本次共计算 1 个题目:每一题对 x 求 2 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数\frac{(b + 1)x}{(b + {x}^{b}*2)} 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{bx}{(b + 2{x}^{b})} + \frac{x}{(b + 2{x}^{b})}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{bx}{(b + 2{x}^{b})} + \frac{x}{(b + 2{x}^{b})}\right)}{dx}\\=&(\frac{-(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{2}})bx + \frac{b}{(b + 2{x}^{b})} + (\frac{-(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{2}})x + \frac{1}{(b + 2{x}^{b})}\\=&\frac{-2b^{2}{x}^{b}}{(b + 2{x}^{b})^{2}} - \frac{2b{x}^{b}}{(b + 2{x}^{b})^{2}} + \frac{b}{(b + 2{x}^{b})} + \frac{1}{(b + 2{x}^{b})}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-2b^{2}{x}^{b}}{(b + 2{x}^{b})^{2}} - \frac{2b{x}^{b}}{(b + 2{x}^{b})^{2}} + \frac{b}{(b + 2{x}^{b})} + \frac{1}{(b + 2{x}^{b})}\right)}{dx}\\=&-2(\frac{-2(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{3}})b^{2}{x}^{b} - \frac{2b^{2}({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)}))}{(b + 2{x}^{b})^{2}} - 2(\frac{-2(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{3}})b{x}^{b} - \frac{2b({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)}))}{(b + 2{x}^{b})^{2}} + (\frac{-(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{2}})b + 0 + (\frac{-(0 + 2({x}^{b}((0)ln(x) + \frac{(b)(1)}{(x)})))}{(b + 2{x}^{b})^{2}})\\=&\frac{8b^{3}{x}^{(2b)}}{(b + 2{x}^{b})^{3}x} - \frac{2b^{3}{x}^{b}}{(b + 2{x}^{b})^{2}x} + \frac{8b^{2}{x}^{(2b)}}{(b + 2{x}^{b})^{3}x} - \frac{4b^{2}{x}^{b}}{(b + 2{x}^{b})^{2}x} - \frac{2b{x}^{b}}{(b + 2{x}^{b})^{2}x}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!