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

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