There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ ln(1 + i{x}^{2})\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln(ix^{2} + 1)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln(ix^{2} + 1)\right)}{dx}\\=&\frac{(i*2x + 0)}{(ix^{2} + 1)}\\=&\frac{2ix}{(ix^{2} + 1)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2ix}{(ix^{2} + 1)}\right)}{dx}\\=&2(\frac{-(i*2x + 0)}{(ix^{2} + 1)^{2}})ix + \frac{2i}{(ix^{2} + 1)}\\=&\frac{-4i^{2}x^{2}}{(ix^{2} + 1)^{2}} + \frac{2i}{(ix^{2} + 1)}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-4i^{2}x^{2}}{(ix^{2} + 1)^{2}} + \frac{2i}{(ix^{2} + 1)}\right)}{dx}\\=&-4(\frac{-2(i*2x + 0)}{(ix^{2} + 1)^{3}})i^{2}x^{2} - \frac{4i^{2}*2x}{(ix^{2} + 1)^{2}} + 2(\frac{-(i*2x + 0)}{(ix^{2} + 1)^{2}})i + 0\\=&\frac{16i^{3}x^{3}}{(ix^{2} + 1)^{3}} - \frac{12i^{2}x}{(ix^{2} + 1)^{2}}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{16i^{3}x^{3}}{(ix^{2} + 1)^{3}} - \frac{12i^{2}x}{(ix^{2} + 1)^{2}}\right)}{dx}\\=&16(\frac{-3(i*2x + 0)}{(ix^{2} + 1)^{4}})i^{3}x^{3} + \frac{16i^{3}*3x^{2}}{(ix^{2} + 1)^{3}} - 12(\frac{-2(i*2x + 0)}{(ix^{2} + 1)^{3}})i^{2}x - \frac{12i^{2}}{(ix^{2} + 1)^{2}}\\=&\frac{-96i^{4}x^{4}}{(ix^{2} + 1)^{4}} + \frac{96i^{3}x^{2}}{(ix^{2} + 1)^{3}} - \frac{12i^{2}}{(ix^{2} + 1)^{2}}\\ \end{split}\end{equation} \]

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