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

你的问题在这里没有得到解决？请到 热门难题 里面看看吧！