本次共计算 1 个题目:每一题对 h 求 4 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数arcsin(h)h 关于 h 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = harcsin(h)\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( harcsin(h)\right)}{dh}\\=&arcsin(h) + h(\frac{(1)}{((1 - (h)^{2})^{\frac{1}{2}})})\\=&arcsin(h) + \frac{h}{(-h^{2} + 1)^{\frac{1}{2}}}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( arcsin(h) + \frac{h}{(-h^{2} + 1)^{\frac{1}{2}}}\right)}{dh}\\=&(\frac{(1)}{((1 - (h)^{2})^{\frac{1}{2}})}) + (\frac{\frac{-1}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{3}{2}}})h + \frac{1}{(-h^{2} + 1)^{\frac{1}{2}}}\\=&\frac{h^{2}}{(-h^{2} + 1)^{\frac{3}{2}}} + \frac{1}{(-h^{2} + 1)^{\frac{1}{2}}} + \frac{1}{(-h^{2} + 1)^{\frac{1}{2}}}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{h^{2}}{(-h^{2} + 1)^{\frac{3}{2}}} + \frac{1}{(-h^{2} + 1)^{\frac{1}{2}}} + \frac{1}{(-h^{2} + 1)^{\frac{1}{2}}}\right)}{dh}\\=&(\frac{\frac{-3}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{5}{2}}})h^{2} + \frac{2h}{(-h^{2} + 1)^{\frac{3}{2}}} + (\frac{\frac{-1}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{3}{2}}}) + (\frac{\frac{-1}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{3}{2}}})\\=&\frac{3h^{3}}{(-h^{2} + 1)^{\frac{5}{2}}} + \frac{4h}{(-h^{2} + 1)^{\frac{3}{2}}}\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{3h^{3}}{(-h^{2} + 1)^{\frac{5}{2}}} + \frac{4h}{(-h^{2} + 1)^{\frac{3}{2}}}\right)}{dh}\\=&3(\frac{\frac{-5}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{7}{2}}})h^{3} + \frac{3*3h^{2}}{(-h^{2} + 1)^{\frac{5}{2}}} + 4(\frac{\frac{-3}{2}(-2h + 0)}{(-h^{2} + 1)^{\frac{5}{2}}})h + \frac{4}{(-h^{2} + 1)^{\frac{3}{2}}}\\=&\frac{15h^{4}}{(-h^{2} + 1)^{\frac{7}{2}}} + \frac{21h^{2}}{(-h^{2} + 1)^{\frac{5}{2}}} + \frac{4}{(-h^{2} + 1)^{\frac{3}{2}}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!