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