There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ ln(sec(2x) + tan(2x))\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln(sec(2x) + tan(2x))\right)}{dx}\\=&\frac{(sec(2x)tan(2x)*2 + sec^{2}(2x)(2))}{(sec(2x) + tan(2x))}\\=&\frac{2tan(2x)sec(2x)}{(sec(2x) + tan(2x))} + \frac{2sec^{2}(2x)}{(sec(2x) + tan(2x))}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2tan(2x)sec(2x)}{(sec(2x) + tan(2x))} + \frac{2sec^{2}(2x)}{(sec(2x) + tan(2x))}\right)}{dx}\\=&2(\frac{-(sec(2x)tan(2x)*2 + sec^{2}(2x)(2))}{(sec(2x) + tan(2x))^{2}})tan(2x)sec(2x) + \frac{2sec^{2}(2x)(2)sec(2x)}{(sec(2x) + tan(2x))} + \frac{2tan(2x)sec(2x)tan(2x)*2}{(sec(2x) + tan(2x))} + 2(\frac{-(sec(2x)tan(2x)*2 + sec^{2}(2x)(2))}{(sec(2x) + tan(2x))^{2}})sec^{2}(2x) + \frac{2*2sec^{2}(2x)tan(2x)*2}{(sec(2x) + tan(2x))}\\=&\frac{-4tan^{2}(2x)sec^{2}(2x)}{(sec(2x) + tan(2x))^{2}} - \frac{8tan(2x)sec^{3}(2x)}{(sec(2x) + tan(2x))^{2}} + \frac{4sec^{3}(2x)}{(sec(2x) + tan(2x))} + \frac{4tan^{2}(2x)sec(2x)}{(sec(2x) + tan(2x))} - \frac{4sec^{4}(2x)}{(sec(2x) + tan(2x))^{2}} + \frac{8tan(2x)sec^{2}(2x)}{(sec(2x) + tan(2x))}\\ \end{split}\end{equation} \]

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