$$$\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}$$$ 的導數

套用常數倍法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$，使用 $$$c = e^{- \frac{1}{10}}$$$ 與 $$$f{\left(x \right)} = x - 10 + e^{\frac{1}{10}}$$$：

$${\color{red}\left(\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(x - 10 + e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}}\right)}$$

和/差的導數等於導數的和/差：

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x - 10 + e^{\frac{1}{10}}\right)\right)}}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)\right)}}{e^{\frac{1}{10}}}$$

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(1\right)} - \frac{d}{dx} \left(10\right) + \frac{d}{dx} \left(e^{\frac{1}{10}}\right)}{e^{\frac{1}{10}}}$$

常數的導數為$$$0$$$：

$$\frac{- {\color{red}\left(\frac{d}{dx} \left(10\right)\right)} + \frac{d}{dx} \left(e^{\frac{1}{10}}\right) + 1}{e^{\frac{1}{10}}} = \frac{- {\color{red}\left(0\right)} + \frac{d}{dx} \left(e^{\frac{1}{10}}\right) + 1}{e^{\frac{1}{10}}}$$

常數的導數為$$$0$$$：

$$\frac{{\color{red}\left(\frac{d}{dx} \left(e^{\frac{1}{10}}\right)\right)} + 1}{e^{\frac{1}{10}}} = \frac{{\color{red}\left(0\right)} + 1}{e^{\frac{1}{10}}}$$

因此，$$$\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right) = e^{- \frac{1}{10}}$$$。