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。\[ \begin{equation}\begin{split}【1/1】求函数\frac{xsqrt({x}^{2} + 1)}{2} + \frac{ln(x + sqrt({x}^{2} + 1))}{2} - \frac{(3 - x)sqrt({(3 - x)}^{2} + 4)}{2} - 2ln(\frac{(3 - x)}{2} + \frac{sqrt({(3 - x)}^{2} + 4)}{2}) 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{1}{2}xsqrt(x^{2} + 1) + \frac{1}{2}ln(x + sqrt(x^{2} + 1)) - \frac{3}{2}sqrt(x^{2} - 6x + 13) + \frac{1}{2}xsqrt(x^{2} - 6x + 13) - 2ln(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{1}{2}xsqrt(x^{2} + 1) + \frac{1}{2}ln(x + sqrt(x^{2} + 1)) - \frac{3}{2}sqrt(x^{2} - 6x + 13) + \frac{1}{2}xsqrt(x^{2} - 6x + 13) - 2ln(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})\right)}{dx}\\=&\frac{1}{2}sqrt(x^{2} + 1) + \frac{\frac{1}{2}x(2x + 0)*\frac{1}{2}}{(x^{2} + 1)^{\frac{1}{2}}} + \frac{\frac{1}{2}(1 + \frac{(2x + 0)*\frac{1}{2}}{(x^{2} + 1)^{\frac{1}{2}}})}{(x + sqrt(x^{2} + 1))} - \frac{\frac{3}{2}(2x - 6 + 0)*\frac{1}{2}}{(x^{2} - 6x + 13)^{\frac{1}{2}}} + \frac{1}{2}sqrt(x^{2} - 6x + 13) + \frac{\frac{1}{2}x(2x - 6 + 0)*\frac{1}{2}}{(x^{2} - 6x + 13)^{\frac{1}{2}}} - \frac{2(\frac{\frac{1}{2}(2x - 6 + 0)*\frac{1}{2}}{(x^{2} - 6x + 13)^{\frac{1}{2}}} - \frac{1}{2} + 0)}{(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})}\\=&\frac{sqrt(x^{2} + 1)}{2} + \frac{x^{2}}{2(x^{2} + 1)^{\frac{1}{2}}} + \frac{x}{2(x + sqrt(x^{2} + 1))(x^{2} + 1)^{\frac{1}{2}}} + \frac{x^{2}}{2(x^{2} - 6x + 13)^{\frac{1}{2}}} - \frac{3x}{(x^{2} - 6x + 13)^{\frac{1}{2}}} - \frac{x}{(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})(x^{2} - 6x + 13)^{\frac{1}{2}}} + \frac{sqrt(x^{2} - 6x + 13)}{2} + \frac{3}{(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})(x^{2} - 6x + 13)^{\frac{1}{2}}} + \frac{1}{2(x + sqrt(x^{2} + 1))} + \frac{9}{2(x^{2} - 6x + 13)^{\frac{1}{2}}} + \frac{1}{(\frac{1}{2}sqrt(x^{2} - 6x + 13) - \frac{1}{2}x + \frac{3}{2})}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!