There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ ln(sin(x)){\frac{1}{(π - 2x)}}^{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{ln(sin(x))}{(π - 2x)^{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{ln(sin(x))}{(π - 2x)^{2}}\right)}{dx}\\=&(\frac{-2(0 - 2)}{(π - 2x)^{3}})ln(sin(x)) + \frac{cos(x)}{(π - 2x)^{2}(sin(x))}\\=&\frac{4ln(sin(x))}{(π - 2x)^{3}} + \frac{cos(x)}{(π - 2x)^{2}sin(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{4ln(sin(x))}{(π - 2x)^{3}} + \frac{cos(x)}{(π - 2x)^{2}sin(x)}\right)}{dx}\\=&4(\frac{-3(0 - 2)}{(π - 2x)^{4}})ln(sin(x)) + \frac{4cos(x)}{(π - 2x)^{3}(sin(x))} + \frac{(\frac{-2(0 - 2)}{(π - 2x)^{3}})cos(x)}{sin(x)} + \frac{-cos(x)cos(x)}{(π - 2x)^{2}sin^{2}(x)} + \frac{-sin(x)}{(π - 2x)^{2}sin(x)}\\=&\frac{24ln(sin(x))}{(π - 2x)^{4}} + \frac{8cos(x)}{(π - 2x)^{3}sin(x)} - \frac{cos^{2}(x)}{(π - 2x)^{2}sin^{2}(x)} - \frac{1}{(π - 2x)^{2}}\\ \end{split}\end{equation} \]

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