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$$$x^{3} + 5 x^{2} + 7 x + 4$$$の導関数
入力内容
$$$\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right)$$$ を求めよ。
解答
和/差の導関数は、導関数の和/差である:
$${\color{red}\left(\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(4\right)\right)}$$定数の導数は$$$0$$$です:
$${\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right)$$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 $$$c = 5$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用します:
$${\color{red}\left(\frac{d}{dx} \left(5 x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right) = {\color{red}\left(5 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right)$$冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:
$$5 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right) = 5 {\color{red}\left(2 x\right)} + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(x^{3}\right)$$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:
$$10 x + {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} + \frac{d}{dx} \left(7 x\right) = 10 x + {\color{red}\left(3 x^{2}\right)} + \frac{d}{dx} \left(7 x\right)$$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 $$$c = 7$$$ と $$$f{\left(x \right)} = x$$$ に対して適用します:
$$3 x^{2} + 10 x + {\color{red}\left(\frac{d}{dx} \left(7 x\right)\right)} = 3 x^{2} + 10 x + {\color{red}\left(7 \frac{d}{dx} \left(x\right)\right)}$$$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$3 x^{2} + 10 x + 7 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 3 x^{2} + 10 x + 7 {\color{red}\left(1\right)}$$簡単化せよ:
$$3 x^{2} + 10 x + 7 = \left(x + 1\right) \left(3 x + 7\right)$$したがって、$$$\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right) = \left(x + 1\right) \left(3 x + 7\right)$$$。
解答
$$$\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right) = \left(x + 1\right) \left(3 x + 7\right)$$$A