x^{2} + 1)}\right)}{dx}\\=&-2(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})xln(-2x + 1) - \frac{2ln(-2x + 1)}{(x^{2} + 1)^{2}} - \frac{2x(-2 + 0)}{(x^{2} + 1)^{2}(-2x + 1)} - \frac{2(\frac{-(-2 + 0)}{(-2x + 1)^{2}})}{(x^{2} + 1)} - \frac{2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})}{(-2x + 1)}\\=&\frac{8x^{2}ln(-2x + 1)}{(x^{2} + 1)^{3}} - \frac{2ln(-2x + 1)}{(x^{2} + 1)^{2}} + \frac{8x}{(-2x + 1)(x^{2} + 1)^{2}} - \frac{4}{(-2x + 1)^{2}(x^{2} + 1)}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{8x^{2}ln(-2x + 1)}{(x^{2} + 1)^{3}} - \frac{2ln(-2x + 1)}{(x^{2} + 1)^{2}} + \frac{8x}{(-2x + 1)(x^{2} + 1)^{2}} - \frac{4}{(-2x + 1)^{2}(x^{2} + 1)}\right)}{dx}\\=&8(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})x^{2}ln(-2x + 1) + \frac{8*2xln(-2x + 1)}{(x^{2} + 1)^{3}} + \frac{8x^{2}(-2 + 0)}{(x^{2} + 1)^{3}(-2x + 1)} - 2(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})ln(-2x + 1) - \frac{2(-2 + 0)}{(x^{2} + 1)^{2}(-2x + 1)} + \frac{8(\frac{-(-2 + 0)}{(-2x + 1)^{2}})x}{(x^{2} + 1)^{2}} + \frac{8(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x}{(-2x + 1)} + \frac{8}{(-2x + 1)(x^{2} + 1)^{2}} - \frac{4(\frac{-2(-2 + 0)}{(-2x + 1)^{3}})}{(x^{2} + 1)} - \frac{4(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})}{(-2x + 1)^{2}}\\=&\frac{-48x^{3}ln(-2x + 1)}{(x^{2} + 1)^{4}} + \frac{24xln(-2x + 1)}{(x^{2} + 1)^{3}} - \frac{48x^{2}}{(-2x + 1)(x^{2} + 1)^{3}} + \frac{24x}{(-2x + 1)^{2}(x^{2} + 1)^{2}} - \frac{16}{(-2x + 1)^{3}(x^{2} + 1)} + \frac{12}{(-2x + 1)(x^{2} + 1)^{2}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-48x^{3}ln(-2x + 1)}{(x^{2} + 1)^{4}} + \frac{24xln(-2x + 1)}{(x^{2} + 1)^{3}} - \frac{48x^{2}}{(-2x + 1)(x^{2} + 1)^{3}} + \frac{24x}{(-2x + 1)^{2}(x^{2} + 1)^{2}} - \frac{16}{(-2x + 1)^{3}(x^{2} + 1)} + \frac{12}{(-2x + 1)(x^{2} + 1)^{2}}\right)}{dx}\\=&-48(\frac{-4(2x + 0)}{(x^{2} + 1)^{5}})x^{3}ln(-2x + 1) - \frac{48*3x^{2}ln(-2x + 1)}{(x^{2} + 1)^{4}} - \frac{48x^{3}(-2 + 0)}{(x^{2} + 1)^{4}(-2x + 1)} + 24(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})xln(-2x + 1) + \frac{24ln(-2x + 1)}{(x^{2} + 1)^{3}} + \frac{24x(-2 + 0)}{(x^{2} + 1)^{3}(-2x + 1)} - \frac{48(\frac{-(-2 + 0)}{(-2x + 1)^{2}})x^{2}}{(x^{2} + 1)^{3}} - \frac{48(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})x^{2}}{(-2x + 1)} - \frac{48*2x}{(-2x + 1)(x^{2} + 1)^{3}} + \frac{24(\frac{-2(-2 + 0)}{(-2x + 1)^{3}})x}{(x^{2} + 1)^{2}} + \frac{24(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x}{(-2x + 1)^{2}} + \frac{24}{(-2x + 1)^{2}(x^{2} + 1)^{2}} - \frac{16(\frac{-3(-2 + 0)}{(-2x + 1)^{4}})}{(x^{2} + 1)} - \frac{16(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})}{(-2x + 1)^{3}} + \frac{12(\frac{-(-2 + 0)}{(-2x + 1)^{2}})}{(x^{2} + 1)^{2}} + \frac{12(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})}{(-2x + 1)}\\=&\frac{384x^{4}ln(-2x + 1)}{(x^{2} + 1)^{5}} - \frac{288x^{2}ln(-2x + 1)}{(x^{2} + 1)^{4}} + \frac{384x^{3}}{(-2x + 1)(x^{2} + 1)^{4}} + \frac{24ln(-2x + 1)}{(x^{2} + 1)^{3}} - \frac{96x}{(-2x + 1)(x^{2} + 1)^{3}} - \frac{192x^{2}}{(-2x + 1)^{2}(x^{2} + 1)^{3}} - \frac{96x}{(x^{2} + 1)^{3}(-2x + 1)} + \frac{128x}{(-2x + 1)^{3}(x^{2} + 1)^{2}} - \frac{96}{(-2x + 1)^{4}(x^{2} + 1)} + \frac{48}{(-2x + 1)^{2}(x^{2} + 1)^{2}}\\ \end{split}\end{equation} \]

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