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

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