There are 1 questions in this calculation: for each question, the 1 derivative of y is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ first\ derivative\ of\ function\ \frac{(-2x{y}^{4} - 2x)}{({x}^{2}{y}^{4} + 2{y}^{4} + {x}^{2} + 2)}\ with\ respect\ to\ y：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = \frac{-2xy^{4}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)} - \frac{2x}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)}\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( \frac{-2xy^{4}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)} - \frac{2x}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)}\right)}{dy}\\=&-2(\frac{-(x^{2}*4y^{3} + 2*4y^{3} + 0 + 0)}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}})xy^{4} - \frac{2x*4y^{3}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)} - 2(\frac{-(x^{2}*4y^{3} + 2*4y^{3} + 0 + 0)}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}})x + 0\\=&\frac{8x^{3}y^{7}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}} + \frac{16xy^{7}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}} - \frac{8xy^{3}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)} + \frac{8x^{3}y^{3}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}} + \frac{16xy^{3}}{(x^{2}y^{4} + 2y^{4} + x^{2} + 2)^{2}}\\ \end{split}\end{equation} \]

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