There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {x}^{3}cos(2x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = x^{3}cos(2x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( x^{3}cos(2x)\right)}{dx}\\=&3x^{2}cos(2x) + x^{3}*-sin(2x)*2\\=&3x^{2}cos(2x) - 2x^{3}sin(2x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 3x^{2}cos(2x) - 2x^{3}sin(2x)\right)}{dx}\\=&3*2xcos(2x) + 3x^{2}*-sin(2x)*2 - 2*3x^{2}sin(2x) - 2x^{3}cos(2x)*2\\=&6xcos(2x) - 12x^{2}sin(2x) - 4x^{3}cos(2x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 6xcos(2x) - 12x^{2}sin(2x) - 4x^{3}cos(2x)\right)}{dx}\\=&6cos(2x) + 6x*-sin(2x)*2 - 12*2xsin(2x) - 12x^{2}cos(2x)*2 - 4*3x^{2}cos(2x) - 4x^{3}*-sin(2x)*2\\=&6cos(2x) - 36xsin(2x) - 36x^{2}cos(2x) + 8x^{3}sin(2x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( 6cos(2x) - 36xsin(2x) - 36x^{2}cos(2x) + 8x^{3}sin(2x)\right)}{dx}\\=&6*-sin(2x)*2 - 36sin(2x) - 36xcos(2x)*2 - 36*2xcos(2x) - 36x^{2}*-sin(2x)*2 + 8*3x^{2}sin(2x) + 8x^{3}cos(2x)*2\\=&-48sin(2x) - 144xcos(2x) + 96x^{2}sin(2x) + 16x^{3}cos(2x)\\ \end{split}\end{equation} \]

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