There are 1 questions in this calculation: for each question, the 4 derivative of m is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ log_{ari}^{th(m)}\ with\ respect\ to\ m：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( log_{ari}^{th(m)}\right)}{dm}\\=&(\frac{(\frac{((1 - th^{2}(m)))}{(th(m))} - \frac{(0)log_{ari}^{th(m)}}{(ari)})}{(ln(ari))})\\=&\frac{-th(m)}{ln(ari)} + \frac{1}{ln(ari)th(m)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-th(m)}{ln(ari)} + \frac{1}{ln(ari)th(m)}\right)}{dm}\\=&\frac{--0th(m)}{ln^{2}(ari)(ari)} - \frac{(1 - th^{2}(m))}{ln(ari)} + \frac{-0}{ln^{2}(ari)(ari)th(m)} + \frac{-(1 - th^{2}(m))}{ln(ari)th^{2}(m)}\\=&\frac{-1}{ln(ari)th^{2}(m)} + \frac{th^{2}(m)}{ln(ari)}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-1}{ln(ari)th^{2}(m)} + \frac{th^{2}(m)}{ln(ari)}\right)}{dm}\\=&\frac{--0}{ln^{2}(ari)(ari)th^{2}(m)} - \frac{-2(1 - th^{2}(m))}{ln(ari)th^{3}(m)} + \frac{-0th^{2}(m)}{ln^{2}(ari)(ari)} + \frac{2th(m)(1 - th^{2}(m))}{ln(ari)}\\=&\frac{2}{ln(ari)th^{3}(m)} - \frac{2}{ln(ari)th(m)} + \frac{2th(m)}{ln(ari)} - \frac{2th^{3}(m)}{ln(ari)}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{2}{ln(ari)th^{3}(m)} - \frac{2}{ln(ari)th(m)} + \frac{2th(m)}{ln(ari)} - \frac{2th^{3}(m)}{ln(ari)}\right)}{dm}\\=&\frac{2*-0}{ln^{2}(ari)(ari)th^{3}(m)} + \frac{2*-3(1 - th^{2}(m))}{ln(ari)th^{4}(m)} - \frac{2*-0}{ln^{2}(ari)(ari)th(m)} - \frac{2*-(1 - th^{2}(m))}{ln(ari)th^{2}(m)} + \frac{2*-0th(m)}{ln^{2}(ari)(ari)} + \frac{2(1 - th^{2}(m))}{ln(ari)} - \frac{2*-0th^{3}(m)}{ln^{2}(ari)(ari)} - \frac{2*3th^{2}(m)(1 - th^{2}(m))}{ln(ari)}\\=&\frac{-6}{ln(ari)th^{4}(m)} + \frac{8}{ln(ari)th^{2}(m)} - \frac{8th^{2}(m)}{ln(ari)} + \frac{6th^{4}(m)}{ln(ari)}\\ \end{split}\end{equation} \]

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