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。\[ \begin{equation}\begin{split}【1/1】求函数sin(x)sin(x) - {x}^{2}cos(x)cos(x) 关于 x 的 4 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = sin^{2}(x) - x^{2}cos^{2}(x)\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( sin^{2}(x) - x^{2}cos^{2}(x)\right)}{dx}\\=&2sin(x)cos(x) - 2xcos^{2}(x) - x^{2}*-2cos(x)sin(x)\\=&2sin(x)cos(x) - 2xcos^{2}(x) + 2x^{2}sin(x)cos(x)\\\\ &\color{blue}{函数的第 2 阶导数:} \\&\frac{d\left( 2sin(x)cos(x) - 2xcos^{2}(x) + 2x^{2}sin(x)cos(x)\right)}{dx}\\=&2cos(x)cos(x) + 2sin(x)*-sin(x) - 2cos^{2}(x) - 2x*-2cos(x)sin(x) + 2*2xsin(x)cos(x) + 2x^{2}cos(x)cos(x) + 2x^{2}sin(x)*-sin(x)\\=&-2sin^{2}(x) + 8xsin(x)cos(x) + 2x^{2}cos^{2}(x) - 2x^{2}sin^{2}(x)\\\\ &\color{blue}{函数的第 3 阶导数:} \\&\frac{d\left( -2sin^{2}(x) + 8xsin(x)cos(x) + 2x^{2}cos^{2}(x) - 2x^{2}sin^{2}(x)\right)}{dx}\\=&-2*2sin(x)cos(x) + 8sin(x)cos(x) + 8xcos(x)cos(x) + 8xsin(x)*-sin(x) + 2*2xcos^{2}(x) + 2x^{2}*-2cos(x)sin(x) - 2*2xsin^{2}(x) - 2x^{2}*2sin(x)cos(x)\\=&4sin(x)cos(x) + 12xcos^{2}(x) - 8x^{2}sin(x)cos(x) - 12xsin^{2}(x)\\\\ &\color{blue}{函数的第 4 阶导数:} \\&\frac{d\left( 4sin(x)cos(x) + 12xcos^{2}(x) - 8x^{2}sin(x)cos(x) - 12xsin^{2}(x)\right)}{dx}\\=&4cos(x)cos(x) + 4sin(x)*-sin(x) + 12cos^{2}(x) + 12x*-2cos(x)sin(x) - 8*2xsin(x)cos(x) - 8x^{2}cos(x)cos(x) - 8x^{2}sin(x)*-sin(x) - 12sin^{2}(x) - 12x*2sin(x)cos(x)\\=&16cos^{2}(x) - 16sin^{2}(x) - 64xsin(x)cos(x) - 8x^{2}cos^{2}(x) + 8x^{2}sin^{2}(x)\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!