There are 1 questions in this calculation: for each question, the 2 derivative of y is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ ln(\frac{(1 + {y}^{4})}{(2 + {x}^{2})})\ with\ respect\ to\ y：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = ln(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( ln(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})\right)}{dy}\\=&\frac{((\frac{-(0 + 0)}{(x^{2} + 2)^{2}})y^{4} + \frac{4y^{3}}{(x^{2} + 2)} + (\frac{-(0 + 0)}{(x^{2} + 2)^{2}}))}{(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})}\\=&\frac{4y^{3}}{(x^{2} + 2)(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( \frac{4y^{3}}{(x^{2} + 2)(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})}\right)}{dy}\\=&\frac{4(\frac{-(0 + 0)}{(x^{2} + 2)^{2}})y^{3}}{(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})} + \frac{4(\frac{-((\frac{-(0 + 0)}{(x^{2} + 2)^{2}})y^{4} + \frac{4y^{3}}{(x^{2} + 2)} + (\frac{-(0 + 0)}{(x^{2} + 2)^{2}}))}{(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})^{2}})y^{3}}{(x^{2} + 2)} + \frac{4*3y^{2}}{(x^{2} + 2)(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})}\\=&\frac{-16y^{6}}{(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})^{2}(x^{2} + 2)^{2}} + \frac{12y^{2}}{(\frac{y^{4}}{(x^{2} + 2)} + \frac{1}{(x^{2} + 2)})(x^{2} + 2)}\\ \end{split}\end{equation} \]

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