There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {x}^{2}ln(2 + x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = x^{2}ln(x + 2)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( x^{2}ln(x + 2)\right)}{dx}\\=&2xln(x + 2) + \frac{x^{2}(1 + 0)}{(x + 2)}\\=&2xln(x + 2) + \frac{x^{2}}{(x + 2)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 2xln(x + 2) + \frac{x^{2}}{(x + 2)}\right)}{dx}\\=&2ln(x + 2) + \frac{2x(1 + 0)}{(x + 2)} + (\frac{-(1 + 0)}{(x + 2)^{2}})x^{2} + \frac{2x}{(x + 2)}\\=&2ln(x + 2) + \frac{4x}{(x + 2)} - \frac{x^{2}}{(x + 2)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( 2ln(x + 2) + \frac{4x}{(x + 2)} - \frac{x^{2}}{(x + 2)^{2}}\right)}{dx}\\=&\frac{2(1 + 0)}{(x + 2)} + 4(\frac{-(1 + 0)}{(x + 2)^{2}})x + \frac{4}{(x + 2)} - (\frac{-2(1 + 0)}{(x + 2)^{3}})x^{2} - \frac{2x}{(x + 2)^{2}}\\=& - \frac{6x}{(x + 2)^{2}} + \frac{2x^{2}}{(x + 2)^{3}} + \frac{6}{(x + 2)}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - \frac{6x}{(x + 2)^{2}} + \frac{2x^{2}}{(x + 2)^{3}} + \frac{6}{(x + 2)}\right)}{dx}\\=& - 6(\frac{-2(1 + 0)}{(x + 2)^{3}})x - \frac{6}{(x + 2)^{2}} + 2(\frac{-3(1 + 0)}{(x + 2)^{4}})x^{2} + \frac{2*2x}{(x + 2)^{3}} + 6(\frac{-(1 + 0)}{(x + 2)^{2}})\\=&\frac{16x}{(x + 2)^{3}} - \frac{6x^{2}}{(x + 2)^{4}} - \frac{12}{(x + 2)^{2}}\\ \end{split}\end{equation} \]

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