There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {sin(x)}^{2}{cos(x)}^{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = sin^{2}(x)cos^{2}(x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( sin^{2}(x)cos^{2}(x)\right)}{dx}\\=&2sin(x)cos(x)cos^{2}(x) + sin^{2}(x)*-2cos(x)sin(x)\\=&2sin(x)cos^{3}(x) - 2sin^{3}(x)cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 2sin(x)cos^{3}(x) - 2sin^{3}(x)cos(x)\right)}{dx}\\=&2cos(x)cos^{3}(x) + 2sin(x)*-3cos^{2}(x)sin(x) - 2*3sin^{2}(x)cos(x)cos(x) - 2sin^{3}(x)*-sin(x)\\=&2cos^{4}(x) - 12sin^{2}(x)cos^{2}(x) + 2sin^{4}(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 2cos^{4}(x) - 12sin^{2}(x)cos^{2}(x) + 2sin^{4}(x)\right)}{dx}\\=&2*-4cos^{3}(x)sin(x) - 12*2sin(x)cos(x)cos^{2}(x) - 12sin^{2}(x)*-2cos(x)sin(x) + 2*4sin^{3}(x)cos(x)\\=& - 32sin(x)cos^{3}(x) + 32sin^{3}(x)cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - 32sin(x)cos^{3}(x) + 32sin^{3}(x)cos(x)\right)}{dx}\\=& - 32cos(x)cos^{3}(x) - 32sin(x)*-3cos^{2}(x)sin(x) + 32*3sin^{2}(x)cos(x)cos(x) + 32sin^{3}(x)*-sin(x)\\=& - 32cos^{4}(x) + 192sin^{2}(x)cos^{2}(x) - 32sin^{4}(x)\\ \end{split}\end{equation} \]

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