本次共计算 1 个题目:每一题对 a 求 1 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数(2d{\frac{1}{(4d(r + p) - {a}^{2})}}^{\frac{1}{2}})arctan(\frac{{(4d(r + p) - {a}^{2})}^{\frac{1}{2}}}{a}) 关于 a 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{2darctan(\frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}}{a})}{(4dr + 4dp - a^{2})^{\frac{1}{2}}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{2darctan(\frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}}{a})}{(4dr + 4dp - a^{2})^{\frac{1}{2}}}\right)}{da}\\=&2(\frac{\frac{-1}{2}(0 + 0 - 2a)}{(4dr + 4dp - a^{2})^{\frac{3}{2}}})darctan(\frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}}{a}) + \frac{2d(\frac{(\frac{(\frac{\frac{1}{2}(0 + 0 - 2a)}{(4dr + 4dp - a^{2})^{\frac{1}{2}}})}{a} + \frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}*-1}{a^{2}})}{(1 + (\frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}}{a})^{2})})}{(4dr + 4dp - a^{2})^{\frac{1}{2}}}\\=&\frac{2daarctan(\frac{(4dr + 4dp - a^{2})^{\frac{1}{2}}}{a})}{(4dr + 4dp - a^{2})^{\frac{3}{2}}} - \frac{2d}{(\frac{4dr}{a^{2}} + \frac{4dp}{a^{2}})a^{2}} - \frac{2d}{(4dr + 4dp - a^{2})(\frac{4dr}{a^{2}} + \frac{4dp}{a^{2}})}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!