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。\[ \begin{equation}\begin{split}【1/1】求函数-arctan((\frac{(\frac{({A}^{2}C{sin(x)}^{2})}{({A}^{2}{cos(x)}^{2} - {B}^{2})} - {D}^{2})}{(\frac{ACDtan(x)}{2} + \frac{(2ADcos(x)sin(x))}{({A}^{2}{cos(x)}^{2} - B)})})) 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = -arctan(\frac{A^{2}Csin^{2}(x)}{(A^{2}cos^{2}(x) - B^{2})(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})} - \frac{D^{2}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( -arctan(\frac{A^{2}Csin^{2}(x)}{(A^{2}cos^{2}(x) - B^{2})(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})} - \frac{D^{2}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})})\right)}{dx}\\=&-(\frac{(\frac{(\frac{-(A^{2}*-2cos(x)sin(x) + 0)}{(A^{2}cos^{2}(x) - B^{2})^{2}})A^{2}Csin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})} + \frac{(\frac{-(\frac{1}{2}ACDsec^{2}(x)(1) + 2(\frac{-(A^{2}*-2cos(x)sin(x) + 0)}{(A^{2}cos^{2}(x) - B)^{2}})ADsin(x)cos(x) + \frac{2ADcos(x)cos(x)}{(A^{2}cos^{2}(x) - B)} + \frac{2ADsin(x)*-sin(x)}{(A^{2}cos^{2}(x) - B)})}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}})A^{2}Csin^{2}(x)}{(A^{2}cos^{2}(x) - B^{2})} + \frac{A^{2}C*2sin(x)cos(x)}{(A^{2}cos^{2}(x) - B^{2})(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})} - (\frac{-(\frac{1}{2}ACDsec^{2}(x)(1) + 2(\frac{-(A^{2}*-2cos(x)sin(x) + 0)}{(A^{2}cos^{2}(x) - B)^{2}})ADsin(x)cos(x) + \frac{2ADcos(x)cos(x)}{(A^{2}cos^{2}(x) - B)} + \frac{2ADsin(x)*-sin(x)}{(A^{2}cos^{2}(x) - B)})}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}})D^{2} + 0)}{(1 + (\frac{A^{2}Csin^{2}(x)}{(A^{2}cos^{2}(x) - B^{2})(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})} - \frac{D^{2}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})})^{2})})\\=&\frac{-2A^{4}Csin^{3}(x)cos(x)}{(A^{2}cos^{2}(x) - B^{2})^{2}(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} + \frac{A^{3}C^{2}Dsin^{2}(x)sec^{2}(x)}{2(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} + \frac{4A^{5}CDsin^{4}(x)cos^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})(A^{2}cos^{2}(x) - B)^{2}(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} + \frac{2A^{3}CDsin^{2}(x)cos^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})(A^{2}cos^{2}(x) - B)(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} - \frac{2A^{3}CDsin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})(A^{2}cos^{2}(x) - B)(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} - \frac{2A^{2}Csin(x)cos(x)}{(A^{2}cos^{2}(x) - B^{2})(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} - \frac{ACD^{3}sec^{2}(x)}{2(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} - \frac{4A^{3}D^{3}sin^{2}(x)cos^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B)^{2}(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} - \frac{2AD^{3}cos^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B)(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)} + \frac{2AD^{3}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B)(\frac{A^{4}C^{2}sin^{4}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})^{2}} - \frac{2A^{2}CD^{2}sin^{2}(x)}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}(A^{2}cos^{2}(x) - B^{2})} + \frac{D^{4}}{(\frac{1}{2}ACDtan(x) + \frac{2ADsin(x)cos(x)}{(A^{2}cos^{2}(x) - B)})^{2}} + 1)}\\ \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!