本次共计算 1 个题目:每一题对 z 求 3 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数({(z - \frac{a{e}^{i}pai}{4})}^{4}){({z}^{4} + {a}^{4})}^{-1} 关于 z 的 3 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{z^{4}}{(z^{4} + a^{4})} - \frac{a^{2}ipz^{3}{e}^{i}}{(z^{4} + a^{4})} + \frac{\frac{3}{8}a^{4}i^{2}p^{2}z^{2}{e}^{(2(i))}}{(z^{4} + a^{4})} - \frac{\frac{1}{16}a^{6}i^{3}p^{3}z{e}^{(3(i))}}{(z^{4} + a^{4})} + \frac{\frac{1}{256}a^{8}i^{4}p^{4}{e}^{(4(i))}}{(z^{4} + a^{4})}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{z^{4}}{(z^{4} + a^{4})} - \frac{a^{2}ipz^{3}{e}^{i}}{(z^{4} + a^{4})} + \frac{\frac{3}{8}a^{4}i^{2}p^{2}z^{2}{e}^{(2(i))}}{(z^{4} + a^{4})} - \frac{\frac{1}{16}a^{6}i^{3}p^{3}z{e}^{(3(i))}}{(z^{4} + a^{4})} + \frac{\frac{1}{256}a^{8}i^{4}p^{4}{e}^{(4(i))}}{(z^{4} + a^{4})}\right)}{dz}\\=&(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})z^{4} + \frac{4z^{3}}{(z^{4} + a^{4})} - (\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{2}ipz^{3}{e}^{i} - \frac{a^{2}ip*3z^{2}{e}^{i}}{(z^{4} + a^{4})} - \frac{a^{2}ipz^{3}({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})} + \frac{3}{8}(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{4}i^{2}p^{2}z^{2}{e}^{(2(i))} + \frac{\frac{3}{8}a^{4}i^{2}p^{2}*2z{e}^{(2(i))}}{(z^{4} + a^{4})} + \frac{\frac{3}{8}a^{4}i^{2}p^{2}z^{2}({e}^{(2(i))}((2(0))ln(e) + \frac{(2(i))(0)}{(e)}))}{(z^{4} + a^{4})} - \frac{1}{16}(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{6}i^{3}p^{3}z{e}^{(3(i))} - \frac{\frac{1}{16}a^{6}i^{3}p^{3}{e}^{(3(i))}}{(z^{4} + a^{4})} - \frac{\frac{1}{16}a^{6}i^{3}p^{3}z({e}^{(3(i))}((3(0))ln(e) + \frac{(3(i))(0)}{(e)}))}{(z^{4} + a^{4})} + \frac{1}{256}(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{8}i^{4}p^{4}{e}^{(4(i))} + \frac{\frac{1}{256}a^{8}i^{4}p^{4}({e}^{(4(i))}((4(0))ln(e) + \frac{(4(i))(0)}{(e)}))}{(z^{4} + a^{4})}\\=&\frac{-4z^{7}}{(z^{4} + a^{4})^{2}} + \frac{4z^{3}}{(z^{4} + a^{4})} + \frac{4a^{2}ipz^{6}{e}^{i}}{(z^{4} + a^{4})^{2}} - \frac{3a^{2}ipz^{2}{e}^{i}}{(z^{4} + a^{4})} - \frac{3a^{4}i^{2}p^{2}z^{5}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} + \frac{3a^{4}i^{2}p^{2}z{e}^{(2i)}}{4(z^{4} + a^{4})} + \frac{a^{6}i^{3}p^{3}z^{4}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} - \frac{a^{6}i^{3}p^{3}{e}^{(3i)}}{16(z^{4} + a^{4})} - \frac{a^{8}i^{4}p^{4}z^{3}{e}^{(4i)}}{64(z^{4} + a^{4})^{2}}\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{-4z^{7}}{(z^{4} + a^{4})^{2}} + \frac{4z^{3}}{