求导函数：
输入一个原函数（即需要求导的函数），然后设置需要求导的变量和求导的阶数，点击“下一步”按钮，即可获得该函数相应阶数的导函数。
注意，输入的函数支持数学函数和其它常量。
输入一个原函数（即需要求导的函数），然后设置需要求导的变量和求导的阶数，点击“下一步”按钮，即可获得该函数相应阶数的导函数。
注意，输入的函数支持数学函数和其它常量。
本次共计算 1 个题目：每一题对 r 求 3 阶导数。
注意，变量是区分大小写的。
\[ \begin{equation}\begin{split}【1/1】求函数(-PA{r}^{3} + (PA + C){r}^{2} - (C + 2mn{P}^{2}A)r){\frac{1}{(2rPA - C)}}^{2} 关于 r 的 3 阶导数：\\\end{split}\end{equation} \]\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{-PAr^{3}}{(2PAr - C)^{2}} + \frac{PAr^{2}}{(2PAr - C)^{2}} + \frac{Cr^{2}}{(2PAr - C)^{2}} - \frac{Cr}{(2PAr - C)^{2}} - \frac{2P^{2}Amnr}{(2PAr - C)^{2}}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( \frac{-PAr^{3}}{(2PAr - C)^{2}} + \frac{PAr^{2}}{(2PAr - C)^{2}} + \frac{Cr^{2}}{(2PAr - C)^{2}} - \frac{Cr}{(2PAr - C)^{2}} - \frac{2P^{2}Amnr}{(2PAr - C)^{2}}\right)}{dr}\\=&-(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PAr^{3} - \frac{PA*3r^{2}}{(2PAr - C)^{2}} + (\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PAr^{2} + \frac{PA*2r}{(2PAr - C)^{2}} + (\frac{-2(2PA + 0)}{(2PAr - C)^{3}})Cr^{2} + \frac{C*2r}{(2PAr - C)^{2}} - (\frac{-2(2PA + 0)}{(2PAr - C)^{3}})Cr - \frac{C}{(2PAr - C)^{2}} - 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})P^{2}Amnr - \frac{2P^{2}Amn}{(2PAr - C)^{2}}\\=&\frac{4P^{2}A^{2}r^{3}}{(2PAr - C)^{3}} - \frac{3PAr^{2}}{(2PAr - C)^{2}} - \frac{4P^{2}A^{2}r^{2}}{(2PAr - C)^{3}} + \frac{2PAr}{(2PAr - C)^{2}} - \frac{4PACr^{2}}{(2PAr - C)^{3}} + \frac{2Cr}{(2PAr - C)^{2}} + \frac{4PACr}{(2PAr - C)^{3}} - \frac{C}{(2PAr - C)^{2}} + \frac{8P^{3}A^{2}mnr}{(2PAr - C)^{3}} - \frac{2P^{2}Amn}{(2PAr - C)^{2}}\\\\ &\color{blue}{函数的第 2 阶导数：} \\&\frac{d\left( \frac{4P^{2}A^{2}r^{3}}{(2PAr - C)^{3}} - \frac{3PAr^{2}}{(2PAr - C)^{2}} - \frac{4P^{2}A^{2}r^{2}}{(2PAr - C)^{3}} + \frac{2PAr}{(2PAr - C)^{2}} - \frac{4PACr^{2}}{(2PAr - C)^{3}} + \frac{2Cr}{(2PAr - C)^{2}} + \frac{4PACr}{(2PAr - C)^{3}} - \frac{C}{(2PAr - C)^{2}} + \frac{8P^{3}A^{2}mnr}{(2PAr - C)^{3}} - \frac{2P^{2}Amn}{(2PAr - C)^{2}}\right)}{dr}\\=&4(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{2}A^{2}r^{3} + \frac{4P^{2}A^{2}*3r^{2}}{(2PAr - C)^{3}} - 3(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PAr^{2} - \frac{3PA*2r}{(2PAr - C)^{2}} - 4(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{2}A^{2}r^{2} - \frac{4P^{2}A^{2}*2r}{(2PAr - C)^{3}} + 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PAr + \frac{2PA}{(2PAr - C)^{2}} - 4(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})PACr^{2} - \frac{4PAC*2r}{(2PAr - C)^{3}} + 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})Cr + \frac{2C}{(2PAr - C)^{2}} + 4(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})PACr + \frac{4PAC}{(2PAr - C)^{3}} - (\frac{-2(2PA + 0)}{(2PAr - C)^{3}})C + 0 + 8(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{3}A^{2}mnr + \frac{8P^{3}A^{2}mn}{(2PAr - C)^{3}} - 