There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
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\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {e}^{(-x + 2)}{(2x + 1)}^{4}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = 16x^{4}{e}^{(-x + 2)} + 32x^{3}{e}^{(-x + 2)} + 24x^{2}{e}^{(-x + 2)} + 8x{e}^{(-x + 2)} + {e}^{(-x + 2)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( 16x^{4}{e}^{(-x + 2)} + 32x^{3}{e}^{(-x + 2)} + 24x^{2}{e}^{(-x + 2)} + 8x{e}^{(-x + 2)} + {e}^{(-x + 2)}\right)}{dx}\\=&16*4x^{3}{e}^{(-x + 2)} + 16x^{4}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 32*3x^{2}{e}^{(-x + 2)} + 32x^{3}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 24*2x{e}^{(-x + 2)} + 24x^{2}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 8{e}^{(-x + 2)} + 8x({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + ({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)}))\\=&32x^{3}{e}^{(-x + 2)} - 16x^{4}{e}^{(-x + 2)} + 72x^{2}{e}^{(-x + 2)} + 40x{e}^{(-x + 2)} + 7{e}^{(-x + 2)}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 32x^{3}{e}^{(-x + 2)} - 16x^{4}{e}^{(-x + 2)} + 72x^{2}{e}^{(-x + 2)} + 40x{e}^{(-x + 2)} + 7{e}^{(-x + 2)}\right)}{dx}\\=&32*3x^{2}{e}^{(-x + 2)} + 32x^{3}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) - 16*4x^{3}{e}^{(-x + 2)} - 16x^{4}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 72*2x{e}^{(-x + 2)} + 72x^{2}({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 40{e}^{(-x + 2)} + 40x({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)})) + 7({e}^{(-x + 2)}((-1 + 0)ln(e) + \frac{(-x + 2)(0)}{(e)}))\\=&24x^{2}{e}^{(-x + 2)} - 96x^{3}{e}^{(-x + 2)} + 16x^{4}{e}^{(-x + 2)} + 104x{e}^{(-x + 2)} + 33{e}^{(-x + 2)}\\ \end{split}\end{equation} \]

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