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

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