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

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