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。\[ \begin{equation}\begin{split}【1/1】求函数\frac{dnt{4}^{(e^{\frac{1}{5}} + 2t)}}{d} 关于 t 的 3 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = nt{4}^{(e^{\frac{1}{5}} + 2t)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( nt{4}^{(e^{\frac{1}{5}} + 2t)}\right)}{dt}\\=&n{4}^{(e^{\frac{1}{5}} + 2t)} + nt({4}^{(e^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(e^{\frac{1}{5}} + 2t)(0)}{(4)}))\\=&n{4}^{(e^{\frac{1}{5}} + 2t)} + 2nt{4}^{(e^{\frac{1}{5}} + 2t)}ln(4)\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( n{4}^{(e^{\frac{1}{5}} + 2t)} + 2nt{4}^{(e^{\frac{1}{5}} + 2t)}ln(4)\right)}{dt}\\=&n({4}^{(e^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(e^{\frac{1}{5}} + 2t)(0)}{(4)})) + 2n{4}^{(e^{\frac{1}{5}} + 2t)}ln(4) + 2nt({4}^{(e^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(e^{\frac{1}{5}} + 2t)(0)}{(4)}))ln(4) + \frac{2nt{4}^{(e^{\frac{1}{5}} + 2t)}*0}{(4)}\\=&4n{4}^{(e^{\frac{1}{5}} + 2t)}ln(4) + 4nt{4}^{(e^{\frac{1}{5}} + 2t)}ln^{2}(4)\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( 4n{4}^{(e^{\frac{1}{5}} + 2t)}ln(4) + 4nt{4}^{(e^{\frac{1}{5}} + 2t)}ln^{2}(4)\right)}{dt}\\=&4n({4}^{(e^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(e^{\frac{1}{5}} + 2t)(0)}{(4)}))ln(4) + \frac{4n{4}^{(e^{\frac{1}{5}} + 2t)}*0}{(4)} + 4n{4}^{(e^{\frac{1}{5}} + 2t)}ln^{2}(4) + 4nt({4}^{(e^{\frac{1}{5}} + 2t)}((e^{\frac{1}{5}}*0 + 2)ln(4) + \frac{(e^{\frac{1}{5}} + 2t)(0)}{(4)}))ln^{2}(4) + \frac{4nt{4}^{(e^{\frac{1}{5}} + 2t)}*2ln(4)*0}{(4)}\\=&12n{4}^{(e^{\frac{1}{5}} + 2t)}ln^{2}(4) + 8nt{4}^{(e^{\frac{1}{5}} + 2t)}ln^{3}(4)\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!