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

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