There are 1 questions in this calculation: for each question, the 3 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ third\ derivative\ of\ function\ \frac{{x}^{4}}{(1 + x)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{x^{4}}{(x + 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{x^{4}}{(x + 1)}\right)}{dx}\\=&(\frac{-(1 + 0)}{(x + 1)^{2}})x^{4} + \frac{4x^{3}}{(x + 1)}\\=&\frac{-x^{4}}{(x + 1)^{2}} + \frac{4x^{3}}{(x + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-x^{4}}{(x + 1)^{2}} + \frac{4x^{3}}{(x + 1)}\right)}{dx}\\=&-(\frac{-2(1 + 0)}{(x + 1)^{3}})x^{4} - \frac{4x^{3}}{(x + 1)^{2}} + 4(\frac{-(1 + 0)}{(x + 1)^{2}})x^{3} + \frac{4*3x^{2}}{(x + 1)}\\=&\frac{2x^{4}}{(x + 1)^{3}} - \frac{8x^{3}}{(x + 1)^{2}} + \frac{12x^{2}}{(x + 1)}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{2x^{4}}{(x + 1)^{3}} - \frac{8x^{3}}{(x + 1)^{2}} + \frac{12x^{2}}{(x + 1)}\right)}{dx}\\=&2(\frac{-3(1 + 0)}{(x + 1)^{4}})x^{4} + \frac{2*4x^{3}}{(x + 1)^{3}} - 8(\frac{-2(1 + 0)}{(x + 1)^{3}})x^{3} - \frac{8*3x^{2}}{(x + 1)^{2}} + 12(\frac{-(1 + 0)}{(x + 1)^{2}})x^{2} + \frac{12*2x}{(x + 1)}\\=&\frac{-6x^{4}}{(x + 1)^{4}} + \frac{24x^{3}}{(x + 1)^{3}} - \frac{36x^{2}}{(x + 1)^{2}} + \frac{24x}{(x + 1)}\\ \end{split}\end{equation} \]

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