$$$x^{3} + 5 x^{2} + 7 x + 4$$$ 的導數

和/差的導數等於導數的和/差：

$${\color{red}\left(\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(4\right)\right)}$$

常數的導數為$$$0$$$：

$${\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right)$$

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 $$$c = 5$$$ 與 $$$f{\left(x \right)} = x^{2}$$$：

$${\color{red}\left(\frac{d}{dx} \left(5 x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right) = {\color{red}\left(5 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right)$$

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

$$5 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right) = 5 {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right)$$

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 3$$$：

$$10 x + {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} + \frac{d}{dx} \left(7 x\right) = 10 x + {\color{red}\left(3 x^{2}\right)} + \frac{d}{dx} \left(7 x\right)$$

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 $$$c = 7$$$ 與 $$$f{\left(x \right)} = x$$$：

$$3 x^{2} + 10 x + {\color{red}\left(\frac{d}{dx} \left(7 x\right)\right)} = 3 x^{2} + 10 x + {\color{red}\left(7 \frac{d}{dx} \left(x\right)\right)}$$

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

$$3 x^{2} + 10 x + 7 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 3 x^{2} + 10 x + 7 {\color{red}\left(1\right)}$$

化簡：

$$3 x^{2} + 10 x + 7 = \left(x + 1\right) \left(3 x + 7\right)$$

因此，$$$\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right) = \left(x + 1\right) \left(3 x + 7\right)$$$。