本次共计算 2 个题目：每一题对 x 求 1 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/2】求函数\frac{5{e}^{(\frac{4x}{5})}(5sin(2x) + 2cos(2x))}{58} + C 关于 x 的 1 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{25}{58}{e}^{(\frac{4}{5}x)}sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x) + C\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{25}{58}{e}^{(\frac{4}{5}x)}sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x) + C\right)}{dx}\\=&\frac{25}{58}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))sin(2x) + \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x)*2 + \frac{5}{29}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))cos(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}*-sin(2x)*2 + 0\\=&{e}^{(\frac{4}{5}x)}cos(2x)\\ \end{split}\end{equation} \]

\[ \begin{equation}\begin{split}【2/2】求函数\frac{5{e}^{(\frac{4x}{5})}(2sin(2x) - 5cos(2x))}{58} + C 关于 x 的 1 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{5}{29}{e}^{(\frac{4}{5}x)}sin(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x) + C\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{5}{29}{e}^{(\frac{4}{5}x)}sin(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}cos(2x) + C\right)}{dx}\\=&\frac{5}{29}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))sin(2x) + \frac{5}{29}{e}^{(\frac{4}{5}x)}cos(2x)*2 - \frac{25}{58}({e}^{(\frac{4}{5}x)}((\frac{4}{5})ln(e) + \frac{(\frac{4}{5}x)(0)}{(e)}))cos(2x) - \frac{25}{58}{e}^{(\frac{4}{5}x)}*-sin(2x)*2 + 0\\=&{e}^{(\frac{4}{5}x)}sin(2x)\\ \end{split}\end{equation} \]

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