本次共计算 1 个题目：每一题对 x 求 2 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数(\frac{6}{4}){e}^{(\frac{-({(x - 5)}^{2})}{32})} 关于 x 的 2 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{3}{2}{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{3}{2}{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}\right)}{dx}\\=&\frac{3}{2}({e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}((\frac{-1}{32}*2x + \frac{5}{16} + 0)ln(e) + \frac{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})(0)}{(e)}))\\=&\frac{-3x{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{32} + \frac{15{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{32}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{-3x{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{32} + \frac{15{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{32}\right)}{dx}\\=&\frac{-3{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{32} - \frac{3x({e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}((\frac{-1}{32}*2x + \frac{5}{16} + 0)ln(e) + \frac{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})(0)}{(e)}))}{32} + \frac{15({e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}((\frac{-1}{32}*2x + \frac{5}{16} + 0)ln(e) + \frac{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})(0)}{(e)}))}{32}\\=&\frac{27{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{512} + \frac{3x^{2}{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{512} - \frac{15x{e}^{(\frac{-1}{32}x^{2} + \frac{5}{16}x - \frac{25}{32})}}{256}\\ \end{split}\end{equation} \]

你的问题在这里没有得到解决？请到 热门难题 里面看看吧！