There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{1}{(1 + {x}^{3})}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{(x^{3} + 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{(x^{3} + 1)}\right)}{dx}\\=&(\frac{-(3x^{2} + 0)}{(x^{3} + 1)^{2}})\\=&\frac{-3x^{2}}{(x^{3} + 1)^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-3x^{2}}{(x^{3} + 1)^{2}}\right)}{dx}\\=&-3(\frac{-2(3x^{2} + 0)}{(x^{3} + 1)^{3}})x^{2} - \frac{3*2x}{(x^{3} + 1)^{2}}\\=&\frac{18x^{4}}{(x^{3} + 1)^{3}} - \frac{6x}{(x^{3} + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{18x^{4}}{(x^{3} + 1)^{3}} - \frac{6x}{(x^{3} + 1)^{2}}\right)}{dx}\\=&18(\frac{-3(3x^{2} + 0)}{(x^{3} + 1)^{4}})x^{4} + \frac{18*4x^{3}}{(x^{3} + 1)^{3}} - 6(\frac{-2(3x^{2} + 0)}{(x^{3} + 1)^{3}})x - \frac{6}{(x^{3} + 1)^{2}}\\=&\frac{-162x^{6}}{(x^{3} + 1)^{4}} + \frac{108x^{3}}{(x^{3} + 1)^{3}} - \frac{6}{(x^{3} + 1)^{2}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-162x^{6}}{(x^{3} + 1)^{4}} + \frac{108x^{3}}{(x^{3} + 1)^{3}} - \frac{6}{(x^{3} + 1)^{2}}\right)}{dx}\\=&-162(\frac{-4(3x^{2} + 0)}{(x^{3} + 1)^{5}})x^{6} - \frac{162*6x^{5}}{(x^{3} + 1)^{4}} + 108(\frac{-3(3x^{2} + 0)}{(x^{3} + 1)^{4}})x^{3} + \frac{108*3x^{2}}{(x^{3} + 1)^{3}} - 6(\frac{-2(3x^{2} + 0)}{(x^{3} + 1)^{3}})\\=&\frac{1944x^{8}}{(x^{3} + 1)^{5}} - \frac{1944x^{5}}{(x^{3} + 1)^{4}} + \frac{360x^{2}}{(x^{3} + 1)^{3}}\\ \end{split}\end{equation} \]

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