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

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