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