$$$\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}$$$의 도함수

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = e^{- \frac{1}{10}}$$$와 $$$f{\left(x \right)} = x - 10 + e^{\frac{1}{10}}$$$에 적용합니다:

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

합/차의 도함수는 도함수들의 합/차이다:

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

상수의 도함수는 $$$0$$$입니다:

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

멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

상수의 도함수는 $$$0$$$입니다:

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

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