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

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