$$$2 x^{2} - 2^{\frac{2}{3}} x + \sqrt[3]{2}$$$的导数

和/差的导数等于导数的和/差：

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

对 $$$c = 2^{\frac{2}{3}}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

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

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

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

对 $$$c = 2$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

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

应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，其中 $$$n = 2$$$:

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

常数的导数是$$$0$$$:

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

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