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

Your problem has not been solved here? Please take a look at the  hot problems ！