There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {e}^{(2x)}cos(3x) - ln(2x + 1)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( {e}^{(2x)}cos(3x) - ln(2x + 1)\right)}{dx}\\=&({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(3x) + {e}^{(2x)}*-sin(3x)*3 - \frac{(2 + 0)}{(2x + 1)}\\=&2{e}^{(2x)}cos(3x) - 3{e}^{(2x)}sin(3x) - \frac{2}{(2x + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 2{e}^{(2x)}cos(3x) - 3{e}^{(2x)}sin(3x) - \frac{2}{(2x + 1)}\right)}{dx}\\=&2({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(3x) + 2{e}^{(2x)}*-sin(3x)*3 - 3({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(3x) - 3{e}^{(2x)}cos(3x)*3 - 2(\frac{-(2 + 0)}{(2x + 1)^{2}})\\=&-5{e}^{(2x)}cos(3x) - 12{e}^{(2x)}sin(3x) + \frac{4}{(2x + 1)^{2}}\\ \end{split}\end{equation} \]

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