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。\[ \begin{equation}\begin{split}【1/1】求函数sech(\frac{k(2x - t - i)}{(t - i)}) 关于 x 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})\right)}{dx}\\=&-sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})tanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})(2(\frac{-(0 + 0)}{(t - i)^{2}})kx + \frac{2k}{(t - i)} - (\frac{-(0 + 0)}{(t - i)^{2}})kt + 0 - (\frac{-(0 + 0)}{(t - i)^{2}})ki + 0)\\=&\frac{-2ktanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})}{(t - i)}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-2ktanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})}{(t - i)}\right)}{dx}\\=&-2(\frac{-(0 + 0)}{(t - i)^{2}})ktanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)}) - \frac{2ksech^{2}(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})(2(\frac{-(0 + 0)}{(t - i)^{2}})kx + \frac{2k}{(t - i)} - (\frac{-(0 + 0)}{(t - i)^{2}})kt + 0 - (\frac{-(0 + 0)}{(t - i)^{2}})ki + 0)sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})}{(t - i)} - \frac{2ktanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})*-sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})tanh(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})(2(\frac{-(0 + 0)}{(t - i)^{2}})kx + \frac{2k}{(t - i)} - (\frac{-(0 + 0)}{(t - i)^{2}})kt + 0 - (\frac{-(0 + 0)}{(t - i)^{2}})ki + 0)}{(t - i)}\\=& - \frac{4k^{2}sech^{3}(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})}{(t - i)^{2}} + \frac{4k^{2}tanh^{2}(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})sech(\frac{2kx}{(t - i)} - \frac{kt}{(t - i)} - \frac{ki}{(t - i)})}{(t - i)^{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!