$$$\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}$$$の導関数

定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = e^{- \frac{1}{10}}$$$ と $$$f{\left(x \right)} = x - 10 + e^{\frac{1}{10}}$$$ に対して適用します:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

定数の導数は$$$0$$$です:

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

したがって、$$$\frac{d}{dx} \left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right) = e^{- \frac{1}{10}}$$$。