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。\[ \begin{equation}\begin{split}【1/1】求函数e^{cos(27xx + 27tt + 8pi - 24)}(\frac{xx}{2} + \frac{1}{2})(\frac{tt}{2} + \frac{1}{2}) 关于 x 的 1 阶导数:\\\end{split}\end{equation} \]
\[ \begin{equation}\begin{split}\\解:&\\ &原函数 = \frac{1}{4}t^{2}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}t^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}\\&\color{blue}{函数的第 1 阶导数:}\\&\frac{d\left( \frac{1}{4}t^{2}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}t^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}\right)}{dx}\\=&\frac{1}{4}t^{2}*2xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}t^{2}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}*-sin(27x^{2} + 27t^{2} + 8pi - 24)(27*2x + 0 + 0 + 0) + \frac{1}{4}*2xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)} + \frac{1}{4}x^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}*-sin(27x^{2} + 27t^{2} + 8pi - 24)(27*2x + 0 + 0 + 0) + \frac{1}{4}t^{2}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}*-sin(27x^{2} + 27t^{2} + 8pi - 24)(27*2x + 0 + 0 + 0) + \frac{1}{4}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}*-sin(27x^{2} + 27t^{2} + 8pi - 24)(27*2x + 0 + 0 + 0)\\=&\frac{-27t^{2}x^{3}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}sin(27x^{2} + 27t^{2} + 8pi - 24)}{2} - \frac{27t^{2}xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)}sin(27x^{2} + 27t^{2} + 8pi - 24)}{2} - \frac{27x^{3}e^{cos(27x^{2} + 27t^{2} + 8pi - 24)}sin(27x^{2} + 27t^{2} + 8pi - 24)}{2} - \frac{27xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)}sin(27x^{2} + 27t^{2} + 8pi - 24)}{2} + \frac{t^{2}xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)}}{2} + \frac{xe^{cos(27x^{2} + 27t^{2} + 8pi - 24)}}{2}\\ \end{split}\end{equation} \]你的问题在这里没有得到解决?请到 热门难题 里面看看吧!