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。\[ \begin{equation}\begin{split}【1/1】求函数(2xcos({x}^{2} + {y}^{2}) - \frac{1}{10}(2x + y + 2)sin({x}^{2} + {y}^{2})){e}^{(\frac{-1}{10}({x}^{2} + {y}^{2} + xy + 2x))} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = 2x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2}) - \frac{1}{5}x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2}) - \frac{1}{10}y{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2}) - \frac{1}{5}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( 2x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2}) - \frac{1}{5}x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2}) - \frac{1}{10}y{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2}) - \frac{1}{5}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})\right)}{dx}\\=&2{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2}) + 2x({e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}((\frac{-1}{10}*2x - \frac{1}{10}y + 0 - \frac{1}{5})ln(e) + \frac{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)(0)}{(e)}))cos(x^{2} + y^{2}) + 2x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}*-sin(x^{2} + y^{2})(2x + 0) - \frac{1}{5}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2}) - \frac{1}{5}x({e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}((\frac{-1}{10}*2x - \frac{1}{10}y + 0 - \frac{1}{5})ln(e) + \frac{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)(0)}{(e)}))sin(x^{2} + y^{2}) - \frac{1}{5}x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})(2x + 0) - \frac{1}{10}y({e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}((\frac{-1}{10}*2x - \frac{1}{10}y + 0 - \frac{1}{5})ln(e) + \frac{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)(0)}{(e)}))sin(x^{2} + y^{2}) - \frac{1}{10}y{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})(2x + 0) - \frac{1}{5}({e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}((\frac{-1}{10}*2x - \frac{1}{10}y + 0 - \frac{1}{5})ln(e) + \frac{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)(0)}{(e)}))sin(x^{2} + y^{2}) - \frac{1}{5}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})(2x + 0)\\=&2{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2}) - \frac{4x^{2}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})}{5} - \frac{2yx{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})}{5} - \frac{4x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}cos(x^{2} + y^{2})}{5} - \frac{99x^{2}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{25} - \frac{4{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{25} + \frac{yx{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{25} + \frac{2x{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{25} + \frac{y^{2}{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{100} + \frac{y{e}^{(\frac{-1}{10}x^{2} - \frac{1}{10}yx - \frac{1}{10}y^{2} - \frac{1}{5}x)}sin(x^{2} + y^{2})}{25}\\ \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!