$$$\alpha \left(\beta + x\right)$$$ 對 $$$x$$$ 的導數
您的輸入
求$$$\frac{d}{dx} \left(\alpha \left(\beta + x\right)\right)$$$。
解答
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 $$$c = \alpha$$$ 與 $$$f{\left(x \right)} = \beta + x$$$：
$${\color{red}\left(\frac{d}{dx} \left(\alpha \left(\beta + x\right)\right)\right)} = {\color{red}\left(\alpha \frac{d}{dx} \left(\beta + x\right)\right)}$$和/差的導數等於導數的和/差：
$$\alpha {\color{red}\left(\frac{d}{dx} \left(\beta + x\right)\right)} = \alpha {\color{red}\left(\frac{d\beta}{dx} + \frac{d}{dx} \left(x\right)\right)}$$套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：
$$\alpha \left({\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d\beta}{dx}\right) = \alpha \left({\color{red}\left(1\right)} + \frac{d\beta}{dx}\right)$$常數的導數為$$$0$$$：
$$\alpha \left({\color{red}\left(\frac{d\beta}{dx}\right)} + 1\right) = \alpha \left({\color{red}\left(0\right)} + 1\right)$$因此，$$$\frac{d}{dx} \left(\alpha \left(\beta + x\right)\right) = \alpha$$$。
答案
$$$\frac{d}{dx} \left(\alpha \left(\beta + x\right)\right) = \alpha$$$A