There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ (\frac{1}{(ab)}){e}^{(\frac{-({(x - c)}^{2})({a}^{2})}{2})}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{ab}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{ab}\right)}{dx}\\=&\frac{({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{ab}\\=&\frac{-ax{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{ac{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-ax{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{ac{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\right)}{dx}\\=&\frac{-a{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} - \frac{ax({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{b} + \frac{ac({e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}((\frac{-1}{2}a^{2}*2x + a^{2}c + 0)ln(e) + \frac{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})(0)}{(e)}))}{b}\\=&\frac{-a{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{a^{3}x^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} - \frac{2a^{3}cx{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b} + \frac{a^{3}c^{2}{e}^{(\frac{-1}{2}a^{2}x^{2} + a^{2}cx - \frac{1}{2}a^{2}c^{2})}}{b}\\ \end{split}\end{equation} \]

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