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。
$$\begin{equation}\begin{split}【1/1】求函数4{x}^{((π - 1)(1 - sin(π)x)(π + 1))} + 4{(1 - x)}^{((π - 1)(1 - sin(π)x)(π + 1))} + {9}^{x} + {9}^{(1 - x)} 关于 x 的 1 阶导数：\\\end{split}\end{equation}$$ $$\begin{equation}\begin{split}\\解:&\\ &原函数 = 4{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)} + 4(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)} + {9}^{x} + {9}^{(-x + 1)}\\&\color{blue}{函数的第 1 阶导数：}\\&\frac{d\left( 4{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)} + 4(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)} + {9}^{x} + {9}^{(-x + 1)}\right)}{dx}\\=&4({x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}((-π^{2}sin(π) - π^{2}xcos(π)*0 + 0 + sin(π) + xcos(π)*0 + 0)ln(x) + \frac{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)(1)}{(x)})) + 4((-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}((-π^{2}sin(π) - π^{2}xcos(π)*0 + 0 + sin(π) + xcos(π)*0 + 0)ln(-x + 1) + \frac{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)(-1 + 0)}{(-x + 1)})) + ({9}^{x}((1)ln(9) + \frac{(x)(0)}{(9)})) + ({9}^{(-x + 1)}((-1 + 0)ln(9) + \frac{(-x + 1)(0)}{(9)}))\\=&-4π^{2}{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}ln(x)sin(π) + 4{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}ln(x)sin(π) - 4π^{2}{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}sin(π) + \frac{4π^{2}{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}}{x} + 4{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}sin(π) - \frac{4{x}^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}}{x} - 4π^{2}(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}ln(-x + 1)sin(π) + 4(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}ln(-x + 1)sin(π) + \frac{4π^{2}x(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}sin(π)}{(-x + 1)} - \frac{4π^{2}(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}}{(-x + 1)} - \frac{4x(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}sin(π)}{(-x + 1)} + \frac{4(-x + 1)^{(-π^{2}xsin(π) + π^{2} + xsin(π) - 1)}}{(-x + 1)} + {9}^{x}ln(9) - {9}^{(-x + 1)}ln(9)\\ \end{split}\end{equation}$$

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