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