There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {ln(x)}^{2} + sin(x)\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln^{2}(x) + sin(x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln^{2}(x) + sin(x)\right)}{dx}\\=&\frac{2ln(x)}{(x)} + cos(x)\\=&\frac{2ln(x)}{x} + cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{2ln(x)}{x} + cos(x)\right)}{dx}\\=&\frac{2*-ln(x)}{x^{2}} + \frac{2}{x(x)} + -sin(x)\\=&\frac{-2ln(x)}{x^{2}} + \frac{2}{x^{2}} - sin(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2ln(x)}{x^{2}} + \frac{2}{x^{2}} - sin(x)\right)}{dx}\\=&\frac{-2*-2ln(x)}{x^{3}} - \frac{2}{x^{2}(x)} + \frac{2*-2}{x^{3}} - cos(x)\\=&\frac{4ln(x)}{x^{3}} - \frac{6}{x^{3}} - cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{4ln(x)}{x^{3}} - \frac{6}{x^{3}} - cos(x)\right)}{dx}\\=&\frac{4*-3ln(x)}{x^{4}} + \frac{4}{x^{3}(x)} - \frac{6*-3}{x^{4}} - -sin(x)\\=&\frac{-12ln(x)}{x^{4}} + \frac{22}{x^{4}} + sin(x)\\ \end{split}\end{equation} \]

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