There are 1 questions in this calculation: for each question, the 1 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ ({x}^{2} + 2{y}^{2} + 3{z}^{2}){e}^{(-(x*2 + y*2 + z*2))}\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = x^{2}{e}^{(-2x - 2y - 2z)} + 2y^{2}{e}^{(-2x - 2y - 2z)} + 3z^{2}{e}^{(-2x - 2y - 2z)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( x^{2}{e}^{(-2x - 2y - 2z)} + 2y^{2}{e}^{(-2x - 2y - 2z)} + 3z^{2}{e}^{(-2x - 2y - 2z)}\right)}{dx}\\=&2x{e}^{(-2x - 2y - 2z)} + x^{2}({e}^{(-2x - 2y - 2z)}((-2 + 0 + 0)ln(e) + \frac{(-2x - 2y - 2z)(0)}{(e)})) + 2y^{2}({e}^{(-2x - 2y - 2z)}((-2 + 0 + 0)ln(e) + \frac{(-2x - 2y - 2z)(0)}{(e)})) + 3z^{2}({e}^{(-2x - 2y - 2z)}((-2 + 0 + 0)ln(e) + \frac{(-2x - 2y - 2z)(0)}{(e)}))\\=&2x{e}^{(-2x - 2y - 2z)} - 2x^{2}{e}^{(-2x - 2y - 2z)} - 4y^{2}{e}^{(-2x - 2y - 2z)} - 6z^{2}{e}^{(-2x - 2y - 2z)}\\ \end{split}\end{equation} \]

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