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

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