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

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