本次共计算 1 个题目：每一题对 t 求 3 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数\frac{dnt{4}^{(e^{\frac{1}{5}}t + 2t)}}{d} 关于 t 的 3 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = nt{4}^{(te^{\frac{1}{5}} + 2t)}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( nt{4}^{(te^{\frac{1}{5}} + 2t)}\right)}{dt}\\=&n{4}^{(te^{\frac{1}{5}} + 2t)} + nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))\\=&n{4}^{(te^{\frac{1}{5}} + 2t)} + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln(4) + 2nt{4}^{(te^{\frac{1}{5}} + 2t)}ln(4)\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( n{4}^{(te^{\frac{1}{5}} + 2t)} + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln(4) + 2nt{4}^{(te^{\frac{1}{5}} + 2t)}ln(4)\right)}{dt}\\=&n({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)})) + n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln(4) + nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))e^{\frac{1}{5}}ln(4) + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*0ln(4) + \frac{nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*0}{(4)} + 2n{4}^{(te^{\frac{1}{5}} + 2t)}ln(4) + 2nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))ln(4) + \frac{2nt{4}^{(te^{\frac{1}{5}} + 2t)}*0}{(4)}\\=&2n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln(4) + 4n{4}^{(te^{\frac{1}{5}} + 2t)}ln(4) + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}ln^{2}(4) + 4nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln^{2}(4) + 4nt{4}^{(te^{\frac{1}{5}} + 2t)}ln^{2}(4)\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( 2n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln(4) + 4n{4}^{(te^{\frac{1}{5}} + 2t)}ln(4) + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}ln^{2}(4) + 4nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln^{2}(4) + 4nt{4}^{(te^{\frac{1}{5}} + 2t)}ln^{2}(4)\right)}{dt}\\=&2n({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))e^{\frac{1}{5}}ln(4) + 2n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*0ln(4) + \frac{2n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*0}{(4)} + 4n({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))ln(4) + \frac{4n{4}^{(te^{\frac{1}{5}} + 2t)}*0}{(4)} + n{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}ln^{2}(4) + nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))e^{{\frac{1}{5}}*{2}}ln^{2}(4) + nt{4}^{(te^{\frac{1}{5}} + 2t)}*2e^{\frac{1}{5}}e^{\frac{1}{5}}*0ln^{2}(4) + \frac{nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}*2ln(4)*0}{(4)} + 4n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln^{2}(4) + 4nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))e^{\frac{1}{5}}ln^{2}(4) + 4nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*0ln^{2}(4) + \frac{4nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}*2ln(4)*0}{(4)} + 4n{4}^{(te^{\frac{1}{5}} + 2t)}ln^{2}(4) + 4nt({4}^{(te^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}} + te^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(te^{\frac{1}{5}} + 2t)(0)}{(4)}))ln^{2}(4) + \frac{4nt{4}^{(te^{\frac{1}{5}} + 2t)}*2ln(4)*0}{(4)}\\=&3n{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}ln^{2}(4) + 12n{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln^{2}(4) + 12n{4}^{(te^{\frac{1}{5}} + 2t)}ln^{2}(4) + nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{3}}ln^{3}(4) + 6nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{{\frac{1}{5}}*{2}}ln^{3}(4) + 12nt{4}^{(te^{\frac{1}{5}} + 2t)}e^{\frac{1}{5}}ln^{3}(4) + 8nt{4}^{(te^{\frac{1}{5}} + 2t)}ln^{3}(4)\\ \end{split}\end{equation} \]

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