$$$x^{4} - 6 x^{2}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(x^{4} - 6 x^{2}\right)$$$。
解答
和/差的导数等于导数的和/差:
$${\color{red}\left(\frac{d}{dx} \left(x^{4} - 6 x^{2}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{4}\right) - \frac{d}{dx} \left(6 x^{2}\right)\right)}$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,其中 $$$n = 4$$$:
$${\color{red}\left(\frac{d}{dx} \left(x^{4}\right)\right)} - \frac{d}{dx} \left(6 x^{2}\right) = {\color{red}\left(4 x^{3}\right)} - \frac{d}{dx} \left(6 x^{2}\right)$$对 $$$c = 6$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$:
$$4 x^{3} - {\color{red}\left(\frac{d}{dx} \left(6 x^{2}\right)\right)} = 4 x^{3} - {\color{red}\left(6 \frac{d}{dx} \left(x^{2}\right)\right)}$$应用幂次法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$,其中 $$$n = 2$$$:
$$4 x^{3} - 6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 4 x^{3} - 6 {\color{red}\left(2 x\right)}$$化简:
$$4 x^{3} - 12 x = 4 x \left(x^{2} - 3\right)$$因此,$$$\frac{d}{dx} \left(x^{4} - 6 x^{2}\right) = 4 x \left(x^{2} - 3\right)$$$。
答案
$$$\frac{d}{dx} \left(x^{4} - 6 x^{2}\right) = 4 x \left(x^{2} - 3\right)$$$A
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