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

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