本次共计算 1 个题目:每一题对 x 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数{(h(\frac{x}{A} - \frac{sin(\frac{2Bx}{A})}{(2B)}))}^{2} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{h^{2}x^{2}}{A^{2}} - \frac{h^{2}xsin(\frac{2Bx}{A})}{AB} + \frac{\frac{1}{4}h^{2}sin^{2}(\frac{2Bx}{A})}{B^{2}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{h^{2}x^{2}}{A^{2}} - \frac{h^{2}xsin(\frac{2Bx}{A})}{AB} + \frac{\frac{1}{4}h^{2}sin^{2}(\frac{2Bx}{A})}{B^{2}}\right)}{dx}\\=&\frac{h^{2}*2x}{A^{2}} - \frac{h^{2}sin(\frac{2Bx}{A})}{AB} - \frac{h^{2}xcos(\frac{2Bx}{A})*2B}{ABA} + \frac{\frac{1}{4}h^{2}*2sin(\frac{2Bx}{A})cos(\frac{2Bx}{A})*2B}{B^{2}A}\\=& - \frac{2h^{2}xcos(\frac{2Bx}{A})}{A^{2}} + \frac{h^{2}sin(\frac{2Bx}{A})cos(\frac{2Bx}{A})}{AB} + \frac{2h^{2}x}{A^{2}} - \frac{h^{2}sin(\frac{2Bx}{A})}{AB}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!