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

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