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

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