There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {(1 + 2x)}^{(\frac{3}{sin(x)})}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = (2x + 1)^{(\frac{3}{sin(x)})}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( (2x + 1)^{(\frac{3}{sin(x)})}\right)}{dx}\\=&((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))\\=&\frac{-3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)sin(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)sin(x)}\right)}{dx}\\=&\frac{-3((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))ln(2x + 1)cos(x)}{sin^{2}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}(2 + 0)cos(x)}{(2x + 1)sin^{2}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)*-2cos(x)cos(x)}{sin^{3}(x)} - \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)*-sin(x)}{sin^{2}(x)} + \frac{6(\frac{-(2 + 0)}{(2x + 1)^{2}})(2x + 1)^{(\frac{3}{sin(x)})}}{sin(x)} + \frac{6((2x + 1)^{(\frac{3}{sin(x)})}((\frac{3*-cos(x)}{sin^{2}(x)})ln(2x + 1) + \frac{(\frac{3}{sin(x)})(2 + 0)}{(2x + 1)}))}{(2x + 1)sin(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}*-cos(x)}{(2x + 1)sin^{2}(x)}\\=&\frac{9(2x + 1)^{(\frac{3}{sin(x)})}ln^{2}(2x + 1)cos^{2}(x)}{sin^{4}(x)} - \frac{36(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos(x)}{(2x + 1)sin^{3}(x)} - \frac{12(2x + 1)^{(\frac{3}{sin(x)})}cos(x)}{(2x + 1)sin^{2}(x)} + \frac{6(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)cos^{2}(x)}{sin^{3}(x)} + \frac{3(2x + 1)^{(\frac{3}{sin(x)})}ln(2x + 1)}{sin(x)} + \frac{36(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)^{2}sin^{2}(x)} - \frac{12(2x + 1)^{(\frac{3}{sin(x)})}}{(2x + 1)^{2}sin(x)}\\ \end{split}\end{equation} \]

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