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

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