$$$x^{2} + 6 x + 25$$$の導関数
入力内容
$$$\frac{d}{dx} \left(x^{2} + 6 x + 25\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dx} \left(x^{2} + 6 x + 25\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(25\right)\right)}$$定数の導数は$$$0$$$です:
$${\color{red}\left(\frac{d}{dx} \left(25\right)\right)} + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(6 x\right) + \frac{d}{dx} \left(x^{2}\right)$$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 $$$c = 6$$$ と $$$f{\left(x \right)} = x$$$ に対して適用します:
$${\color{red}\left(\frac{d}{dx} \left(6 x\right)\right)} + \frac{d}{dx} \left(x^{2}\right) = {\color{red}\left(6 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{2}\right)$$$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$6 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(x^{2}\right) = 6 {\color{red}\left(1\right)} + \frac{d}{dx} \left(x^{2}\right)$$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:
$${\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 6 = {\color{red}\left(2 x\right)} + 6$$したがって、$$$\frac{d}{dx} \left(x^{2} + 6 x + 25\right) = 2 x + 6$$$。
解答
$$$\frac{d}{dx} \left(x^{2} + 6 x + 25\right) = 2 x + 6$$$A