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

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