There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ (\frac{1}{(x + 2)}) + (lg(\frac{(1 - x)}{(1 + x)}))\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{(x + 2)} + lg(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{(x + 2)} + lg(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})\right)}{dx}\\=&(\frac{-(1 + 0)}{(x + 2)^{2}}) + \frac{(-(\frac{-(1 + 0)}{(x + 1)^{2}})x - \frac{1}{(x + 1)} + (\frac{-(1 + 0)}{(x + 1)^{2}}))}{ln{10}(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})}\\=&\frac{x}{(x + 1)^{2}(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})ln{10}} - \frac{1}{(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})(x + 1)ln{10}} - \frac{1}{(x + 1)^{2}(\frac{-x}{(x + 1)} + \frac{1}{(x + 1)})ln{10}} - \frac{1}{(x + 2)^{2}}\\ \end{split}\end{equation} \]

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