本次共计算 1 个题目:每一题对 t 求 2 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数x + ln(tan(th(e^{t})a) + sec(th(e^{t})a)) 关于 t 的 2 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = x + ln(tan(ath(e^{t})) + sec(ath(e^{t})))\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( x + ln(tan(ath(e^{t})) + sec(ath(e^{t})))\right)}{dt}\\=&0 + \frac{(sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t}) + sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t})}{(tan(ath(e^{t})) + sec(ath(e^{t})))}\\=&\frac{-ae^{t}th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}tan(ath(e^{t}))sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-ae^{t}th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}tan(ath(e^{t}))sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))}\right)}{dt}\\=&-(\frac{-(sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t}) + sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t})}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}})ae^{t}th^{2}(e^{t})sec^{2}(ath(e^{t})) - \frac{ae^{t}th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}*2th(e^{t})(1 - th^{2}(e^{t}))e^{t}sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}th^{2}(e^{t})*2sec^{2}(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t}}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + (\frac{-(sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t}) + sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t})}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}})ae^{t}sec^{2}(ath(e^{t})) + \frac{ae^{t}sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}*2sec^{2}(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t}}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + (\frac{-(sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t}) + sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t})}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}})ae^{t}tan(ath(e^{t}))sec(ath(e^{t})) + \frac{ae^{t}tan(ath(e^{t}))sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{ae^{t}tan(ath(e^{t}))sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t}}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - (\frac{-(sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t}) + sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t})}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}})ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t})) - \frac{ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}sec^{2}(ath(e^{t}))(a(1 - th^{2}(e^{t}))e^{t})th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}tan(ath(e^{t}))*2th(e^{t})(1 - th^{2}(e^{t}))e^{t}sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))tan(ath(e^{t}))a(1 - th^{2}(e^{t}))e^{t}}{(tan(ath(e^{t})) + sec(ath(e^{t})))}\\=&\frac{-a^{2}e^{{t}*{2}}th^{4}(e^{t})sec^{4}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} + \frac{2a^{2}e^{{t}*{2}}th^{2}(e^{t})sec^{4}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} - \frac{2a^{2}e^{{t}*{2}}tan(ath(e^{t}))th^{4}(e^{t})sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} + \frac{4a^{2}e^{{t}*{2}}tan(ath(e^{t}))th^{2}(e^{t})sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} - \frac{ae^{t}th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{2ae^{{t}*{2}}th(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{2ae^{{t}*{2}}th^{3}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{4a^{2}e^{{t}*{2}}tan(ath(e^{t}))th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{2a^{2}e^{{t}*{2}}tan(ath(e^{t}))th^{4}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{a^{2}e^{{t}*{2}}sec^{4}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} - \frac{2a^{2}e^{{t}*{2}}tan(ath(e^{t}))sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} + \frac{ae^{t}sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{2a^{2}e^{{t}*{2}}tan(ath(e^{t}))sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{2a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))th^{2}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} - \frac{a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} + \frac{ae^{t}tan(ath(e^{t}))sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{2a^{2}e^{{t}*{2}}th^{2}(e^{t})sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{a^{2}e^{{t}*{2}}sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{2a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))th^{4}(e^{t})sec^{2}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))^{2}} - \frac{ae^{t}tan(ath(e^{t}))th^{2}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{a^{2}e^{{t}*{2}}th^{4}(e^{t})sec^{3}(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} - \frac{2ae^{{t}*{2}}tan(ath(e^{t}))th(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{2ae^{{t}*{2}}tan(ath(e^{t}))th^{3}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))} + \frac{a^{2}e^{{t}*{2}}tan^{2}(ath(e^{t}))th^{4}(e^{t})sec(ath(e^{t}))}{(tan(ath(e^{t})) + sec(ath(e^{t})))}\\ \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!