There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ \frac{sin(2x)cos(3x)x}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{2}xsin(2x)cos(3x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{2}xsin(2x)cos(3x)\right)}{dx}\\=&\frac{1}{2}sin(2x)cos(3x) + \frac{1}{2}xcos(2x)*2cos(3x) + \frac{1}{2}xsin(2x)*-sin(3x)*3\\=&\frac{sin(2x)cos(3x)}{2} + xcos(2x)cos(3x) - \frac{3xsin(3x)sin(2x)}{2}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{sin(2x)cos(3x)}{2} + xcos(2x)cos(3x) - \frac{3xsin(3x)sin(2x)}{2}\right)}{dx}\\=&\frac{cos(2x)*2cos(3x)}{2} + \frac{sin(2x)*-sin(3x)*3}{2} + cos(2x)cos(3x) + x*-sin(2x)*2cos(3x) + xcos(2x)*-sin(3x)*3 - \frac{3sin(3x)sin(2x)}{2} - \frac{3xcos(3x)*3sin(2x)}{2} - \frac{3xsin(3x)cos(2x)*2}{2}\\=&2cos(2x)cos(3x) - 3sin(3x)sin(2x) - \frac{13xsin(2x)cos(3x)}{2} - 6xsin(3x)cos(2x)\\ \end{split}\end{equation} \]

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