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})P^{2}Amn + 0\\=&\frac{-24P^{3}A^{3}r^{3}}{(2PAr - C)^{4}} + \frac{24P^{2}A^{2}r^{2}}{(2PAr - C)^{3}} - \frac{6PAr}{(2PAr - C)^{2}} + \frac{24P^{3}A^{3}r^{2}}{(2PAr - C)^{4}} - \frac{16P^{2}A^{2}r}{(2PAr - C)^{3}} - \frac{16PACr}{(2PAr - C)^{3}} + \frac{24P^{2}A^{2}Cr^{2}}{(2PAr - C)^{4}} - \frac{24P^{2}A^{2}Cr}{(2PAr - C)^{4}} + \frac{2C}{(2PAr - C)^{2}} + \frac{8PAC}{(2PAr - C)^{3}} - \frac{48P^{4}A^{3}mnr}{(2PAr - C)^{4}} + \frac{16P^{3}A^{2}mn}{(2PAr - C)^{3}} + \frac{2PA}{(2PAr - C)^{2}}\\\\ &\color{blue}{函数的第 3 阶导数：} \\&\frac{d\left( \frac{-24P^{3}A^{3}r^{3}}{(2PAr - C)^{4}} + \frac{24P^{2}A^{2}r^{2}}{(2PAr - C)^{3}} - \frac{6PAr}{(2PAr - C)^{2}} + \frac{24P^{3}A^{3}r^{2}}{(2PAr - C)^{4}} - \frac{16P^{2}A^{2}r}{(2PAr - C)^{3}} - \frac{16PACr}{(2PAr - C)^{3}} + \frac{24P^{2}A^{2}Cr^{2}}{(2PAr - C)^{4}} - \frac{24P^{2}A^{2}Cr}{(2PAr - C)^{4}} + \frac{2C}{(2PAr - C)^{2}} + \frac{8PAC}{(2PAr - C)^{3}} - \frac{48P^{4}A^{3}mnr}{(2PAr - C)^{4}} + \frac{16P^{3}A^{2}mn}{(2PAr - C)^{3}} + \frac{2PA}{(2PAr - C)^{2}}\right)}{dr}\\=&-24(\frac{-4(2PA + 0)}{(2PAr - C)^{5}})P^{3}A^{3}r^{3} - \frac{24P^{3}A^{3}*3r^{2}}{(2PAr - C)^{4}} + 24(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{2}A^{2}r^{2} + \frac{24P^{2}A^{2}*2r}{(2PAr - C)^{3}} - 6(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PAr - \frac{6PA}{(2PAr - C)^{2}} + 24(\frac{-4(2PA + 0)}{(2PAr - C)^{5}})P^{3}A^{3}r^{2} + \frac{24P^{3}A^{3}*2r}{(2PAr - C)^{4}} - 16(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{2}A^{2}r - \frac{16P^{2}A^{2}}{(2PAr - C)^{3}} - 16(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})PACr - \frac{16PAC}{(2PAr - C)^{3}} + 24(\frac{-4(2PA + 0)}{(2PAr - C)^{5}})P^{2}A^{2}Cr^{2} + \frac{24P^{2}A^{2}C*2r}{(2PAr - C)^{4}} - 24(\frac{-4(2PA + 0)}{(2PAr - C)^{5}})P^{2}A^{2}Cr - \frac{24P^{2}A^{2}C}{(2PAr - C)^{4}} + 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})C + 0 + 8(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})PAC + 0 - 48(\frac{-4(2PA + 0)}{(2PAr - C)^{5}})P^{4}A^{3}mnr - \frac{48P^{4}A^{3}mn}{(2PAr - C)^{4}} + 16(\frac{-3(2PA + 0)}{(2PAr - C)^{4}})P^{3}A^{2}mn + 0 + 2(\frac{-2(2PA + 0)}{(2PAr - C)^{3}})PA + 0\\=&\frac{192P^{4}A^{4}r^{3}}{(2PAr - C)^{5}} - \frac{216P^{3}A^{3}r^{2}}{(2PAr - C)^{4}} + \frac{72P^{2}A^{2}r}{(2PAr - C)^{3}} + \frac{144P^{2}A^{2}Cr}{(2PAr - C)^{4}} - \frac{192P^{4}A^{4}r^{2}}{(2PAr - C)^{5}} + \frac{144P^{3}A^{3}r}{(2PAr - C)^{4}} - \frac{192P^{3}A^{3}Cr^{2}}{(2PAr - C)^{5}} + \frac{192P^{3}A^{3}Cr}{(2PAr - C)^{5}} + \frac{384P^{5}A^{4}mnr}{(2PAr - C)^{5}} - \frac{24PAC}{(2PAr - C)^{3}} - \frac{72P^{2}A^{2}C}{(2PAr - C)^{4}} - \frac{144P^{4}A^{3}mn}{(2PAr - C)^{4}} - \frac{24P^{2}A^{2}}{(2PAr - C)^{3}} - \frac{6PA}{(2PAr - C)^{2}}\\ \end{split}\end{equation} \]
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