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$$$x^{3} - 3 x^{2}$$$の導関数
入力内容
$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dx} \left(x^{3} - 3 x^{2}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(3 x^{2}\right)\right)}$$定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 3$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用します:
$$- {\color{red}\left(\frac{d}{dx} \left(3 x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - {\color{red}\left(3 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right)$$冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:
$$- 3 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(x^{3}\right) = - 3 {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(x^{3}\right)$$冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 3$$$ に対して適用する:
$$- 6 x + {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} = - 6 x + {\color{red}\left(3 x^{2}\right)}$$簡単化せよ:
$$3 x^{2} - 6 x = 3 x \left(x - 2\right)$$したがって、$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right) = 3 x \left(x - 2\right)$$$。
解答
$$$\frac{d}{dx} \left(x^{3} - 3 x^{2}\right) = 3 x \left(x - 2\right)$$$A
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