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。
\[ \begin{equation}\begin{split}【1/1】求函数(\frac{b}{a}){e}^{(\frac{-({(x - c)}^{2})({a}^{2})}{2})} 关于 x 的 2 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{b{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{a}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{b{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{a}\right)}{dx}\\=&\frac{b({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{a}\\=&-abx{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} + abc{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( -abx{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} + abc{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}\right)}{dx}\\=&-ab{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} - abx({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)})) + abc({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))\\=&-ab{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} + a^{3}bx^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} - 2a^{3}bcx{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})} + a^{3}bc^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}\\ \end{split}\end{equation} \]

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