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

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