There are 1 questions in this calculation: for each question, the 4 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ 4th\ derivative\ of\ function\ a + bx - {e}^{(2x)}((f)cos(x) + (g)sin(x))\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = a + bx - f{e}^{(2x)}cos(x) - g{e}^{(2x)}sin(x)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( a + bx - f{e}^{(2x)}cos(x) - g{e}^{(2x)}sin(x)\right)}{dx}\\=&0 + b - f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - f{e}^{(2x)}*-sin(x) - g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) - g{e}^{(2x)}cos(x)\\=&b - 2f{e}^{(2x)}cos(x) + f{e}^{(2x)}sin(x) - 2g{e}^{(2x)}sin(x) - g{e}^{(2x)}cos(x)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( b - 2f{e}^{(2x)}cos(x) + f{e}^{(2x)}sin(x) - 2g{e}^{(2x)}sin(x) - g{e}^{(2x)}cos(x)\right)}{dx}\\=&0 - 2f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - 2f{e}^{(2x)}*-sin(x) + f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) + f{e}^{(2x)}cos(x) - 2g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) - 2g{e}^{(2x)}cos(x) - g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - g{e}^{(2x)}*-sin(x)\\=& - 3f{e}^{(2x)}cos(x) + 4f{e}^{(2x)}sin(x) - 3g{e}^{(2x)}sin(x) - 4g{e}^{(2x)}cos(x)\\\\ &\color{blue}{The\ third\ derivative\ of\ function:} \\&\frac{d\left( - 3f{e}^{(2x)}cos(x) + 4f{e}^{(2x)}sin(x) - 3g{e}^{(2x)}sin(x) - 4g{e}^{(2x)}cos(x)\right)}{dx}\\=& - 3f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - 3f{e}^{(2x)}*-sin(x) + 4f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) + 4f{e}^{(2x)}cos(x) - 3g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) - 3g{e}^{(2x)}cos(x) - 4g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - 4g{e}^{(2x)}*-sin(x)\\=& - 2f{e}^{(2x)}cos(x) + 11f{e}^{(2x)}sin(x) - 2g{e}^{(2x)}sin(x) - 11g{e}^{(2x)}cos(x)\\\\ &\color{blue}{The\ 4th\ derivative\ of\ function:} \\&\frac{d\left( - 2f{e}^{(2x)}cos(x) + 11f{e}^{(2x)}sin(x) - 2g{e}^{(2x)}sin(x) - 11g{e}^{(2x)}cos(x)\right)}{dx}\\=& - 2f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - 2f{e}^{(2x)}*-sin(x) + 11f({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) + 11f{e}^{(2x)}cos(x) - 2g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))sin(x) - 2g{e}^{(2x)}cos(x) - 11g({e}^{(2x)}((2)ln(e) + \frac{(2x)(0)}{(e)}))cos(x) - 11g{e}^{(2x)}*-sin(x)\\=&7f{e}^{(2x)}cos(x) + 24f{e}^{(2x)}sin(x) + 7g{e}^{(2x)}sin(x) - 24g{e}^{(2x)}cos(x)\\ \end{split}\end{equation} \]

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