There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{({e}^{(2x)} + {e}^{(-2x)})}{({e}^{(2x)} - {e}^{(-2x)})}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{{e}^{(2x)}}{({e}^{(2x)} - {e}^{(-2x)})} + \frac{{e}^{(-2x)}}{({e}^{(2x)} - {e}^{(-2x)})}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{{e}^{(2x)}}{({e}^{(2x)} - {e}^{(-2x)})} + \frac{{e}^{(-2x)}}{({e}^{(2x)} - {e}^{(-2x)})}\right)}{dx}\\=&(\frac{-(({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)})) - ({e}^{(-2x)}((-2)ln(e) + \frac{(-2x)(0)}{(e)})))}{({e}^{(2x)} - {e}^{(-2x)})^{2}}){e}^{(2x)} + \frac{({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))}{({e}^{(2x)} - {e}^{(-2x)})} + (\frac{-(({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)})) - ({e}^{(-2x)}((-2)ln(e) + \frac{(-2x)(0)}{(e)})))}{({e}^{(2x)} - {e}^{(-2x)})^{2}}){e}^{(-2x)} + \frac{({e}^{(-2x)}((-2)ln(e) + \frac{(-2x)(0)}{(e)}))}{({e}^{(2x)} - {e}^{(-2x)})}\\=&\frac{-2{e}^{(4x)}}{({e}^{(2x)} - {e}^{(-2x)})^{2}} + \frac{2{e}^{(2x)}}{({e}^{(2x)} - {e}^{(-2x)})} - \frac{2{e}^{(-4x)}}{({e}^{(2x)} - {e}^{(-2x)})^{2}} - \frac{2{e}^{(-2x)}}{({e}^{(2x)} - {e}^{(-2x)})} - \frac{4}{({e}^{(2x)} - {e}^{(-2x)})^{2}}\\ \end{split}\end{equation} \]

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