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

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