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

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