There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{x}{({x}^{4} + 1)}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{x}{(x^{4} + 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{x}{(x^{4} + 1)}\right)}{dx}\\=&(\frac{-(4x^{3} + 0)}{(x^{4} + 1)^{2}})x + \frac{1}{(x^{4} + 1)}\\=&\frac{-4x^{4}}{(x^{4} + 1)^{2}} + \frac{1}{(x^{4} + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-4x^{4}}{(x^{4} + 1)^{2}} + \frac{1}{(x^{4} + 1)}\right)}{dx}\\=&-4(\frac{-2(4x^{3} + 0)}{(x^{4} + 1)^{3}})x^{4} - \frac{4*4x^{3}}{(x^{4} + 1)^{2}} + (\frac{-(4x^{3} + 0)}{(x^{4} + 1)^{2}})\\=&\frac{32x^{7}}{(x^{4} + 1)^{3}} - \frac{20x^{3}}{(x^{4} + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{32x^{7}}{(x^{4} + 1)^{3}} - \frac{20x^{3}}{(x^{4} + 1)^{2}}\right)}{dx}\\=&32(\frac{-3(4x^{3} + 0)}{(x^{4} + 1)^{4}})x^{7} + \frac{32*7x^{6}}{(x^{4} + 1)^{3}} - 20(\frac{-2(4x^{3} + 0)}{(x^{4} + 1)^{3}})x^{3} - \frac{20*3x^{2}}{(x^{4} + 1)^{2}}\\=&\frac{-384x^{10}}{(x^{4} + 1)^{4}} + \frac{384x^{6}}{(x^{4} + 1)^{3}} - \frac{60x^{2}}{(x^{4} + 1)^{2}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-384x^{10}}{(x^{4} + 1)^{4}} + \frac{384x^{6}}{(x^{4} + 1)^{3}} - \frac{60x^{2}}{(x^{4} + 1)^{2}}\right)}{dx}\\=&-384(\frac{-4(4x^{3} + 0)}{(x^{4} + 1)^{5}})x^{10} - \frac{384*10x^{9}}{(x^{4} + 1)^{4}} + 384(\frac{-3(4x^{3} + 0)}{(x^{4} + 1)^{4}})x^{6} + \frac{384*6x^{5}}{(x^{4} + 1)^{3}} - 60(\frac{-2(4x^{3} + 0)}{(x^{4} + 1)^{3}})x^{2} - \frac{60*2x}{(x^{4} + 1)^{2}}\\=&\frac{6144x^{13}}{(x^{4} + 1)^{5}} - \frac{8448x^{9}}{(x^{4} + 1)^{4}} + \frac{2784x^{5}}{(x^{4} + 1)^{3}} - \frac{120x}{(x^{4} + 1)^{2}}\\ \end{split}\end{equation} \]

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