$$$\left(x - 1\right)^{2}$$$の導関数

関数$$$\left(x - 1\right)^{2}$$$は、2つの関数$$$f{\left(u \right)} = u^{2}$$$と$$$g{\left(x \right)} = x - 1$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:

$${\color{red}\left(\frac{d}{dx} \left(\left(x - 1\right)^{2}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(x - 1\right)\right)}$$

冪法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(x - 1\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(x - 1\right)$$

元の変数に戻す:

$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(x - 1\right) = 2 {\color{red}\left(x - 1\right)} \frac{d}{dx} \left(x - 1\right)$$

和/差の導関数は、導関数の和/差である:

$$2 \left(x - 1\right) {\color{red}\left(\frac{d}{dx} \left(x - 1\right)\right)} = 2 \left(x - 1\right) {\color{red}\left(\frac{d}{dx} \left(x\right) - \frac{d}{dx} \left(1\right)\right)}$$

$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$2 \left(x - 1\right) \left({\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \frac{d}{dx} \left(1\right)\right) = 2 \left(x - 1\right) \left({\color{red}\left(1\right)} - \frac{d}{dx} \left(1\right)\right)$$

定数の導数は$$$0$$$です:

$$2 \left(1 - {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}\right) \left(x - 1\right) = 2 \left(1 - {\color{red}\left(0\right)}\right) \left(x - 1\right)$$

したがって、$$$\frac{d}{dx} \left(\left(x - 1\right)^{2}\right) = 2 x - 2$$$。