$$$\left(x - 1\right)^{2}$$$의 도함수

함수 $$$\left(x - 1\right)^{2}$$$는 두 함수 $$$f{\left(u \right)} = u^{2}$$$와 $$$g{\left(x \right)} = x - 1$$$의 합성함수 $$$f{\left(g{\left(x \right)} \right)}$$$이다.

연쇄법칙 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을(를) 적용하십시오:

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

거듭제곱법칙 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$을 $$$n = 2$$$에 적용합니다:

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

역치환:

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

합/차의 도함수는 도함수들의 합/차이다:

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

멱법칙 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$을 $$$n = 1$$$에 대해 적용하면, 즉 $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

상수의 도함수는 $$$0$$$입니다:

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

따라서, $$$\frac{d}{dx} \left(\left(x - 1\right)^{2}\right) = 2 x - 2$$$.