$$$2 x^{2} - 2^{\frac{2}{3}} x + \sqrt[3]{2}$$$ 的導數

和/差的導數等於導數的和/差：

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

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 $$$c = 2^{\frac{2}{3}}$$$ 與 $$$f{\left(x \right)} = x$$$：

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

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

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

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

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

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 2$$$：

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

常數的導數為$$$0$$$：

$$4 x + {\color{red}\left(\frac{d}{dx} \left(\sqrt[3]{2}\right)\right)} - 2^{\frac{2}{3}} = 4 x + {\color{red}\left(0\right)} - 2^{\frac{2}{3}}$$

因此，$$$\frac{d}{dx} \left(2 x^{2} - 2^{\frac{2}{3}} x + \sqrt[3]{2}\right) = 4 x - 2^{\frac{2}{3}}$$$。