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

Your problem has not been solved here? Please take a look at the  hot problems ！