(z^{4} + a^{4})} + \frac{4a^{2}ipz^{6}{e}^{i}}{(z^{4} + a^{4})^{2}} - \frac{3a^{2}ipz^{2}{e}^{i}}{(z^{4} + a^{4})} - \frac{3a^{4}i^{2}p^{2}z^{5}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} + \frac{3a^{4}i^{2}p^{2}z{e}^{(2i)}}{4(z^{4} + a^{4})} + \frac{a^{6}i^{3}p^{3}z^{4}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} - \frac{a^{6}i^{3}p^{3}{e}^{(3i)}}{16(z^{4} + a^{4})} - \frac{a^{8}i^{4}p^{4}z^{3}{e}^{(4i)}}{64(z^{4} + a^{4})^{2}}\right)}{dz}\\=&-4(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})z^{7} - \frac{4*7z^{6}}{(z^{4} + a^{4})^{2}} + 4(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})z^{3} + \frac{4*3z^{2}}{(z^{4} + a^{4})} + 4(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{2}ipz^{6}{e}^{i} + \frac{4a^{2}ip*6z^{5}{e}^{i}}{(z^{4} + a^{4})^{2}} + \frac{4a^{2}ipz^{6}({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})^{2}} - 3(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{2}ipz^{2}{e}^{i} - \frac{3a^{2}ip*2z{e}^{i}}{(z^{4} + a^{4})} - \frac{3a^{2}ipz^{2}({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})} - \frac{3(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{4}i^{2}p^{2}z^{5}{e}^{(2i)}}{2} - \frac{3a^{4}i^{2}p^{2}*5z^{4}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} - \frac{3a^{4}i^{2}p^{2}z^{5}({e}^{(2i)}((0)ln(e) + \frac{(2i)(0)}{(e)}))}{2(z^{4} + a^{4})^{2}} + \frac{3(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{4}i^{2}p^{2}z{e}^{(2i)}}{4} + \frac{3a^{4}i^{2}p^{2}{e}^{(2i)}}{4(z^{4} + a^{4})} + \frac{3a^{4}i^{2}p^{2}z({e}^{(2i)}((0)ln(e) + \frac{(2i)(0)}{(e)}))}{4(z^{4} + a^{4})} + \frac{(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{6}i^{3}p^{3}z^{4}{e}^{(3i)}}{4} + \frac{a^{6}i^{3}p^{3}*4z^{3}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} + \frac{a^{6}i^{3}p^{3}z^{4}({e}^{(3i)}((0)ln(e) + \frac{(3i)(0)}{(e)}))}{4(z^{4} + a^{4})^{2}} - \frac{(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{6}i^{3}p^{3}{e}^{(3i)}}{16} - \frac{a^{6}i^{3}p^{3}({e}^{(3i)}((0)ln(e) + \frac{(3i)(0)}{(e)}))}{16(z^{4} + a^{4})} - \frac{(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{8}i^{4}p^{4}z^{3}{e}^{(4i)}}{64} - \frac{a^{8}i^{4}p^{4}*3z^{2}{e}^{(4i)}}{64(z^{4} + a^{4})^{2}} - \frac{a^{8}i^{4}p^{4}z^{3}({e}^{(4i)}((0)ln(e) + \frac{(4i)(0)}{(e)}))}{64(z^{4} + a^{4})^{2}}\\=&\frac{32z^{10}}{(z^{4} + a^{4})^{3}} - \frac{44z^{6}}{(z^{4} + a^{4})^{2}} + \frac{12z^{2}}{(z^{4} + a^{4})} - \frac{32a^{2}ipz^{9}{e}^{i}}{(z^{4} + a^{4