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

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