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