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

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