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。\[ \begin{equation}\begin{split}【1/1】求函数{({(rsin(x) - x)}^{2} + {(rcos(x) - y)}^{2})}^{\frac{1}{2}} + {({(rsin(x) - x*2)}^{2} + {(rcos(x) - y*2)}^{2})}^{\frac{1}{2}} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = (r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}} + (r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( (r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}} + (r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}\right)}{dx}\\=&(\frac{\frac{1}{2}(r^{2}*2sin(x)cos(x) - 2rsin(x) - 2rxcos(x) + 2x + r^{2}*-2cos(x)sin(x) - 2ry*-sin(x) + 0)}{(r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}}}) + (\frac{\frac{1}{2}(r^{2}*2sin(x)cos(x) - 4rsin(x) - 4rxcos(x) + 4*2x + r^{2}*-2cos(x)sin(x) - 4ry*-sin(x) + 0)}{(r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}})\\=& - \frac{rxcos(x)}{(r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}}} + \frac{x}{(r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}}} + \frac{rysin(x)}{(r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}}} - \frac{rsin(x)}{(r^{2}sin^{2}(x) - 2rxsin(x) + x^{2} + r^{2}cos^{2}(x) - 2rycos(x) + y^{2})^{\frac{1}{2}}} - \frac{2rsin(x)}{(r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}} - \frac{2rxcos(x)}{(r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}} + \frac{4x}{(r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}} + \frac{2rysin(x)}{(r^{2}sin^{2}(x) - 4rxsin(x) + 4x^{2} + r^{2}cos^{2}(x) - 4rycos(x) + 4y^{2})^{\frac{1}{2}}}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!