There are 1 questions in this calculation: for each question, the 2 derivative of r is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {e}^{(-30{r}^{2})}(1 + rsin(5m))cos(t)\ with\ respect\ to\ r：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = {e}^{(-30r^{2})}cos(t) + r{e}^{(-30r^{2})}sin(5m)cos(t)\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( {e}^{(-30r^{2})}cos(t) + r{e}^{(-30r^{2})}sin(5m)cos(t)\right)}{dr}\\=&({e}^{(-30r^{2})}((-30*2r)ln(e) + \frac{(-30r^{2})(0)}{(e)}))cos(t) + {e}^{(-30r^{2})}*-sin(t)*0 + {e}^{(-30r^{2})}sin(5m)cos(t) + r({e}^{(-30r^{2})}((-30*2r)ln(e) + \frac{(-30r^{2})(0)}{(e)}))sin(5m)cos(t) + r{e}^{(-30r^{2})}cos(5m)*0cos(t) + r{e}^{(-30r^{2})}sin(5m)*-sin(t)*0\\=&-60r{e}^{(-30r^{2})}cos(t) + {e}^{(-30r^{2})}sin(5m)cos(t) - 60r^{2}{e}^{(-30r^{2})}sin(5m)cos(t)\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( -60r{e}^{(-30r^{2})}cos(t) + {e}^{(-30r^{2})}sin(5m)cos(t) - 60r^{2}{e}^{(-30r^{2})}sin(5m)cos(t)\right)}{dr}\\=&-60{e}^{(-30r^{2})}cos(t) - 60r({e}^{(-30r^{2})}((-30*2r)ln(e) + \frac{(-30r^{2})(0)}{(e)}))cos(t) - 60r{e}^{(-30r^{2})}*-sin(t)*0 + ({e}^{(-30r^{2})}((-30*2r)ln(e) + \frac{(-30r^{2})(0)}{(e)}))sin(5m)cos(t) + {e}^{(-30r^{2})}cos(5m)*0cos(t) + {e}^{(-30r^{2})}sin(5m)*-sin(t)*0 - 60*2r{e}^{(-30r^{2})}sin(5m)cos(t) - 60r^{2}({e}^{(-30r^{2})}((-30*2r)ln(e) + \frac{(-30r^{2})(0)}{(e)}))sin(5m)cos(t) - 60r^{2}{e}^{(-30r^{2})}cos(5m)*0cos(t) - 60r^{2}{e}^{(-30r^{2})}sin(5m)*-sin(t)*0\\=&3600r^{2}{e}^{(-30r^{2})}cos(t) - 180r{e}^{(-30r^{2})}sin(5m)cos(t) - 60{e}^{(-30r^{2})}cos(t) + 3600r^{3}{e}^{(-30r^{2})}sin(5m)cos(t)\\ \end{split}\end{equation} \]

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