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

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