There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ \frac{({x}^{2} - 2x + 4)}{({x}^{2} + 2x + 4)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{x^{2}}{(x^{2} + 2x + 4)} - \frac{2x}{(x^{2} + 2x + 4)} + \frac{4}{(x^{2} + 2x + 4)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{x^{2}}{(x^{2} + 2x + 4)} - \frac{2x}{(x^{2} + 2x + 4)} + \frac{4}{(x^{2} + 2x + 4)}\right)}{dx}\\=&(\frac{-(2x + 2 + 0)}{(x^{2} + 2x + 4)^{2}})x^{2} + \frac{2x}{(x^{2} + 2x + 4)} - 2(\frac{-(2x + 2 + 0)}{(x^{2} + 2x + 4)^{2}})x - \frac{2}{(x^{2} + 2x + 4)} + 4(\frac{-(2x + 2 + 0)}{(x^{2} + 2x + 4)^{2}})\\=&\frac{-2x^{3}}{(x^{2} + 2x + 4)^{2}} + \frac{2x^{2}}{(x^{2} + 2x + 4)^{2}} + \frac{2x}{(x^{2} + 2x + 4)} - \frac{4x}{(x^{2} + 2x + 4)^{2}} - \frac{2}{(x^{2} + 2x + 4)} - \frac{8}{(x^{2} + 2x + 4)^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2x^{3}}{(x^{2} + 2x + 4)^{2}} + \frac{2x^{2}}{(x^{2} + 2x + 4)^{2}} + \frac{2x}{(x^{2} + 2x + 4)} - \frac{4x}{(x^{2} + 2x + 4)^{2}} - \frac{2}{(x^{2} + 2x + 4)} - \frac{8}{(x^{2} + 2x + 4)^{2}}\right)}{dx}\\=&-2(\frac{-2(2x + 2 + 0)}{(x^{2} + 2x + 4)^{3}})x^{3} - \frac{2*3x^{2}}{(x^{2} + 2x + 4)^{2}} + 2(\frac{-2(2x + 2 + 0)}{(x^{2} + 2x + 4)^{3}})x^{2} + \frac{2*2x}{(x^{2} + 2x + 4)^{2}} + 2(\frac{-(2x + 2 + 0)}{(x^{2} + 2x + 4)^{2}})x + \frac{2}{(x^{2} + 2x + 4)} - 4(\frac{-2(2x + 2 + 0)}{(x^{2} + 2x + 4)^{3}})x - \frac{4}{(x^{2} + 2x + 4)^{2}} - 2(\frac{-(2x + 2 + 0)}{(x^{2} + 2x + 4)^{2}}) - 8(\frac{-2(2x + 2 + 0)}{(x^{2} + 2x + 4)^{3}})\\=&\frac{8x^{4}}{(x^{2} + 2x + 4)^{3}} - \frac{10x^{2}}{(x^{2} + 2x + 4)^{2}} + \frac{8x^{2}}{(x^{2} + 2x + 4)^{3}} + \frac{4x}{(x^{2} + 2x + 4)^{2}} + \frac{48x}{(x^{2} + 2x + 4)^{3}} + \frac{2}{(x^{2} + 2x + 4)} + \frac{32}{(x^{2} + 2x + 4)^{3}}\\ \end{split}\end{equation} \]

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