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