There are 1 questions in this calculation: for each question, the 3 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ third\ derivative\ of\ function\ {e}^{x}{\frac{1}{(1 + x)}}^{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{{e}^{x}}{(x + 1)^{2}}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{{e}^{x}}{(x + 1)^{2}}\right)}{dx}\\=&(\frac{-2(1 + 0)}{(x + 1)^{3}}){e}^{x} + \frac{({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{2}}\\=&\frac{-2{e}^{x}}{(x + 1)^{3}} + \frac{{e}^{x}}{(x + 1)^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2{e}^{x}}{(x + 1)^{3}} + \frac{{e}^{x}}{(x + 1)^{2}}\right)}{dx}\\=&-2(\frac{-3(1 + 0)}{(x + 1)^{4}}){e}^{x} - \frac{2({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{3}} + (\frac{-2(1 + 0)}{(x + 1)^{3}}){e}^{x} + \frac{({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{2}}\\=&\frac{6{e}^{x}}{(x + 1)^{4}} - \frac{4{e}^{x}}{(x + 1)^{3}} + \frac{{e}^{x}}{(x + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{6{e}^{x}}{(x + 1)^{4}} - \frac{4{e}^{x}}{(x + 1)^{3}} + \frac{{e}^{x}}{(x + 1)^{2}}\right)}{dx}\\=&6(\frac{-4(1 + 0)}{(x + 1)^{5}}){e}^{x} + \frac{6({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{4}} - 4(\frac{-3(1 + 0)}{(x + 1)^{4}}){e}^{x} - \frac{4({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{3}} + (\frac{-2(1 + 0)}{(x + 1)^{3}}){e}^{x} + \frac{({e}^{x}((1)ln(e) + \frac{(x)(0)}{(e)}))}{(x + 1)^{2}}\\=&\frac{-24{e}^{x}}{(x + 1)^{5}} + \frac{18{e}^{x}}{(x + 1)^{4}} - \frac{6{e}^{x}}{(x + 1)^{3}} + \frac{{e}^{x}}{(x + 1)^{2}}\\ \end{split}\end{equation} \]

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