There are 1 questions in this calculation: for each question, the 2 derivative of t is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ sin(\frac{1}{(1 + {t}^{2})})\ with\ respect\ to\ t：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = sin(\frac{1}{(t^{2} + 1)})\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( sin(\frac{1}{(t^{2} + 1)})\right)}{dt}\\=&cos(\frac{1}{(t^{2} + 1)})(\frac{-(2t + 0)}{(t^{2} + 1)^{2}})\\=&\frac{-2tcos(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{2}}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{-2tcos(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{2}}\right)}{dt}\\=&-2(\frac{-2(2t + 0)}{(t^{2} + 1)^{3}})tcos(\frac{1}{(t^{2} + 1)}) - \frac{2cos(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{2}} - \frac{2t*-sin(\frac{1}{(t^{2} + 1)})(\frac{-(2t + 0)}{(t^{2} + 1)^{2}})}{(t^{2} + 1)^{2}}\\=&\frac{8t^{2}cos(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{3}} - \frac{2cos(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{2}} - \frac{4t^{2}sin(\frac{1}{(t^{2} + 1)})}{(t^{2} + 1)^{4}}\\ \end{split}\end{equation} \]

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