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。\[ \begin{equation}\begin{split}【1/1】求函数({\frac{1}{({a}^{2} - {b}^{2})}}^{\frac{1}{2}})arcsin(\frac{(arcsin(x) + b)}{(a + bsin(x))}) 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{arcsin(\frac{arcsin(x)}{(a + bsin(x))} + \frac{b}{(a + bsin(x))})}{(a^{2} - b^{2})^{\frac{1}{2}}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{arcsin(\frac{arcsin(x)}{(a + bsin(x))} + \frac{b}{(a + bsin(x))})}{(a^{2} - b^{2})^{\frac{1}{2}}}\right)}{dx}\\=&(\frac{\frac{-1}{2}(0 + 0)}{(a^{2} - b^{2})^{\frac{3}{2}}})arcsin(\frac{arcsin(x)}{(a + bsin(x))} + \frac{b}{(a + bsin(x))}) + \frac{(\frac{((\frac{-(0 + bcos(x))}{(a + bsin(x))^{2}})arcsin(x) + \frac{(\frac{(1)}{((1 - (x)^{2})^{\frac{1}{2}})})}{(a + bsin(x))} + (\frac{-(0 + bcos(x))}{(a + bsin(x))^{2}})b + 0)}{((1 - (\frac{arcsin(x)}{(a + bsin(x))} + \frac{b}{(a + bsin(x))})^{2})^{\frac{1}{2}})})}{(a^{2} - b^{2})^{\frac{1}{2}}}\\=&\frac{-bcos(x)arcsin(x)}{(\frac{-arcsin^{2}(x)}{(a + bsin(x))^{2}} - \frac{2barcsin(x)}{(a + bsin(x))^{2}} - \frac{b^{2}}{(a + bsin(x))^{2}} + 1)^{\frac{1}{2}}(a^{2} - b^{2})^{\frac{1}{2}}(a + bsin(x))^{2}} - \frac{b^{2}cos(x)}{(\frac{-arcsin^{2}(x)}{(a + bsin(x))^{2}} - \frac{2barcsin(x)}{(a + bsin(x))^{2}} - \frac{b^{2}}{(a + bsin(x))^{2}} + 1)^{\frac{1}{2}}(a^{2} - b^{2})^{\frac{1}{2}}(a + bsin(x))^{2}} + \frac{1}{(\frac{-arcsin^{2}(x)}{(a + bsin(x))^{2}} - \frac{2barcsin(x)}{(a + bsin(x))^{2}} - \frac{b^{2}}{(a + bsin(x))^{2}} + 1)^{\frac{1}{2}}(a^{2} - b^{2})^{\frac{1}{2}}(-x^{2} + 1)^{\frac{1}{2}}(a + bsin(x))}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!