There are 1 questions in this calculation: for each question, the 3 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ third\ derivative\ of\ function\ \frac{{e}^{\frac{1}{2}}x{cos(2x)}^{1}}{2}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{1}{2}xe^{\frac{1}{2}}cos(2x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{1}{2}xe^{\frac{1}{2}}cos(2x)\right)}{dx}\\=&\frac{1}{2}e^{\frac{1}{2}}cos(2x) + \frac{\frac{1}{2}x*\frac{1}{2}*0cos(2x)}{e^{\frac{1}{2}}} + \frac{1}{2}xe^{\frac{1}{2}}*-sin(2x)*2\\=&\frac{e^{\frac{1}{2}}cos(2x)}{2} - xe^{\frac{1}{2}}sin(2x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{e^{\frac{1}{2}}cos(2x)}{2} - xe^{\frac{1}{2}}sin(2x)\right)}{dx}\\=&\frac{\frac{1}{2}*0cos(2x)}{2e^{\frac{1}{2}}} + \frac{e^{\frac{1}{2}}*-sin(2x)*2}{2} - e^{\frac{1}{2}}sin(2x) - \frac{x*\frac{1}{2}*0sin(2x)}{e^{\frac{1}{2}}} - xe^{\frac{1}{2}}cos(2x)*2\\=&-2e^{\frac{1}{2}}sin(2x) - 2xe^{\frac{1}{2}}cos(2x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( -2e^{\frac{1}{2}}sin(2x) - 2xe^{\frac{1}{2}}cos(2x)\right)}{dx}\\=&\frac{-2*\frac{1}{2}*0sin(2x)}{e^{\frac{1}{2}}} - 2e^{\frac{1}{2}}cos(2x)*2 - 2e^{\frac{1}{2}}cos(2x) - \frac{2x*\frac{1}{2}*0cos(2x)}{e^{\frac{1}{2}}} - 2xe^{\frac{1}{2}}*-sin(2x)*2\\=& - 6e^{\frac{1}{2}}cos(2x) + 4xe^{\frac{1}{2}}sin(2x)\\ \end{split}\end{equation} \]

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