There are 1 questions in this calculation: for each question, the 3 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ third\ derivative\ of\ function\ ln(1 + x + xx)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln(x^{2} + x + 1)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln(x^{2} + x + 1)\right)}{dx}\\=&\frac{(2x + 1 + 0)}{(x^{2} + x + 1)}\\=&\frac{2x}{(x^{2} + x + 1)} + \frac{1}{(x^{2} + x + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2x}{(x^{2} + x + 1)} + \frac{1}{(x^{2} + x + 1)}\right)}{dx}\\=&2(\frac{-(2x + 1 + 0)}{(x^{2} + x + 1)^{2}})x + \frac{2}{(x^{2} + x + 1)} + (\frac{-(2x + 1 + 0)}{(x^{2} + x + 1)^{2}})\\=&\frac{-4x^{2}}{(x^{2} + x + 1)^{2}} - \frac{4x}{(x^{2} + x + 1)^{2}} + \frac{2}{(x^{2} + x + 1)} - \frac{1}{(x^{2} + x + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-4x^{2}}{(x^{2} + x + 1)^{2}} - \frac{4x}{(x^{2} + x + 1)^{2}} + \frac{2}{(x^{2} + x + 1)} - \frac{1}{(x^{2} + x + 1)^{2}}\right)}{dx}\\=&-4(\frac{-2(2x + 1 + 0)}{(x^{2} + x + 1)^{3}})x^{2} - \frac{4*2x}{(x^{2} + x + 1)^{2}} - 4(\frac{-2(2x + 1 + 0)}{(x^{2} + x + 1)^{3}})x - \frac{4}{(x^{2} + x + 1)^{2}} + 2(\frac{-(2x + 1 + 0)}{(x^{2} + x + 1)^{2}}) - (\frac{-2(2x + 1 + 0)}{(x^{2} + x + 1)^{3}})\\=&\frac{16x^{3}}{(x^{2} + x + 1)^{3}} + \frac{24x^{2}}{(x^{2} + x + 1)^{3}} - \frac{12x}{(x^{2} + x + 1)^{2}} + \frac{12x}{(x^{2} + x + 1)^{3}} - \frac{6}{(x^{2} + x + 1)^{2}} + \frac{2}{(x^{2} + x + 1)^{3}}\\ \end{split}\end{equation} \]

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