There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ \frac{sin(x)}{x}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{sin(x)}{x}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{sin(x)}{x}\right)}{dx}\\=&\frac{-sin(x)}{x^{2}} + \frac{cos(x)}{x}\\=&\frac{-sin(x)}{x^{2}} + \frac{cos(x)}{x}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-sin(x)}{x^{2}} + \frac{cos(x)}{x}\right)}{dx}\\=&\frac{--2sin(x)}{x^{3}} - \frac{cos(x)}{x^{2}} + \frac{-cos(x)}{x^{2}} + \frac{-sin(x)}{x}\\=&\frac{2sin(x)}{x^{3}} - \frac{2cos(x)}{x^{2}} - \frac{sin(x)}{x}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{2sin(x)}{x^{3}} - \frac{2cos(x)}{x^{2}} - \frac{sin(x)}{x}\right)}{dx}\\=&\frac{2*-3sin(x)}{x^{4}} + \frac{2cos(x)}{x^{3}} - \frac{2*-2cos(x)}{x^{3}} - \frac{2*-sin(x)}{x^{2}} - \frac{-sin(x)}{x^{2}} - \frac{cos(x)}{x}\\=&\frac{-6sin(x)}{x^{4}} + \frac{6cos(x)}{x^{3}} + \frac{3sin(x)}{x^{2}} - \frac{cos(x)}{x}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-6sin(x)}{x^{4}} + \frac{6cos(x)}{x^{3}} + \frac{3sin(x)}{x^{2}} - \frac{cos(x)}{x}\right)}{dx}\\=&\frac{-6*-4sin(x)}{x^{5}} - \frac{6cos(x)}{x^{4}} + \frac{6*-3cos(x)}{x^{4}} + \frac{6*-sin(x)}{x^{3}} + \frac{3*-2sin(x)}{x^{3}} + \frac{3cos(x)}{x^{2}} - \frac{-cos(x)}{x^{2}} - \frac{-sin(x)}{x}\\=&\frac{24sin(x)}{x^{5}} - \frac{24cos(x)}{x^{4}} - \frac{12sin(x)}{x^{3}} + \frac{4cos(x)}{x^{2}} + \frac{sin(x)}{x}\\ \end{split}\end{equation} \]

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