There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ arctan(x) - \frac{acrcos(\frac{2x}{(1 + {x}^{2})})}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = arctan(x) - \frac{1}{2}acrcos(\frac{2x}{(x^{2} + 1)})\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( arctan(x) - \frac{1}{2}acrcos(\frac{2x}{(x^{2} + 1)})\right)}{dx}\\=&(\frac{(1)}{(1 + (x)^{2})}) - \frac{1}{2}acr*-sin(\frac{2x}{(x^{2} + 1)})(2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x + \frac{2}{(x^{2} + 1)})\\=& - \frac{2acrx^{2}sin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} + \frac{acrsin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( - \frac{2acrx^{2}sin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} + \frac{acrsin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)}\right)}{dx}\\=& - 2(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})acrx^{2}sin(\frac{2x}{(x^{2} + 1)}) - \frac{2acr*2xsin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} - \frac{2acrx^{2}cos(\frac{2x}{(x^{2} + 1)})(2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x + \frac{2}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} + (\frac{-(2x + 0)}{(x^{2} + 1)^{2}})acrsin(\frac{2x}{(x^{2} + 1)}) + \frac{acrcos(\frac{2x}{(x^{2} + 1)})(2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x + \frac{2}{(x^{2} + 1)})}{(x^{2} + 1)} + (\frac{-(2x + 0)}{(x^{2} + 1)^{2}})\\=&\frac{8acrx^{3}sin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{3}} - \frac{6acrxsin(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} + \frac{8acrx^{4}cos(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{4}} - \frac{8acrx^{2}cos(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{3}} + \frac{2acrcos(\frac{2x}{(x^{2} + 1)})}{(x^{2} + 1)^{2}} - \frac{2x}{(x^{2} + 1)^{2}}\\ \end{split}\end{equation} \]

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