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