$$$- \sqrt{3} x + \sqrt{2} x$$$的导数
您的输入
求$$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right)$$$。
解答
和/差的导数等于导数的和/差:
$${\color{red}\left(\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(\sqrt{3} x\right) + \frac{d}{dx} \left(\sqrt{2} x\right)\right)}$$对 $$$c = \sqrt{3}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(\sqrt{3} x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - {\color{red}\left(\sqrt{3} \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- \sqrt{3} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - \sqrt{3} {\color{red}\left(1\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$对 $$$c = \sqrt{2}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\sqrt{2} x\right)\right)} - \sqrt{3} = {\color{red}\left(\sqrt{2} \frac{d}{dx} \left(x\right)\right)} - \sqrt{3}$$应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,取 $$$n = 1$$$,也就是说,$$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\sqrt{2} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \sqrt{3} = \sqrt{2} {\color{red}\left(1\right)} - \sqrt{3}$$因此,$$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right) = - \sqrt{3} + \sqrt{2}$$$。
答案
$$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right) = - \sqrt{3} + \sqrt{2}$$$A