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

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