There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ cos(x)tan(x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( cos(x)tan(x)\right)}{dx}\\=&-sin(x)tan(x) + cos(x)sec^{2}(x)(1)\\=&-sin(x)tan(x) + cos(x)sec^{2}(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( -sin(x)tan(x) + cos(x)sec^{2}(x)\right)}{dx}\\=&-cos(x)tan(x) - sin(x)sec^{2}(x)(1) + -sin(x)sec^{2}(x) + cos(x)*2sec^{2}(x)tan(x)\\=&2cos(x)tan(x)sec^{2}(x) - 2sin(x)sec^{2}(x) - cos(x)tan(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 2cos(x)tan(x)sec^{2}(x) - 2sin(x)sec^{2}(x) - cos(x)tan(x)\right)}{dx}\\=&2*-sin(x)tan(x)sec^{2}(x) + 2cos(x)sec^{2}(x)(1)sec^{2}(x) + 2cos(x)tan(x)*2sec^{2}(x)tan(x) - 2cos(x)sec^{2}(x) - 2sin(x)*2sec^{2}(x)tan(x) - -sin(x)tan(x) - cos(x)sec^{2}(x)(1)\\=&-6sin(x)tan(x)sec^{2}(x) + 2cos(x)sec^{4}(x) + 4cos(x)tan^{2}(x)sec^{2}(x) - 3cos(x)sec^{2}(x) + sin(x)tan(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( -6sin(x)tan(x)sec^{2}(x) + 2cos(x)sec^{4}(x) + 4cos(x)tan^{2}(x)sec^{2}(x) - 3cos(x)sec^{2}(x) + sin(x)tan(x)\right)}{dx}\\=&-6cos(x)tan(x)sec^{2}(x) - 6sin(x)sec^{2}(x)(1)sec^{2}(x) - 6sin(x)tan(x)*2sec^{2}(x)tan(x) + 2*-sin(x)sec^{4}(x) + 2cos(x)*4sec^{4}(x)tan(x) + 4*-sin(x)tan^{2}(x)sec^{2}(x) + 4cos(x)*2tan(x)sec^{2}(x)(1)sec^{2}(x) + 4cos(x)tan^{2}(x)*2sec^{2}(x)tan(x) - 3*-sin(x)sec^{2}(x) - 3cos(x)*2sec^{2}(x)tan(x) + cos(x)tan(x) + sin(x)sec^{2}(x)(1)\\=&-12cos(x)tan(x)sec^{2}(x) - 8sin(x)sec^{4}(x) - 16sin(x)tan^{2}(x)sec^{2}(x) + 16cos(x)tan(x)sec^{4}(x) + 8cos(x)tan^{3}(x)sec^{2}(x) + 4sin(x)sec^{2}(x) + cos(x)tan(x)\\ \end{split}\end{equation} \]

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