There are 1 questions in this calculation: for each question, the 2 derivative of x is calculated.
    Note that variables are case sensitive.
\[ \begin{equation}\begin{split}[1/1]Find\ the\ second\ derivative\ of\ function\ {x}^{4} + 3{{x}^{3}}^{y} + {y}^{2} - 1\ with\ respect\ to\ x：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\Solution:&\\ &Primitive\ function\ = x^{4} + 3{x^{3}}^{y} + y^{2} - 1\\&\color{blue}{The\ first\ derivative\ function：}\\&\frac{d\left( x^{4} + 3{x^{3}}^{y} + y^{2} - 1\right)}{dx}\\=&4x^{3} + 3({x^{3}}^{y}((0)ln(x^{3}) + \frac{(y)(3x^{2})}{(x^{3})})) + 0 + 0\\=&4x^{3} + \frac{9y{x^{3}}^{y}}{x}\\\\ &\color{blue}{The\ second\ derivative\ of\ function:} \\&\frac{d\left( 4x^{3} + \frac{9y{x^{3}}^{y}}{x}\right)}{dx}\\=&4*3x^{2} + \frac{9y*-{x^{3}}^{y}}{x^{2}} + \frac{9y({x^{3}}^{y}((0)ln(x^{3}) + \frac{(y)(3x^{2})}{(x^{3})}))}{x}\\=&12x^{2} - \frac{9y{x^{3}}^{y}}{x^{2}} + \frac{27y^{2}{x^{3}}^{y}}{x^{2}}\\ \end{split}\end{equation} \]

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