$$$x \left(2 - 3 x\right)$$$의 도함수

$$$f{\left(x \right)} = x$$$와 $$$g{\left(x \right)} = 2 - 3 x$$$에 대해 곱의 미분법칙 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$을 적용하십시오:

$${\color{red}\left(\frac{d}{dx} \left(x \left(2 - 3 x\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \left(2 - 3 x\right) + x \frac{d}{dx} \left(2 - 3 x\right)\right)}$$

합/차의 도함수는 도함수들의 합/차이다:

$$x {\color{red}\left(\frac{d}{dx} \left(2 - 3 x\right)\right)} + \left(2 - 3 x\right) \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{d}{dx} \left(2\right) - \frac{d}{dx} \left(3 x\right)\right)} + \left(2 - 3 x\right) \frac{d}{dx} \left(x\right)$$

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

$$x \left(\frac{d}{dx} \left(2\right) - \frac{d}{dx} \left(3 x\right)\right) + \left(2 - 3 x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x \left(\frac{d}{dx} \left(2\right) - \frac{d}{dx} \left(3 x\right)\right) + \left(2 - 3 x\right) {\color{red}\left(1\right)}$$

상수의 도함수는 $$$0$$$입니다:

$$x \left({\color{red}\left(\frac{d}{dx} \left(2\right)\right)} - \frac{d}{dx} \left(3 x\right)\right) - 3 x + 2 = x \left({\color{red}\left(0\right)} - \frac{d}{dx} \left(3 x\right)\right) - 3 x + 2$$

상수배 법칙 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$을 $$$c = 3$$$와 $$$f{\left(x \right)} = x$$$에 적용합니다:

$$- x {\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)} - 3 x + 2 = - x {\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)} - 3 x + 2$$

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

$$- 3 x {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - 3 x + 2 = - 3 x {\color{red}\left(1\right)} - 3 x + 2$$

따라서, $$$\frac{d}{dx} \left(x \left(2 - 3 x\right)\right) = 2 - 6 x$$$.