There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ {sin(x)}^{-1}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{sin(x)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{sin(x)}\right)}{dx}\\=&\frac{-cos(x)}{sin^{2}(x)}\\=&\frac{-cos(x)}{sin^{2}(x)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-cos(x)}{sin^{2}(x)}\right)}{dx}\\=&\frac{--2cos(x)cos(x)}{sin^{3}(x)} - \frac{-sin(x)}{sin^{2}(x)}\\=&\frac{2cos^{2}(x)}{sin^{3}(x)} + \frac{1}{sin(x)}\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( \frac{2cos^{2}(x)}{sin^{3}(x)} + \frac{1}{sin(x)}\right)}{dx}\\=&\frac{2*-3cos(x)cos^{2}(x)}{sin^{4}(x)} + \frac{2*-2cos(x)sin(x)}{sin^{3}(x)} + \frac{-cos(x)}{sin^{2}(x)}\\=&\frac{-6cos^{3}(x)}{sin^{4}(x)} - \frac{5cos(x)}{sin^{2}(x)}\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( \frac{-6cos^{3}(x)}{sin^{4}(x)} - \frac{5cos(x)}{sin^{2}(x)}\right)}{dx}\\=&\frac{-6*-4cos(x)cos^{3}(x)}{sin^{5}(x)} - \frac{6*-3cos^{2}(x)sin(x)}{sin^{4}(x)} - \frac{5*-2cos(x)cos(x)}{sin^{3}(x)} - \frac{5*-sin(x)}{sin^{2}(x)}\\=&\frac{24cos^{4}(x)}{sin^{5}(x)} + \frac{28cos^{2}(x)}{sin^{3}(x)} + \frac{5}{sin(x)}\\ \end{split}\end{equation} \]

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