})^{3}} + \frac{36a^{2}ipz^{5}{e}^{i}}{(z^{4} + a^{4})^{2}} - \frac{6a^{2}ipz{e}^{i}}{(z^{4} + a^{4})} + \frac{12a^{4}i^{2}p^{2}z^{8}{e}^{(2i)}}{(z^{4} + a^{4})^{3}} - \frac{21a^{4}i^{2}p^{2}z^{4}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} + \frac{3a^{4}i^{2}p^{2}{e}^{(2i)}}{4(z^{4} + a^{4})} - \frac{2a^{6}i^{3}p^{3}z^{7}{e}^{(3i)}}{(z^{4} + a^{4})^{3}} + \frac{5a^{6}i^{3}p^{3}z^{3}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} + \frac{a^{8}i^{4}p^{4}z^{6}{e}^{(4i)}}{8(z^{4} + a^{4})^{3}} - \frac{3a^{8}i^{4}p^{4}z^{2}{e}^{(4i)}}{64(z^{4} + a^{4})^{2}}\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{32z^{10}}{(z^{4} + a^{4})^{3}} - \frac{44z^{6}}{(z^{4} + a^{4})^{2}} + \frac{12z^{2}}{(z^{4} + a^{4})} - \frac{32a^{2}ipz^{9}{e}^{i}}{(z^{4} + a^{4})^{3}} + \frac{36a^{2}ipz^{5}{e}^{i}}{(z^{4} + a^{4})^{2}} - \frac{6a^{2}ipz{e}^{i}}{(z^{4} + a^{4})} + \frac{12a^{4}i^{2}p^{2}z^{8}{e}^{(2i)}}{(z^{4} + a^{4})^{3}} - \frac{21a^{4}i^{2}p^{2}z^{4}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} + \frac{3a^{4}i^{2}p^{2}{e}^{(2i)}}{4(z^{4} + a^{4})} - \frac{2a^{6}i^{3}p^{3}z^{7}{e}^{(3i)}}{(z^{4} + a^{4})^{3}} + \frac{5a^{6}i^{3}p^{3}z^{3}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} + \frac{a^{8}i^{4}p^{4}z^{6}{e}^{(4i)}}{8(z^{4} + a^{4})^{3}} - \frac{3a^{8}i^{4}p^{4}z^{2}{e}^{(4i)}}{64(z^{4} + a^{4})^{2}}\right)}{dz}\\=&32(\frac{-3(4z^{3} + 0)}{(z^{4} + a^{4})^{4}})z^{10} + \frac{32*10z^{9}}{(z^{4} + a^{4})^{3}} - 44(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})z^{6} - \frac{44*6z^{5}}{(z^{4} + a^{4})^{2}} + 12(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})z^{2} + \frac{12*2z}{(z^{4} + a^{4})} - 32(\frac{-3(4z^{3} + 0)}{(z^{4} + a^{4})^{4}})a^{2}ipz^{9}{e}^{i} - \frac{32a^{2}ip*9z^{8}{e}^{i}}{(z^{4} + a^{4})^{3}} - \frac{32a^{2}ipz^{9}({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})^{3}} + 36(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{2}ipz^{5}{e}^{i} + \frac{36a^{2}ip*5z^{4}{e}^{i}}{(z^{4} + a^{4})^{2}} + \frac{36a^{2}ipz^{5}({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})^{2}} - 6(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{2}ipz{e}^{i} - \frac{6a^{2}ip{e}^{i}}{(z^{4} + a^{4})} - \frac{6a^{2}ipz({e}^{i}((0)ln(e) + \frac{(i)(0)}{(e)}))}{(z^{4} + a^{4})} + 12(\frac{-3(4z^{3} + 