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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