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$$$256 x^{2} + 16$$$の導関数
入力内容
$$$\frac{d}{dx} \left(256 x^{2} + 16\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2} + 16\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right) + \frac{d}{dx} \left(16\right)\right)}$$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 $$$c = 256$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用します:
$${\color{red}\left(\frac{d}{dx} \left(256 x^{2}\right)\right)} + \frac{d}{dx} \left(16\right) = {\color{red}\left(256 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(16\right)$$定数の導数は$$$0$$$です:
$${\color{red}\left(\frac{d}{dx} \left(16\right)\right)} + 256 \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(0\right)} + 256 \frac{d}{dx} \left(x^{2}\right)$$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:
$$256 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 256 {\color{red}\left(2 x\right)}$$したがって、$$$\frac{d}{dx} \left(256 x^{2} + 16\right) = 512 x$$$。
解答
$$$\frac{d}{dx} \left(256 x^{2} + 16\right) = 512 x$$$A
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