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。\[ \begin{equation}\begin{split}【1/1】求函数\frac{sqrt({a}^{2} + {x}^{2})}{p} + \frac{sqrt({b}^{2} + {(l - x)}^{2})}{q} 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{sqrt(a^{2} + x^{2})}{p} + \frac{sqrt(b^{2} - 2lx + l^{2} + x^{2})}{q}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{sqrt(a^{2} + x^{2})}{p} + \frac{sqrt(b^{2} - 2lx + l^{2} + x^{2})}{q}\right)}{dx}\\=&\frac{(0 + 2x)*\frac{1}{2}}{p(a^{2} + x^{2})^{\frac{1}{2}}} + \frac{(0 - 2l + 0 + 2x)*\frac{1}{2}}{q(b^{2} - 2lx + l^{2} + x^{2})^{\frac{1}{2}}}\\=&\frac{x}{(a^{2} + x^{2})^{\frac{1}{2}}p} - \frac{l}{(b^{2} - 2lx + l^{2} + x^{2})^{\frac{1}{2}}q} + \frac{x}{(b^{2} - 2lx + l^{2} + x^{2})^{\frac{1}{2}}q}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!