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

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