$$$- \sqrt{3} x + \sqrt{2} x$$$ 的導數

和/差的導數等於導數的和/差：

$${\color{red}\left(\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(\sqrt{3} x\right) + \frac{d}{dx} \left(\sqrt{2} x\right)\right)}$$

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 $$$c = \sqrt{3}$$$ 與 $$$f{\left(x \right)} = x$$$：

$$- {\color{red}\left(\frac{d}{dx} \left(\sqrt{3} x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - {\color{red}\left(\sqrt{3} \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$

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

$$- \sqrt{3} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - \sqrt{3} {\color{red}\left(1\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$

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 $$$c = \sqrt{2}$$$ 與 $$$f{\left(x \right)} = x$$$：

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{2} x\right)\right)} - \sqrt{3} = {\color{red}\left(\sqrt{2} \frac{d}{dx} \left(x\right)\right)} - \sqrt{3}$$

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

$$\sqrt{2} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \sqrt{3} = \sqrt{2} {\color{red}\left(1\right)} - \sqrt{3}$$

因此，$$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right) = - \sqrt{3} + \sqrt{2}$$$。