0)}{(z^{4} + a^{4})^{4}})a^{4}i^{2}p^{2}z^{8}{e}^{(2i)} + \frac{12a^{4}i^{2}p^{2}*8z^{7}{e}^{(2i)}}{(z^{4} + a^{4})^{3}} + \frac{12a^{4}i^{2}p^{2}z^{8}({e}^{(2i)}((0)ln(e) + \frac{(2i)(0)}{(e)}))}{(z^{4} + a^{4})^{3}} - \frac{21(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{4}i^{2}p^{2}z^{4}{e}^{(2i)}}{2} - \frac{21a^{4}i^{2}p^{2}*4z^{3}{e}^{(2i)}}{2(z^{4} + a^{4})^{2}} - \frac{21a^{4}i^{2}p^{2}z^{4}({e}^{(2i)}((0)ln(e) + \frac{(2i)(0)}{(e)}))}{2(z^{4} + a^{4})^{2}} + \frac{3(\frac{-(4z^{3} + 0)}{(z^{4} + a^{4})^{2}})a^{4}i^{2}p^{2}{e}^{(2i)}}{4} + \frac{3a^{4}i^{2}p^{2}({e}^{(2i)}((0)ln(e) + \frac{(2i)(0)}{(e)}))}{4(z^{4} + a^{4})} - 2(\frac{-3(4z^{3} + 0)}{(z^{4} + a^{4})^{4}})a^{6}i^{3}p^{3}z^{7}{e}^{(3i)} - \frac{2a^{6}i^{3}p^{3}*7z^{6}{e}^{(3i)}}{(z^{4} + a^{4})^{3}} - \frac{2a^{6}i^{3}p^{3}z^{7}({e}^{(3i)}((0)ln(e) + \frac{(3i)(0)}{(e)}))}{(z^{4} + a^{4})^{3}} + \frac{5(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{6}i^{3}p^{3}z^{3}{e}^{(3i)}}{4} + \frac{5a^{6}i^{3}p^{3}*3z^{2}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} + \frac{5a^{6}i^{3}p^{3}z^{3}({e}^{(3i)}((0)ln(e) + \frac{(3i)(0)}{(e)}))}{4(z^{4} + a^{4})^{2}} + \frac{(\frac{-3(4z^{3} + 0)}{(z^{4} + a^{4})^{4}})a^{8}i^{4}p^{4}z^{6}{e}^{(4i)}}{8} + \frac{a^{8}i^{4}p^{4}*6z^{5}{e}^{(4i)}}{8(z^{4} + a^{4})^{3}} + \frac{a^{8}i^{4}p^{4}z^{6}({e}^{(4i)}((0)ln(e) + \frac{(4i)(0)}{(e)}))}{8(z^{4} + a^{4})^{3}} - \frac{3(\frac{-2(4z^{3} + 0)}{(z^{4} + a^{4})^{3}})a^{8}i^{4}p^{4}z^{2}{e}^{(4i)}}{64} - \frac{3a^{8}i^{4}p^{4}*2z{e}^{(4i)}}{64(z^{4} + a^{4})^{2}} - \frac{3a^{8}i^{4}p^{4}z^{2}({e}^{(4i)}((0)ln(e) + \frac{(4i)(0)}{(e)}))}{64(z^{4} + a^{4})^{2}}\\=&\frac{-384z^{13}}{(z^{4} + a^{4})^{4}} + \frac{672z^{9}}{(z^{4} + a^{4})^{3}} - \frac{312z^{5}}{(z^{4} + a^{4})^{2}} + \frac{24z}{(z^{4} + a^{4})} + \frac{384a^{2}ipz^{12}{e}^{i}}{(z^{4} + a^{4})^{4}} - \frac{576a^{2}ipz^{8}{e}^{i}}{(z^{4} + a^{4})^{3}} + \frac{204a^{2}ipz^{4}{e}^{i}}{(z^{4} + a^{4})^{2}} - \frac{6a^{2}ip{e}^{i}}{(z^{4} + a^{4})} - \frac{144a^{4}i^{2}p^{2}z^{11}{e}^{(2i)}}{(z^{4} + a^{4})^{4}} + \frac{180a^{4}i^{2}p^{2}z^{7}{e}^{(2i)}}{(z^{4} + a^{4})^{3}} - \frac{45a^{4}i^{2}p^{2}z^{3}{e}^{(2i)}}{(z^{4} + a^{4})^{2}} + \frac{24a^{6}i^{3}p^{3}z^{10}{e}^{(3i)}}{(z^{4} + a^{4})^{4}} - \frac{24a^{6}i^{3}p^{3}z^{6}{e}^{(3i)}}{(z^{4} + a^{4})^{3}} + \frac{15a^{6}i^{3}p^{3}z^{2}{e}^{(3i)}}{4(z^{4} + a^{4})^{2}} - \frac{3a^{8}i^{4}p^{4}z^{9}{e}^{(4i)}}{2(z^{4} + a^{4})^{4}} + \frac{9a^{8}i^{4}p^{4}z^{5}{e}^{(4i)}}{8(z^{4} + a^{4})^{3}} - \frac{3a^{8}i^{4}p^{4}z{e}^{(4i)}}{32(z^{4} + a^{4})^{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!