本次共计算 1 个题目:每一题对 x 求 4 阶导数。
注意,变量是区分大小写的。\[ \begin{equation}\begin{split}【1/1】求函数sin(x)ln(1 + xx) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = ln(x^{2} + 1)sin(x)\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( ln(x^{2} + 1)sin(x)\right)}{dx}\\=&\frac{(2x + 0)sin(x)}{(x^{2} + 1)} + ln(x^{2} + 1)cos(x)\\=&\frac{2xsin(x)}{(x^{2} + 1)} + ln(x^{2} + 1)cos(x)\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( \frac{2xsin(x)}{(x^{2} + 1)} + ln(x^{2} + 1)cos(x)\right)}{dx}\\=&2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})xsin(x) + \frac{2sin(x)}{(x^{2} + 1)} + \frac{2xcos(x)}{(x^{2} + 1)} + \frac{(2x + 0)cos(x)}{(x^{2} + 1)} + ln(x^{2} + 1)*-sin(x)\\=&\frac{-4x^{2}sin(x)}{(x^{2} + 1)^{2}} + \frac{2sin(x)}{(x^{2} + 1)} + \frac{4xcos(x)}{(x^{2} + 1)} - ln(x^{2} + 1)sin(x)\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( \frac{-4x^{2}sin(x)}{(x^{2} + 1)^{2}} + \frac{2sin(x)}{(x^{2} + 1)} + \frac{4xcos(x)}{(x^{2} + 1)} - ln(x^{2} + 1)sin(x)\right)}{dx}\\=&-4(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x^{2}sin(x) - \frac{4*2xsin(x)}{(x^{2} + 1)^{2}} - \frac{4x^{2}cos(x)}{(x^{2} + 1)^{2}} + 2(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})sin(x) + \frac{2cos(x)}{(x^{2} + 1)} + 4(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})xcos(x) + \frac{4cos(x)}{(x^{2} + 1)} + \frac{4x*-sin(x)}{(x^{2} + 1)} - \frac{(2x + 0)sin(x)}{(x^{2} + 1)} - ln(x^{2} + 1)cos(x)\\=&\frac{16x^{3}sin(x)}{(x^{2} + 1)^{3}} - \frac{12xsin(x)}{(x^{2} + 1)^{2}} - \frac{12x^{2}cos(x)}{(x^{2} + 1)^{2}} + \frac{6cos(x)}{(x^{2} + 1)} - \frac{6xsin(x)}{(x^{2} + 1)} - ln(x^{2} + 1)cos(x)\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( \frac{16x^{3}sin(x)}{(x^{2} + 1)^{3}} - \frac{12xsin(x)}{(x^{2} + 1)^{2}} - \frac{12x^{2}cos(x)}{(x^{2} + 1)^{2}} + \frac{6cos(x)}{(x^{2} + 1)} - \frac{6xsin(x)}{(x^{2} + 1)} - ln(x^{2} + 1)cos(x)\right)}{dx}\\=&16(\frac{-3(2x + 0)}{(x^{2} + 1)^{4}})x^{3}sin(x) + \frac{16*3x^{2}sin(x)}{(x^{2} + 1)^{3}} + \frac{16x^{3}cos(x)}{(x^{2} + 1)^{3}} - 12(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})xsin(x) - \frac{12sin(x)}{(x^{2} + 1)^{2}} - \frac{12xcos(x)}{(x^{2} + 1)^{2}} - 12(\frac{-2(2x + 0)}{(x^{2} + 1)^{3}})x^{2}cos(x) - \frac{12*2xcos(x)}{(x^{2} + 1)^{2}} - \frac{12x^{2}*-sin(x)}{(x^{2} + 1)^{2}} + 6(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})cos(x) + \frac{6*-sin(x)}{(x^{2} + 1)} - 6(\frac{-(2x + 0)}{(x^{2} + 1)^{2}})xsin(x) - \frac{6sin(x)}{(x^{2} + 1)} - \frac{6xcos(x)}{(x^{2} + 1)} - \frac{(2x + 0)cos(x)}{(x^{2} + 1)} - ln(x^{2} + 1)*-sin(x)\\=&\frac{-96x^{4}sin(x)}{(x^{2} + 1)^{4}} + \frac{96x^{2}sin(x)}{(x^{2} + 1)^{3}} + \frac{64x^{3}cos(x)}{(x^{2} + 1)^{3}} - \frac{12sin(x)}{(x^{2} + 1)^{2}} - \frac{48xcos(x)}{(x^{2} + 1)^{2}} + \frac{24x^{2}sin(x)}{(x^{2} + 1)^{2}} - \frac{12sin(x)}{(x^{2} + 1)} - \frac{8xcos(x)}{(x^{2} + 1)} + ln(x^{2} + 1)sin(x)\\ \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!