There are 1 questions in this calculation: for each question, the 3 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ third\ derivative\ of\ function\ xsin(x)ln(2x + 3y)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = xln(2x + 3y)sin(x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( xln(2x + 3y)sin(x)\right)}{dx}\\=&ln(2x + 3y)sin(x) + \frac{x(2 + 0)sin(x)}{(2x + 3y)} + xln(2x + 3y)cos(x)\\=&ln(2x + 3y)sin(x) + \frac{2xsin(x)}{(2x + 3y)} + xln(2x + 3y)cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( ln(2x + 3y)sin(x) + \frac{2xsin(x)}{(2x + 3y)} + xln(2x + 3y)cos(x)\right)}{dx}\\=&\frac{(2 + 0)sin(x)}{(2x + 3y)} + ln(2x + 3y)cos(x) + 2(\frac{-(2 + 0)}{(2x + 3y)^{2}})xsin(x) + \frac{2sin(x)}{(2x + 3y)} + \frac{2xcos(x)}{(2x + 3y)} + ln(2x + 3y)cos(x) + \frac{x(2 + 0)cos(x)}{(2x + 3y)} + xln(2x + 3y)*-sin(x)\\=&\frac{4sin(x)}{(2x + 3y)} + 2ln(2x + 3y)cos(x) - \frac{4xsin(x)}{(2x + 3y)^{2}} + \frac{4xcos(x)}{(2x + 3y)} - xln(2x + 3y)sin(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{4sin(x)}{(2x + 3y)} + 2ln(2x + 3y)cos(x) - \frac{4xsin(x)}{(2x + 3y)^{2}} + \frac{4xcos(x)}{(2x + 3y)} - xln(2x + 3y)sin(x)\right)}{dx}\\=&4(\frac{-(2 + 0)}{(2x + 3y)^{2}})sin(x) + \frac{4cos(x)}{(2x + 3y)} + \frac{2(2 + 0)cos(x)}{(2x + 3y)} + 2ln(2x + 3y)*-sin(x) - 4(\frac{-2(2 + 0)}{(2x + 3y)^{3}})xsin(x) - \frac{4sin(x)}{(2x + 3y)^{2}} - \frac{4xcos(x)}{(2x + 3y)^{2}} + 4(\frac{-(2 + 0)}{(2x + 3y)^{2}})xcos(x) + \frac{4cos(x)}{(2x + 3y)} + \frac{4x*-sin(x)}{(2x + 3y)} - ln(2x + 3y)sin(x) - \frac{x(2 + 0)sin(x)}{(2x + 3y)} - xln(2x + 3y)cos(x)\\=&\frac{-12sin(x)}{(2x + 3y)^{2}} + \frac{12cos(x)}{(2x + 3y)} - 3ln(2x + 3y)sin(x) + \frac{16xsin(x)}{(2x + 3y)^{3}} - \frac{12xcos(x)}{(2x + 3y)^{2}} - \frac{6xsin(x)}{(2x + 3y)} - xln(2x + 3y)cos(x)\\ \end{split}\end{equation} \]

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