$$$\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}$$$的导数
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您的输入
求$$$\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right)$$$。
解答
对 $$$c = e^{- \frac{1}{10}}$$$ 和 $$$f{\left(x \right)} = x - 10 + e^{\frac{1}{10}}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(x - 10 + e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}}\right)}$$和/差的导数等于导数的和/差:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(x - 10 + e^{\frac{1}{10}}\right)\right)}}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)\right)}}{e^{\frac{1}{10}}}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(1\right)} - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}}$$常数的导数是$$$0$$$:
$$\frac{- {\color{red}\left(\frac{d}{dx} \left(10\right)\right)} + \frac{d}{dx} \left(e^{\frac{1}{10}}\right) + 1}{e^{\frac{1}{10}}} = \frac{- {\color{red}\left(0\right)} + \frac{d}{dx} \left(e^{\frac{1}{10}}\right) + 1}{e^{\frac{1}{10}}}$$常数的导数是$$$0$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(e^{\frac{1}{10}}\right)\right)} + 1}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(0\right)} + 1}{e^{\frac{1}{10}}}$$因此,$$$\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right) = e^{- \frac{1}{10}}$$$。
答案
$$$\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right) = e^{- \frac{1}{10}}$$$A