There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{x{\frac{1}{(1 + {x}^{2})}}^{1}}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{\frac{1}{2}x}{(x^{2} + 1)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{\frac{1}{2}x}{(x^{2} + 1)}\right)}{dx}\\=&\frac{1}{2}(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})x + \frac{\frac{1}{2}}{(x^{2} + 1)}\\=&\frac{-x^{2}}{(x^{2} + 1)^{2}} + \frac{1}{2(x^{2} + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-x^{2}}{(x^{2} + 1)^{2}} + \frac{1}{2(x^{2} + 1)}\right)}{dx}\\=&-(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x^{2} - \frac{2x}{(x^{2} + 1)^{2}} + \frac{(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})}{2}\\=&\frac{4x^{3}}{(x^{2} + 1)^{3}} - \frac{3x}{(x^{2} + 1)^{2}}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{4x^{3}}{(x^{2} + 1)^{3}} - \frac{3x}{(x^{2} + 1)^{2}}\right)}{dx}\\=&4(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})x^{3} + \frac{4*3x^{2}}{(x^{2} + 1)^{3}} - 3(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x - \frac{3}{(x^{2} + 1)^{2}}\\=&\frac{-24x^{4}}{(x^{2} + 1)^{4}} + \frac{24x^{2}}{(x^{2} + 1)^{3}} - \frac{3}{(x^{2} + 1)^{2}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-24x^{4}}{(x^{2} + 1)^{4}} + \frac{24x^{2}}{(x^{2} + 1)^{3}} - \frac{3}{(x^{2} + 1)^{2}}\right)}{dx}\\=&-24(\frac{-4(2x + 0)}{(x^{2} + 1)^{5}})x^{4} - \frac{24*4x^{3}}{(x^{2} + 1)^{4}} + 24(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})x^{2} + \frac{24*2x}{(x^{2} + 1)^{3}} - 3(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})\\=&\frac{192x^{5}}{(x^{2} + 1)^{5}} - \frac{240x^{3}}{(x^{2} + 1)^{4}} + \frac{60x}{(x^{2} + 1)^{3}}\\ \end{split}\end{equation} \]

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