本次共计算 1 个题目：每一题对 y 求 2 阶导数。
    注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数({(\frac{sin(2x + y)}{4})}^{2}){\frac{1}{(2x + y)}}^{2} - ({(\frac{sin(2x - y)}{4})}^{2}){\frac{1}{(2x - y)}}^{2} 关于 y 的 2 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{\frac{1}{0}sin^{2}(2x + y)}{(2x + y)^{2}}\right)}{dy}\\=&\frac{1}{0}(\frac{-2(0 + 1)}{(2x + y)^{3}})sin^{2}(2x + y) + \frac{\frac{1}{0}*2sin(2x + y)cos(2x + y)(0 + 1)}{(2x + y)^{2}}\\=&\frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{2sin(2x + y)cos(2x + y)}{0(2x + y)^{2}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{3}}\right)}{dy}\\=&\frac{2(\frac{-2(0 + 1)}{(2x + y)^{3}})sin(2x + y)cos(2x + y)}{0} + \frac{2cos(2x + y)(0 + 1)cos(2x + y)}{0(2x + y)^{2}} + \frac{2sin(2x + y)*-sin(2x + y)(0 + 1)}{0(2x + y)^{2}} - \frac{2(\frac{-3(0 + 1)}{(2x + y)^{4}})sin^{2}(2x + y)}{0} - \frac{2*2sin(2x + y)cos(2x + y)(0 + 1)}{0(2x + y)^{3}}\\=&\frac{2cos^{2}(2x + y)}{0(2x + y)^{2}} - \frac{4sin(2x + y)cos(2x + y)}{0(2x + y)^{3}} + \frac{6sin^{2}(2x + y)}{0(2x + y)^{4}} - \frac{2sin^{2}(2x + y)}{0(2x + y)^{2}}\\ \end{split}\end{equation} \]

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