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。\[ \begin{equation}\begin{split}【1/1】求函数(\frac{1}{(ab)}){e}^{(\frac{-({(x - c)}^{2})({a}^{2})}{2})} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{ab}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{ab}\right)}{dx}\\=&\frac{({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{ab}\\=&\frac{-ax{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{ac{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-ax{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{ac{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\right)}{dx}\\=&\frac{-a{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} - \frac{ax({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{b} + \frac{ac({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{b}\\=&\frac{-a{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{a^{3}x^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} - \frac{2a^{3}cx{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{a^{3}c^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!