$$$x^{3} - 3 x^{2}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right)$$$。
解答
和/差的导数等于导数的和/差:
$${\color{red}\left(\frac{d}{dx} \left(x^{3} - 3 x^{2}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(3 x^{2}\right)\right)}$$对 $$$c = 3$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(3 x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - {\color{red}\left(3 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right)$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,其中 $$$n = 2$$$:
$$- 3 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - 3 {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(x^{3}\right)$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,其中 $$$n = 3$$$:
$$- 6 x + {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} = - 6 x + {\color{red}\left(3 x^{2}\right)}$$化简:
$$3 x^{2} - 6 x = 3 x \left(x - 2\right)$$因此,$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right) = 3 x \left(x - 2\right)$$$。
答案
$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right) = 3 x \left(x - 2\right)$$$A
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