There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ (x{e}^{x}){\frac{1}{(x + ln(x))}}^{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{x{e}^{x}}{(x + ln(x))^{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{x{e}^{x}}{(x + ln(x))^{2}}\right)}{dx}\\=&(\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}})x{e}^{x} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}}\\=& - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x{e}^{x}}{(x + ln(x))^{3}} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x{e}^{x}}{(x + ln(x))^{3}} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\right)}{dx}\\=& - 2(\frac{-3(1 + \frac{1}{(x)})}{(x + ln(x))^{4}}){e}^{x} - \frac{2({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{3}} - 2(\frac{-3(1 + \frac{1}{(x)})}{(x + ln(x))^{4}})x{e}^{x} - \frac{2{e}^{x}}{(x + ln(x))^{3}} - \frac{2x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{3}} + (\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}}){e}^{x} + \frac{({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}} + (\frac{-2(1 + \frac{1}{(x)})}{(x + ln(x))^{3}})x{e}^{x} + \frac{{e}^{x}}{(x + ln(x))^{2}} + \frac{x({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + ln(x))^{2}}\\=&\frac{6{e}^{x}}{(x + ln(x))^{4}x} + \frac{12{e}^{x}}{(x + ln(x))^{4}} - \frac{8{e}^{x}}{(x + ln(x))^{3}} + \frac{6x{e}^{x}}{(x + ln(x))^{4}} - \frac{4x{e}^{x}}{(x + ln(x))^{3}} - \frac{2{e}^{x}}{(x + ln(x))^{3}x} + \frac{2{e}^{x}}{(x + ln(x))^{2}} + \frac{x{e}^{x}}{(x + ln(x))^{2}}\\ \end{split}\end{equation} \]

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